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Double Integrals over General Regions

Recognize when a function of two variables is integrable over a general region.

Double Integrals over General Regions

This section builds off of the previous two sections and generalizes the horizontal or vertical slicing of the region of integration and demonstrates how this geometric description exactly matches with the algebraic description given by an iterated integral. There are several important conceptual checks that you should pay attention to:

identification of horizontally or vertically simple regions,
understanding when double integrals will be positive/negative/zero without any calculations (), and
identifying when the algebra of an iterated integral does not make sense as a double integral ().

There is a lot for students to practice in this section, so we recommend spending a couple of class periods on this topic. Spending 50 minutes on writing the inequalities that describe a region as vertically simple or horizontally simple (doing the activities through ) is quite reasonable. These ideas will be used throughout every section for this chapter and through most of , so making sure students understand what a double integral measures and and how to set up appropriate iterated integrals is critical.

General Regions of Integration

An example of a general bounded region \(D\) on a plane is shown in . Since \(D\) is bounded on the plane, there must exist a rectangular region \(R\) on the same plane that encloses the region \(D,\) that is, a rectangular region \(R\) exists such that \(D\) is a subset of \(R(D⊆R).\)

Suppose \(z=f(x,y)\) is defined on a general planar bounded region \(D\) as in . In order to develop double integrals of \(f\) over \(D,\) we extend the definition of the function to include all points on the rectangular region \(R\) and then use the concepts and tools from the preceding section. But how do we extend the definition of \(f\) to include all the points on \(R?\) We do this by defining a new function \(g(x,y)\) on \(R\) as follows:

\[g(x,y)=\{\begin{array}{ll}f(x,y) & \text{if}\ (x,y)\ \text{is in}\ D \\ 0 & \text{if}\ (x,y)\ \text{is in}\ R\ \text{but not in}\ D\end{array}\]

Note that we might have some technical difficulties if the boundary of \(D\) is complicated. So we assume the boundary to be a piecewise smooth and continuous simple closed curve. Also, since all the results developed in Double Integrals over Rectangular Regions used an integrable function \(f(x,y),\) we must be careful about \(g(x,y)\) and verify that \(g(x,y)\) is an integrable function over the rectangular region \(R.\) This happens as long as the region \(D\) is bounded by simple closed curves. For now we will concentrate on the descriptions of the regions rather than the function and extend our theory appropriately for integration.

We consider two types of planar bounded regions.

Condensed — the full section is in OpenStax Calculus Volume 3.

Introduction

Recall from Section that we defined the double integral of a continuous function \(f = f(x,y)\) to measure the volume of the region beneath the graph of \(f\) over a region of input points. If we consider the set of points in a rectangle \(R\) with \(a\leq x\leq b\) and \(c\leq y\leq d\), then we can calculate the double integral using a Riemann sum \[\begin{aligned}\end{aligned}\]. Furthermore, we have seen in Fubini's Theorem that we can evaluate a double integral \(\displaystyle{\iint_R f(x,y) \, dA}\) over \(R\) as an iterated integral of either of the forms \[\begin{aligned}\end{aligned}\]

Most applied or theoretical problems involve regions of inputs that are not rectangles. Therefore, we want to extend the idea of using iterated integrals to allow us to evaluate double integrals over nonrectangular regions. We explore one such example in the following preview activity.

In this section, we will explore the ideas introduced by the Preview Activity in greater generality. We begin by understanding how to describe regions in the plane, as this will be essential to writing iterated integrals that can be used to evaluate double integrals.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

Double Integrals over Nonrectangular Regions

To develop the concept and tools for evaluation of a double integral over a general, nonrectangular region, we need to first understand the region and be able to express it as Type I or Type II or a combination of both. Without understanding the regions, we will not be able to decide the limits of integrations in double integrals. As a first step, let us look at the following theorem.

The right-hand side of this equation is what we have seen before, so this theorem is reasonable because \(R\) is a rectangle and \(\underset{R}{∬}g(x,y)dA\) has been discussed in the preceding section. Also, the equality works because the values of \(g(x,y)\) are \(0\) for any point \((x,y)\) that lies outside \(D,\) and hence these points do not add anything to the integral. However, it is important that the rectangle \(R\) contains the region \(D.\)

As a matter of fact, if the region \(D\) is bounded by smooth curves on a plane and we are able to describe it as Type I or Type II or a mix of both, then we can use the following theorem and not have to find a rectangle \(R\) containing the region.

The integral in each of these expressions is an iterated integral, similar to those we have seen before. Notice that, in the inner integral in the first expression, we integrate \(f(x,y)\) with \(x\) being held constant and the limits of integration being \({g}_{1}(x)\ \text{and}\ {g}_{2}(x).\) In the inner integral in the second expression, we integrate \(f(x,y)\) with \(y\) being held constant and the limits of integration are \({h}_{1}(y)\ \text{and}\ {h}_{2}(y).\)

In , we could have looked at the region in another way, such as \(D=\{(x,y)|0\le y\le 1,0\le x\le 2y\}\) ().

This is a Type II region and the integral would then look like

\[\underset{D}{∬}{x}^{2}{e}^{xy}dA=\int _{y=0}^{y=1}\ \int _{x=0}^{x=2y}{x}^{2}{e}^{xy}dx\ dy.\]\[\underset{R}{∬}f(x,y)dA=\underset{S}{∬}f(x,y)dA+\underset{T}{∬}f(x,y)dA.\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Double Integrals over General Regions

So far, we have learned that a double integral over a rectangular region may be interpreted in three ways detailed in : net signed volume, mass, and when divided by the area of the base region, average value.

As we saw in Preview Activity, a function \(f = f(x,y)\) may be considered over regions other than rectangular ones, and thus we want to understand how to set up double integrals as iterated integrals over non-rectangular regions. Note that if we can set up double integrals over non-rectangular regions as iterated integrals, then the three interpretations of the double integral noted above will extend to solid regions with non-rectangular bases.

Remember that the double integral over a a region can be thought of as the signed volume of the solid bounded by the graph of \(f\) and the \(xy\)-plane. The volume above the \(xy\)-plane is counted as positive and the volume below the \(xy\)-plane is counted as negative. As shown in , the total volume above the \(xy\)-plane is greater than the total volume below the \(xy\)-plane, so the double integral of \(f\) over the region \(R\) is positive. This argument comes from both the heights of the rectangular prisms used in the Riemann sum (determined by output values of \(f\)) and the area of the regions where the output of \(f\) is positive versus negative.

In the next activity, you are asked to assess and justify your thoughts about whether the double integrals over non-rectangular regions are positive, negative, or zero. Your justifications should consider both the area of the given region of integration, as well as the output values of the function being integrated on these regions.

Activity

Let \(D\) be the region inside the unit circle centered at the origin, let \(R\) be the right half of \(D\), and let \(B\) be the bottom half of \(D\).

On three separate plots, graph and label the regions \(D\), \(R\), and \(B\).

For each double integral below, decide without calculation whether the double integral is positive, negative, or zero. Write a sentence or two to explain your answer for each part.

\(\iint_D \; 1\; dA\)
\(\iint_D \; x \; dA\)
\(\iint_B \; x \; dA\)
\(\iint_R \; x \; dA\)
\(\iint_D \; x^2 \; dA\)
\(\iint_B \; 2-x^2 \; dA\)
\(\iint_R \; -e^x \; dA\)
\(\iint_B \; 1-e^y \; dA\)
\(\iint_D \; xy \; dA\)

Changing the Order of Integration

As we have already seen when we evaluate an iterated integral, sometimes one order of integration leads to a computation that is significantly simpler than the other order of integration. Sometimes the order of integration does not matter, but it is important to learn to recognize when a change in order will simplify our work.

Example

Try it.

Reverse the order of integration in the iterated integral \(\int _{x=0}^{x=\sqrt{2}}\ \int _{y=0}^{y=2-{x}^{2}}x{e}^{{x}^{2}}dy\ dx.\) Then evaluate the new iterated integral.

Solution

The region as presented is of Type I. To reverse the order of integration, we must first express the region as Type II. Refer to .

We can see from the limits of integration that the region is bounded above by \(y=2-{x}^{2}\) and below by \(y=0,\) where \(x\) is in the interval \([0,\sqrt{2}].\) By reversing the order, we have the region bounded on the left by \(x=0\) and on the right by \(x=\sqrt{2-y}\) where \(y\) is in the interval \([0,2].\) We solved \(y=2-{x}^{2}\) in terms of \(x\) to obtain \(x=\sqrt{2-y}.\)

Hence

\[\begin{array}{lllll}\int _{0}^{\sqrt{2}}\ \int _{0}^{2-{x}^{2}}x{e}^{{x}^{2}}dy\ dx & =\int _{0}^{2}\ \int _{0}^{\sqrt{2-y}}x{e}^{{x}^{2}}dx\ dy & & & \begin{array}{l}\text{Reverse the order of} \\ \text{integration then use} \\ \text{substitution.}\end{array} \\ & =\int _{0}^{2}[\frac{1}{2}{{{e}^{x}}^{2}|}_{0}^{\sqrt{2-y}}]dy=\int _{0}^{2}\frac{1}{2}({e}^{2-y}-1)dy={-\frac{1}{2}({e}^{2-y}+y)|}_{0}^{2} & & & \\ & =\frac{1}{2}({e}^{2}-3). & & & \end{array}\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Horizontally and Vertically Simple Regions

The key idea to calculating double integrals using iterated integrals as shown in and is to slice the region of integration along a trace where one of the coordinates is held constant. We then express the range of the other coordinate in terms of an inequality. In this subsection, we will step away from double integrals for a short while to practice slicing two dimensional regions and stating corresponding inequalities in terms of each coordinate.

In , we considered a tetrahedron, which had as its base a triangular region \(R\) in the \(xy\)-plane. This region is shown in .

Notice how different values of \(a\) provide different ranges of \(y\)-values for the portion of the \(x=a\) line segment that lies inside \(R\). This is intrinsically linked with the fact that \(R\) is not a rectangular region, in contrast to the regions of integration from which were all of the form \(a \leq x \leq b\) and \(c \leq y \leq d\) (with \(a,b,c,d\) constants).

These forms for the inequalities lead to a nice way to write the corresponding double integral as an iterated integral by considering the inner integral to evaluate the cross section area and the outer integral to sum the corresponding cross section volumes as the thickness of the slabs goes to zero.

Not all two-dimensional regions can be immediately expressed using this idea of converting geometric slices to inequalities. The key characteristic that allows us to do this is a region must have the same upper and lower bound expressions for each slice.

A vertically simple region is one that can be split into vertical slices of the form \(x=\text{constant}\) for \(x\) in the interval \([a,b]\) where every vertical slice has the same upper and lower bound function: \(g_1(x) \leq y \leq g_2(x)\). A horizontally simple region is one that can be split into horizontal slices of the form \(y=\text{constant}\) for \(y\) in the interval \([c,d]\) where every horizontal slice has the same upper (right) and lower (left) bound functions: \(h_1(x) \leq y \leq h_2(x)\).

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

Calculating Volumes, Areas, and Average Values

We can use double integrals over general regions to compute volumes, areas, and average values. The methods are the same as those in Double Integrals over Rectangular Regions, but without the restriction to a rectangular region, we can now solve a wider variety of problems.

Example

Try it.

Find the volume of the solid bounded by the planes \(x=0,y=0,z=0,\) and \(2x+3y+z=6.\)

Solution

The solid is a tetrahedron with the base on the \(xy\)-plane and a height \(z=6-2x-3y.\) The base is the region \(D\) bounded by the lines, \(x=0,y=0\) and \(2x+3y=6\) where \(z=0\) (). Note that we can consider the region \(D\) as Type I or as Type II, and we can integrate in both ways.

First, consider \(D\) as a Type I region, and hence \(D=\{(x,y)|0\le x\le 3,0\le y\le 2-\frac{2}{3}x\}.\)

Therefore, the volume is

\[\begin{array}{ll}V & =\int _{x=0}^{x=3}\ \int _{y=0}^{y=2-(2x\text{/}3)}(6-2x-3y)dy\ dx=\int _{x=0}^{x=3}[{(6y-2xy-\frac{3}{2}{y}^{2})|}_{y=0}^{y=2-(2x\text{/}3)}]dx \\ & =\int _{x=0}^{x=3}[\frac{2}{3}{(x-3)}^{2}]dx=6.\end{array}\]

Now consider \(D\) as a Type II region, so \(D=\{(x,y)|0\le y\le 2,0\le x\le 3-\frac{3}{2}y\}.\) In this calculation, the volume is

\[\begin{array}{ll}V & =\int _{y=0}^{y=2}\ \int _{x=0}^{x=3-(3y\text{/}2)}(6-2x-3y)dx\ dy=\int _{y=0}^{y=2}[{(6x-{x}^{2}-3xy)|}_{x=0}^{x=3-(3y\text{/}2)}]dy \\ & =\int _{y=0}^{y=2}[\frac{9}{4}{(y-2)}^{2}]dy=6.\end{array}\]

Therefore, the volume is \(6\) cubic units.

Finding the area of a rectangular region is easy, but finding the area of a nonrectangular region is not so easy. As we have seen, we can use double integrals to find a rectangular area. As a matter of fact, this comes in very handy for finding the area of a general nonrectangular region, as stated in the next definition.

We have already seen how to find areas in terms of single integration. Here we are seeing another way of finding areas by using double integrals, which can be very useful, as we will see in the later sections of this chapter.

We can also use a double integral to find the average value of a function over a general region. The definition is a direct extension of the earlier formula.

Condensed — the full section is in OpenStax Calculus Volume 3.

Double Integrals Evaluated as Iterated Integrals

We now turn our attention to using iterated integrals to evaluate double integrals. This is a powerful algebraic tool that we will use throughout the remainder of this chapter, but as the second part of showed, not every iterated integral that can be written down is meaningful in the sense of evaluating double integrals. As you read the rest of this section, pay careful attention to understanding the meaning behind the computations.

Activity

Let \(D\) be the triangular region with vertices \((0,0)\), \((4,0)\), and \((0,2)\). Consider the double integral \(\displaystyle{\iint_D (4-x-2y) \, dA}\).

Draw and label a plot of \(D\) with relevant cross sections for a vertically simple description. Give the inequalities that show \(D\) is vertically simple.

Write the double integral as an iterated integral of the form \(\displaystyle \iint_D (4-x-2y) \, dy \, dx\), including correct bounds on the integral.

Evaluate the iterated integrals from the previous part. Write a sentence describing at least one interpretation of the meaning of the value of the double integral.

The region we considered as vertically simple in was shown in to be both vertically simple and horizontally simple. We will now see that it is also possible to set up an iterated integral for the double integral using the horizontally simple description of the region.

We have now seen that in a situation where \(D\) can be described as both horizontally simple and vertically simple, we can set up an iterated integral in either order. If we can reasonably evaluate the iterated integral in either order, then the iterated integrals yield the same value in each case. However, there are times where finding an antiderivative to evaluate the inner integral may be easier in one order than the other.

The process of switching between a vertically simple description and a horizontally simple description for the region of integration is often called switching the order of integation because the corresponding iterated integrals used to calculate a double integral will have the order in which you integrate these variables switched.

The next activity asks you to return to the function and triangular region considered in but to set up an iterated integral from the perspective of the region being horizontally simple.

The next activity explores a region of integration with non-linear sides, given as bounds in an iterated integral. You will also switch the order of integration, evaluate whichever order is most convinient, and finally interpret the result as the average value.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

Improper Double Integrals

An improper double integral is an integral \(\underset{D}{∬}f\ dA\) where either \(D\) is an unbounded region or \(f\) is an unbounded function. For example, \(D=\{(x,y)||x-y|\ge 2\}\) is an unbounded region, and the function \(f(x,y)=1\text{/}(1-{x}^{2}-2{y}^{2})\) over the ellipse \({x}^{2}+2{y}^{2}\le 1\) is an unbounded function. Hence, both of the following integrals are improper integrals:

\(\underset{D}{∬}xy\ dA\) where \(D=\{(x,y)||x-y|\ge 2\};\)
\(\underset{D}{∬}\frac{1}{1-{x}^{2}-2{y}^{2}}dA\) where \(D=\{(x,y)|{x}^{2}+2{y}^{2}\le 1\}.\)

In this section we would like to deal with improper integrals of functions over rectangles or simple regions such that \(f\) has only finitely many discontinuities. Not all such improper integrals can be evaluated; however, a form of Fubini’s theorem does apply for some types of improper integrals.

It is very important to note that we required that the function be nonnegative on \(D\) for the theorem to work. We consider only the case where the function has finitely many discontinuities inside \(D.\)

Example

Try it.

Consider the function \(f(x,y)=\frac{{e}^{y}}{y}\) over the region \(D=\{(x,y)\text{:}\ 0\le x\le 1,x\le y\le \sqrt{x}\}.\)

Notice that the function is nonnegative and continuous at all points on \(D\) except \((0,0).\) Use Fubini’s theorem to evaluate the improper integral.

Solution

First we plot the region \(D\) (); then we express it in another way.

The other way to express the same region \(D\) is

\[D=\{(x,y)\text{:}\ 0\le y\le 1,{y}^{2}\le x\le y\}.\]

Thus we can use Fubini’s theorem for improper integrals and evaluate the integral as

\[\int _{y=0}^{y=1}\ \int _{x={y}^{2}}^{x=y}\frac{{e}^{y}}{y}dx\ dy.\]

Therefore, we have

\[\int _{y=0}^{y=1}\ \int _{x={y}^{2}}^{x=y}\frac{{e}^{y}}{y}dx\ dy=\int _{y=0}^{y=1}\frac{{e}^{y}}{y}{x|}_{x={y}^{2}}^{x=y}dy=\int _{y=0}^{y=1}\frac{{e}^{y}}{y}(y-{y}^{2})dy=\int _{0}^{1}({e}^{y}-y{e}^{y})dy=e-2.\]

As mentioned before, we also have an improper integral if the region of integration is unbounded. Suppose now that the function \(f\) is continuous in an unbounded rectangle \(R.\)

Condensed — the full section is in OpenStax Calculus Volume 3.

Key Concepts

A general bounded region \(D\) on the plane is a region that can be enclosed inside a rectangular region. We can use this idea to define a double integral over a general bounded region.
To evaluate an iterated integral of a function over a general nonrectangular region, we sketch the region and express it as a Type I or as a Type II region or as a union of several Type I or Type II regions that overlap only on their boundaries.
We can use double integrals to find volumes, areas, and average values of a function over general regions, similarly to calculations over rectangular regions.
We can use Fubini’s theorem for improper integrals to evaluate some types of improper integrals.

Key Equations

Iterated integral over a Type I region	\(\underset{D}{∬}f(x,y)dA=\underset{D}{∬}f(x,y)dy\ dx=\int _{a}^{b}[\int _{{g}_{1}(x)}^{{g}_{2}(x)}f(x,y)dy]dx\)
Iterated integral over a Type II region	\(\underset{D}{∬}f(x,y)dA=\underset{D}{∬}f(x,y)dx\ dy=\int _{c}^{d}[\int _{{h}_{1}(y)}^{{h}_{2}(y)}f(x,y)dx]dy\)

Double Integrals over General Regions

In the following exercises, specify whether the region is of Type I or Type II.

In the following exercises, evaluate the double integral \(\underset{D}{∬}f(x,y)dA\) over the region \(D.\)

Evaluate the iterated integrals.

In the following exercises, change the order of integration and evaluate the integral.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Consider the region in the first quadrant between the functions \(y=\sqrt{x}\) and \(y={x}^{3}\) (). Describe the region first as Type I and then as Type II.

Odhaliť odpoveď

When describing a region as Type I, we need to identify the function that lies above the region and the function that lies below the region. Here, region \(D\) is bounded above by \(y=\sqrt{x}\) and below by \(y={x}^{3}\) in the interval for \(x\ \text{in}\ [0,1].\) Hence, as Type I, \(D\) is described as the set \(\{(x,y)|0\le x\le 1,{x}^{3}\le y\le \sqrt{x}\}.\)
However, when describing a region as Type II, we need to identify the function that lies on the left of the region and the function that lies on the right of the region. Here, the region \(D\) is bounded on the left by \(x={y}^{2}\) and on the right by \(x=\sqrt[3]{y}\) in the interval for y in \([0,1].\) Hence, as Type II, \(D\) is described as the set \(\{(x,y)|0\le y\le 1,{y}^{2}\le x\le \sqrt[3]{y}\}.\)
Consider the region in the first quadrant between the functions \(y=2x\) and \(y={x}^{2}.\) Describe the region first as Type I and then as Type II.

Odhaliť odpoveď

Type I and Type II are expressed as \(\{(x,y)|0\le x\le 2,{x}^{2}\le y\le 2x\}\) and \(\{(x,y)|0\le y\le 4,\frac{1}{2}y\le x\le \sqrt{y}\},\) respectively.
Evaluate the integral \(\underset{D}{∬}{x}^{2}{e}^{xy}dA\) where \(D\) is shown in .

Odhaliť odpoveď

First construct the region \(D\) as a Type I region (). Here \(D=\{(x,y)|0\le x\le 2,\frac{1}{2}x\le y\le 1\}.\) Then we have
\[\underset{D}{∬}{x}^{2}{e}^{xy}dA=\int _{x=0}^{x=2}\ \int _{y=1\text{/}2x}^{y=1}{x}^{2}{e}^{xy}dy\ dx.\]
Therefore, we have
\[\begin{array}{lllll}\int _{x=0}^{x=2}\ \int _{y=\frac{1}{2}x}^{y=1}{x}^{2}{e}^{xy}dy\ dx & =\int _{x=0}^{x=2}[\int _{y=1\text{/}2x}^{y=1}{x}^{2}{e}^{xy}dy]dx & & & \text{Iterated integral for a Type I region.} \\ & =\int _{x=0}^{x=2}{[{x}^{2}\frac{{e}^{xy}}{x}]|}_{y=1\text{/}2x}^{y=1}dx & & & \begin{array}{l}\text{Integrate with respect to}\ y\ \text{using} \\ u\text{-substitution with}\ u=xy\ \text{where}\ x\ \text{is held} \\ \text{constant.}\end{array} \\ & =\int _{x=0}^{x=2}[x{e}^{x}-x{e}^{{x}^{2}\text{/}2}]dx & & & \begin{array}{l}\text{Integrate with respect to}\ x\ \text{using} \\ u\text{-substitution with}\ u=\frac{1}{2}{x}^{2}.\end{array} \\ & ={[x{e}^{x}-{e}^{x}-{e}^{\frac{1}{2}{x}^{2}}]|}_{x=0}^{x=2}=2 & & & \end{array}\]
Evaluate the integral \(\underset{D}{∬}(3{x}^{2}+{y}^{2})dA\) where \(=\{(x,y)|-2\le y\le 3,{y}^{2}-3\le x\le y+3\}.\)

Odhaliť odpoveď

Notice that \(D\) can be seen as either a Type I or a Type II region, as shown in . However, in this case describing \(D\) as Type \(\text{I}\) is more complicated than describing it as Type II. Therefore, we use \(D\) as a Type II region for the integration.
Choosing this order of integration, we have
\[\begin{array}{lllll}\underset{D}{∬}(3{x}^{2}+{y}^{2})dA & =\int _{y=-2}^{y=3}\ \int _{x={y}^{2}-3}^{x=y+3}(3{x}^{2}+{y}^{2})dx\ dy & & & \text{Iterated integral, Type II region.} \\ & ={\int _{y=-2}^{y=3}({x}^{3}+x{y}^{2})|}_{{y}^{2}-3}^{y+3}dy & & & \text{Integrate with respect to}\ x. \\ & =\int _{y=-2}^{y=3}({(y+3)}^{3}+(y+3){y}^{2}-{({y}^{2}-3)}^{3}-({y}^{2}-3){y}^{2})dy & & & \\ & =\int _{-2}^{3}(54+27y-12{y}^{2}+2{y}^{3}+8{y}^{4}-{y}^{6})dy & & & \text{Integrate with respect to}\ y. \\ & ={[54y+\frac{27{y}^{2}}{2}-4{y}^{3}+\frac{{y}^{4}}{2}+\frac{8{y}^{5}}{5}-\frac{{y}^{7}}{7}]|}_{-2}^{3} & & & \\ & =\frac{2375}{7}. & & & \end{array}\]
Sketch the region \(D\) and evaluate the iterated integral \(\underset{D}{∬}xy\ dy\ dx\) where \(D\) is the region bounded by the curves \(y=\text{cos}\ x\) and \(y=\text{sin}\ x\) in the interval \([-3\pi \text{/}4,\pi \text{/}4].\)

Odhaliť odpoveď

\(\pi \text{/}4\)
Express the region \(D\) shown in as a union of regions of Type I or Type II, and evaluate the integral
\[\underset{D}{∬}(2x+5y)dA.\]

Odhaliť odpoveď

The region \(D\) is not easy to decompose into any one type; it is actually a combination of different types. So we can write it as a union of three regions \({D}_{1},{D}_{2},\text{and}\ {D}_{3}\) where, \({D}_{1}=\{(x,y)|-2\le x\le 0,0\le y\le {(x+2)}^{2}\},\) \({D}_{2}=\{(x,y)|0\le y\le 4,0\le x\le (y-\frac{1}{16}{y}^{3})\},\) \({D}_{3}=\left\{\left(x,y\right)-4\le y\le 0,-2\le x\le y-\frac{{y}^{3}}{16}\right\}.\) These regions are illustrated more clearly in .
Here \({D}_{1}\) is Type \(\text{I}\) and \({D}_{2}\) and \({D}_{3}\) are both of Type II. Hence,
\[\begin{array}{ll}\underset{D}{∬}(2x+5y)dA & =\underset{{D}_{1}}{∬}(2x+5y)dA+\underset{{D}_{2}}{∬}(2x+5y)dA+\underset{{D}_{3}}{∬}(2x+5y)dA \\ & =\int _{x=-2}^{x=0}\ \int _{y=0}^{y={(x+2)}^{2}}(2x+5y)dy\ dx+\int _{y=0}^{y=4}\ \int _{x=0}^{x=y-(1\text{/}16){y}^{3}}(2+5y)dx\ dy+\int _{y=-4}^{y=0}\ \int _{x=-2}^{x=y-(1\text{/}16){y}^{3}}(2x+5y)dx\ dy \\ & =\int _{x=-2}^{x=0}[\frac{1}{2}{(2+x)}^{2}(20+24x+5{x}^{2})]dx+\int _{y=0}^{y=4}[\frac{1}{256}{y}^{6}-\frac{7}{16}{y}^{4}+6{y}^{2}]dy \\ & \ +\int _{y=-4}^{y=0}[\frac{1}{256}{y}^{6}-\frac{7}{16}{y}^{4}+6{y}^{2}+10y-4]dy \\ & =\frac{40}{3}+\frac{1664}{35}-\frac{1696}{35}=\frac{1304}{105}.\end{array}\]
Now we could redo this example using a union of two Type II regions (see the Checkpoint).
Consider the region bounded by the curves \(y=\text{ln}\ x\) and \(y={e}^{x}\) in the interval \([1,2].\) Decompose the region into smaller regions of Type II.

Odhaliť odpoveď

\(\{(x,y)|0\le y\le \ln 2,1\le x\le {e}^{y}\}\cup \{(x,y)|\ln 2\le y\le e,1\le x\le 2\}\cup \{(x,y)|e\le y\le {e}^{2},\text{ln}\ y\le x\le 2\}\)
Redo using a union of two Type II regions.

Odhaliť odpoveď

Same as in the example shown.
Reverse the order of integration in the iterated integral \(\int _{x=0}^{x=\sqrt{2}}\ \int _{y=0}^{y=2-{x}^{2}}x{e}^{{x}^{2}}dy\ dx.\) Then evaluate the new iterated integral.

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The region as presented is of Type I. To reverse the order of integration, we must first express the region as Type II. Refer to .
We can see from the limits of integration that the region is bounded above by \(y=2-{x}^{2}\) and below by \(y=0,\) where \(x\) is in the interval \([0,\sqrt{2}].\) By reversing the order, we have the region bounded on the left by \(x=0\) and on the right by \(x=\sqrt{2-y}\) where \(y\) is in the interval \([0,2].\) We solved \(y=2-{x}^{2}\) in terms of \(x\) to obtain \(x=\sqrt{2-y}.\)
Hence
\[\begin{array}{lllll}\int _{0}^{\sqrt{2}}\ \int _{0}^{2-{x}^{2}}x{e}^{{x}^{2}}dy\ dx & =\int _{0}^{2}\ \int _{0}^{\sqrt{2-y}}x{e}^{{x}^{2}}dx\ dy & & & \begin{array}{l}\text{Reverse the order of} \\ \text{integration then use} \\ \text{substitution.}\end{array} \\ & =\int _{0}^{2}[\frac{1}{2}{{{e}^{x}}^{2}|}_{0}^{\sqrt{2-y}}]dy=\int _{0}^{2}\frac{1}{2}({e}^{2-y}-1)dy={-\frac{1}{2}({e}^{2-y}+y)|}_{0}^{2} & & & \\ & =\frac{1}{2}({e}^{2}-3). & & & \end{array}\]
Consider the iterated integral \(\underset{R}{∬}f(x,y)dx\ dy\) where \(z=f(x,y)=x-2y\) over a triangular region \(R\) that has sides on \(x=0,y=0,\) and the line \(x+y=1.\) Sketch the region, and then evaluate the iterated integral by
1. integrating first with respect to \(y\) and then
2. integrating first with respect to \(x.\)
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A sketch of the region appears in .
We can complete this integration in two different ways.
1. One way to look at it is by first integrating \(y\) from \(y=0\ \text{to}\ y=1-x\) vertically and then integrating \(x\) from \(x=0\ \text{to}\ x=1\text{:}\)
  \[\begin{array}{ll}\underset{R}{∬}f(x,y)dx\ dy & =\int _{x=0}^{x=1}\ \int _{y=0}^{y=1-x}(x-2y)dy\ dx=\int _{x=0}^{x=1}{[xy-{y}^{2}]}_{y=0}^{y=1-x}dx \\ & =\int _{x=0}^{x=1}[x(1-x)-{(1-x)}^{2}]dx=\int _{x=0}^{x=1}[-1+3x-2{x}^{2}]dx={[\text{-}x+\frac{3}{2}{x}^{2}-\frac{2}{3}{x}^{3}]}_{x=0}^{x=1}=-\frac{1}{6}.\end{array}\]
2. The other way to do this problem is by first integrating \(x\) from \(x=0\ \text{to}\ x=1-y\) horizontally and then integrating \(y\) from \(y=0\ \text{to}\ y=1\text{:}\)
  \[\begin{array}{ll}\underset{R}{∬}f(x,y)dx\ dy & =\int _{y=0}^{y=1}\ \int _{x=0}^{x=1-y}(x-2y)dx\ dy=\int _{y=0}^{y=1}{[\frac{1}{2}{x}^{2}-2xy]}_{x=0}^{x=1-y}dy \\ & =\int _{y=0}^{y=1}[\frac{1}{2}{(1-y)}^{2}-2y(1-y)]dy=\int _{y=0}^{y=1}[\frac{1}{2}-3y+\frac{5}{2}{y}^{2}]dy \\ & ={[\frac{1}{2}y-\frac{3}{2}{y}^{2}+\frac{5}{6}{y}^{3}]}_{y=0}^{y=1}=-\frac{1}{6}.\end{array}\]
Evaluate the iterated integral \(\underset{D}{∬}({x}^{2}+{y}^{2})dA\) over the region \(D\) in the first quadrant between the functions \(y=2x\) and \(y={x}^{2}.\) Evaluate the iterated integral by integrating first with respect to \(y\) and then integrating first with resect to \(x.\)

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\(\frac{216}{35}\)
Find the volume of the solid bounded by the planes \(x=0,y=0,z=0,\) and \(2x+3y+z=6.\)

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The solid is a tetrahedron with the base on the \(xy\)-plane and a height \(z=6-2x-3y.\) The base is the region \(D\) bounded by the lines, \(x=0,y=0\) and \(2x+3y=6\) where \(z=0\) (). Note that we can consider the region \(D\) as Type I or as Type II, and we can integrate in both ways.
First, consider \(D\) as a Type I region, and hence \(D=\{(x,y)|0\le x\le 3,0\le y\le 2-\frac{2}{3}x\}.\)
Therefore, the volume is
\[\begin{array}{ll}V & =\int _{x=0}^{x=3}\ \int _{y=0}^{y=2-(2x\text{/}3)}(6-2x-3y)dy\ dx=\int _{x=0}^{x=3}[{(6y-2xy-\frac{3}{2}{y}^{2})|}_{y=0}^{y=2-(2x\text{/}3)}]dx \\ & =\int _{x=0}^{x=3}[\frac{2}{3}{(x-3)}^{2}]dx=6.\end{array}\]
Now consider \(D\) as a Type II region, so \(D=\{(x,y)|0\le y\le 2,0\le x\le 3-\frac{3}{2}y\}.\) In this calculation, the volume is
\[\begin{array}{ll}V & =\int _{y=0}^{y=2}\ \int _{x=0}^{x=3-(3y\text{/}2)}(6-2x-3y)dx\ dy=\int _{y=0}^{y=2}[{(6x-{x}^{2}-3xy)|}_{x=0}^{x=3-(3y\text{/}2)}]dy \\ & =\int _{y=0}^{y=2}[\frac{9}{4}{(y-2)}^{2}]dy=6.\end{array}\]
Therefore, the volume is \(6\) cubic units.
Find the volume of the solid bounded above by \(f(x,y)=10-2x+y\) over the region enclosed by the curves \(y=0\) and \(y={e}^{x},\) where \(x\) is in the interval \([0,1].\)

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\(\frac{{e}^{2}}{4}+10e-\frac{49}{4}\) cubic units
Find the area of the region bounded below by the curve \(y={x}^{2}\) and above by the line \(y=2x\) in the first quadrant ().

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We just have to integrate the constant function \(f(x,y)=1\) over the region. Thus, the area \(A\) of the bounded region is \(\int _{x=0}^{x=2}\ \int _{y={x}^{2}}^{y=2x}dy\ dx\) or \(\int _{y=0}^{x=4}\ \int _{x=y\text{/}2}^{x=\sqrt{y}}dx\ dy\text{:}\)
\[A=\underset{D}{∬}1dx\ dy=\int _{x=0}^{x=2}\ \int _{y={x}^{2}}^{y=2x}1dy\ dx=\int _{x=0}^{x=2}[{y|}_{y={x}^{2}}^{y=2x}]dx=\int _{x=0}^{x=2}(2x-{x}^{2})dx={{x}^{2}-\frac{{x}^{3}}{3}|}_{0}^{2}=\frac{4}{3}.\]
Find the area of a region bounded above by the curve \(y={x}^{3}\) and below by \(y=0\) over the interval \([0,3].\)

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\(\frac{81}{4}\) square units
Find the average value of the function \(f(x,y)=7x{y}^{2}\) on the region bounded by the line \(x=y\) and the curve \(x=\sqrt{y}\) ().

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First find the area \(A(D)\) where the region \(D\) is given by the figure. We have
\[A(D)=\underset{D}{∬}1dA=\int _{y=0}^{y=1}\ \int _{x=y}^{x=\sqrt{y}}1dx\ dy=\int _{y=0}^{y=1}[{x|}_{x=y}^{x=\sqrt{y}}]dy=\int _{y=0}^{y=1}(\sqrt{y}-y)dy=\frac{2}{3}{y}^{3\text{/}2}-{\frac{{y}^{2}}{2}|}_{0}^{1}=\frac{1}{6}.\]
Then the average value of the given function over this region is
\[\begin{array}{ll}{f}_{ave} & =\frac{1}{A(D)}\underset{D}{∬}f(x,y)dA=\frac{1}{A(D)}\int _{y=0}^{y=1}\ \int _{x=y}^{x=\sqrt{y}}7x{y}^{2}dx\ dy=\frac{1}{1\text{/}6}\int _{y=0}^{y=1}[{\frac{7}{2}{x}^{2}{y}^{2}|}_{x=y}^{x=\sqrt{y}}]dy \\ & =6\int _{y=0}^{y=1}[\frac{7}{2}{y}^{2}(y-{y}^{2})]dy=6\int _{y=0}^{y=1}[\frac{7}{2}({y}^{3}-{y}^{4})]dy=\frac{42}{2}{(\frac{{y}^{4}}{4}-\frac{{y}^{5}}{5})|}_{0}^{1}=\frac{42}{40}=\frac{21}{20}.\end{array}\]
Find the average value of the function \(f(x,y)=xy\) over the triangle with vertices \((0,0),(1,0)\ \text{and}\ (1,3).\)

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\(\frac{3}{4}\)
Consider the function \(f(x,y)=\frac{{e}^{y}}{y}\) over the region \(D=\{(x,y)\text{:}\ 0\le x\le 1,x\le y\le \sqrt{x}\}.\)
Notice that the function is nonnegative and continuous at all points on \(D\) except \((0,0).\) Use Fubini’s theorem to evaluate the improper integral.

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First we plot the region \(D\) (); then we express it in another way.
The other way to express the same region \(D\) is
\[D=\{(x,y)\text{:}\ 0\le y\le 1,{y}^{2}\le x\le y\}.\]
Thus we can use Fubini’s theorem for improper integrals and evaluate the integral as
\[\int _{y=0}^{y=1}\ \int _{x={y}^{2}}^{x=y}\frac{{e}^{y}}{y}dx\ dy.\]
Therefore, we have
\[\int _{y=0}^{y=1}\ \int _{x={y}^{2}}^{x=y}\frac{{e}^{y}}{y}dx\ dy=\int _{y=0}^{y=1}\frac{{e}^{y}}{y}{x|}_{x={y}^{2}}^{x=y}dy=\int _{y=0}^{y=1}\frac{{e}^{y}}{y}(y-{y}^{2})dy=\int _{0}^{1}({e}^{y}-y{e}^{y})dy=e-2.\]
Evaluate the integral \(\underset{R}{∬}xy{e}^{\text{-}{x}^{2}-{y}^{2}}dA\) where \(R\) is the first quadrant of the plane.

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The region \(R\) is the first quadrant of the plane, which is unbounded. So
\[\begin{array}{ll}\underset{R}{∬}xy{e}^{\text{-}{x}^{2}-{y}^{2}}dA & =\underset{(b,d)\to (\infty ,\infty )}{\text{lim}}\int _{x=0}^{x=b}(\int _{y=0}^{y=d}xy{e}^{\text{-}{x}^{2}-{y}^{2}}dy)dx=\underset{(b,d)\to (\infty ,\infty )}{\text{lim}}\int _{y=0}^{y=d}(\int _{x=0}^{x=b}xy{e}^{\text{-}{x}^{2}-{y}^{2}}dx)dy \\ & =\underset{(b,d)\to (\infty ,\infty )}{\text{lim}}\frac{1}{4}(1-{e}^{\text{-}{b}^{2}})(1-{e}^{\text{-}{d}^{2}})=\frac{1}{4}\end{array}\]
Thus, \(\underset{R}{∬}xy{e}^{\text{-}{x}^{2}-{y}^{2}}dA\) is convergent and the value is \(\frac{1}{4}.\)
Evaluate the improper integral \(\underset{D}{∬}\frac{y}{\sqrt{1-{x}^{2}-{y}^{2}}}dA\) where \(D=\{(x,y)|x\ge 0,y\ge 0,{x}^{2}+{y}^{2}\le 1\}.\)

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\(\frac{\pi }{4}\)
At Sydney’s Restaurant, customers must wait an average of \(15\) minutes for a table. From the time they are seated until they have finished their meal requires an additional \(40\) minutes, on average. What is the probability that a customer spends less than an hour and a half at the diner, assuming that waiting for a table and completing the meal are independent events?

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Waiting times are mathematically modeled by exponential density functions, with \(m\) being the average waiting time, as
\[f(t)=\{\begin{array}{ll}0 & \text{if}\ t<0, \\ \frac{1}{m}{e}^{\text{-}t\text{/}m} & \text{if}\ t\ge 0.\end{array}\]
If \(X\) and \(Y\) are random variables for ‘waiting for a table’ and ‘completing the meal,’ then the probability density functions are, respectively,
\[{f}_{1}(x)=\{\begin{array}{ll}0 & \text{if}\ x<0, \\ \frac{1}{15}{e}^{\text{-}x\text{/}15} & \text{if}\ x\ge 0.\end{array}\ \text{and}\ {f}_{2}(y)=\{\begin{array}{ll}0 & \text{if}\ y<0, \\ \frac{1}{40}{e}^{\text{-}y\text{/}40} & \text{if}\ y\ge 0.\end{array}\]
Clearly, the events are independent and hence the joint density function is the product of the individual functions
\[f(x,y)={f}_{1}(x){f}_{2}(y)=\{\begin{array}{ll}0 & \text{if}\ x<0\ \text{or}\ y<0, \\ \frac{1}{600}{e}^{\text{-}x\text{/}15}{e}^{\text{-}y\text{/}60} & \text{if}\ x,y\ge 0.\end{array}\]
We want to find the probability that the combined time \(X+Y\) is less than \(90\) minutes. In terms of geometry, it means that the region \(D\) is in the first quadrant bounded by the line \(x+y=90\) ().
Hence, the probability that \((X,Y)\) is in the region \(D\) is
\[P(X+Y\le 90)=P((X,Y)\in D)=\underset{D}{∬}f(x,y)dA=\underset{D}{∬}\frac{1}{600}{e}^{\text{-}x\text{/}15}{e}^{\text{-}y\text{/}40}dA.\]
Since \(x+y=90\) is the same as \(y=90-x,\) we have a region of Type I, so
\[\begin{array}{llll}D & = & \{(x,y)|0\le x\le 90,0\le y\le 90-x\}, \\ P(X+Y\le 90) & = & \begin{array}{ll}\frac{1}{600} & \int _{x=0}^{x=90}\int _{y=0}^{y=90-x}{e}^{\text{-}x\text{/}15}{e}^{\text{-}y\text{/}40}dydx\end{array} \\ & = & \begin{array}{ll}\frac{1}{600} & \int _{x=0}^{x=90}\int _{y=0}^{y=90-x}{e}^{\text{-}(x\text{/}15+y\text{/}40)}dydx=0.8328.\end{array}\end{array}\]
Thus, there is an \(83.28\text{\%}\) chance that a customer spends less than an hour and a half at the restaurant.
Find the expected time for the events ‘waiting for a table’ and ‘completing the meal’ in .

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Using the first quadrant of the rectangular coordinate plane as the sample space, we have improper integrals for \(E(X)\) and \(E(Y).\) The expected time for a table is
\[\begin{array}{ll}E(X) & =\underset{S}{∬}x\frac{1}{600}{e}^{\text{-}x\text{/}15}{e}^{\text{-}y\text{/}40}dA=\frac{1}{600}\int _{x=0}^{x=\infty }\ \int _{y=0}^{y=\infty }x{e}^{\text{-}x\text{/}15}{e}^{\text{-}y\text{/}40}dA \\ & =\frac{1}{600}\underset{(a,b)\to (\infty ,\infty )}{\text{lim}}\int _{x=0}^{x=a}\ \int _{y=0}^{y=b}x{e}^{\text{-}x\text{/}15}{e}^{\text{-}y\text{/}40}dx\ dy \\ & =\frac{1}{600}(\underset{a\to \infty }{\text{lim}}\int _{x=0}^{x=a}x{e}^{\text{-}x\text{/}15}dx)(\underset{b\to \infty }{\text{lim}}\int _{y=0}^{y=b}{e}^{\text{-}y\text{/}40}dy) \\ & =\frac{1}{600}({(\underset{a\to \infty }{\text{lim}}(-15{e}^{\text{-}x\text{/}15}(x+15)))|}_{x=0}^{x=a})({(\underset{b\to \infty }{\text{lim}}(-40{e}^{\text{-}y\text{/}40}))|}_{y=0}^{y=b}) \\ & =\frac{1}{600}(\underset{a\to \infty }{\text{lim}}(-15{e}^{\text{-}a\text{/}15}(x+15)+225))(\underset{b\to \infty }{\text{lim}}(-40{e}^{\text{-}b\text{/}40}+40)) \\ & =\frac{1}{600}(225)(40) \\ & =15.\end{array}\]
A similar calculation shows that \(E(Y)=40.\) This means that the expected values of the two random events are the average waiting time and the average dining time, respectively.
The joint density function for two random variables \(X\) and \(Y\) is given by
\[f(x,y)=\begin{array}{ll}\frac{1}{16250}({x}^{2}+{y}^{2}) & \text{if}\ 0\le x\le 15,0\le y\le 10 \\ 0 & \text{otherwise}\end{array}\]
Find the probability that \(X\) is at most \(10\) and \(Y\) is at least \(5.\)

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\(\frac{11}{39}\approx 0.282\)
The region \(D\) bounded by \(y={x}^{3},\) \(y={x}^{3}+1,\) \(x=0,\) and \(x=1\) as given in the following figure.
Find the average value of the function \(f(x,y)=3xy\) on the region graphed in the previous exercise.

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\(\frac{27}{20}\)
Find the area of the region \(D\) given in the previous exercise.
The region \(D\) bounded by \(y=\text{sin}\ x,y=1+\text{sin}\ x,x=0,\ \text{and}\ x=\frac{\pi }{2}\) as given in the following figure.

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Type I but not Type II
Find the average value of the function \(f(x,y)=\text{cos}\ x\) on the region graphed in the previous exercise.
Find the area of the region \(D\) given in the previous exercise.

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\(\frac{\pi }{2}\)
The region \(D\) bounded by \(x={y}^{2}-1\) and \(x=\sqrt{1-{y}^{2}}\) as given in the following figure.
Find the volume of the solid under the graph of the function \(f(x,y)=xy+1\) and above the region in the figure in the previous exercise.

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\(\frac{1}{6}(8+3\pi )\)
The region \(D\) bounded by \(y=0,x=-10+y,\ \text{and}\ x=10-y\) as given in the following figure.
Find the signed volume of the solid under the graph of the function \(f(x,y)=x+y\) and above the region in the figure from the previous exercise.

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\(\frac{1000}{3}\)
The region \(D\) bounded by \(y=0,x=y-1,\) \(x=\frac{\pi }{2}\) as given in the following figure.
The region \(D\) bounded by \(y=0\) and \(y={x}^{2}-1\) as given in the following figure.

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Type I and Type II
Let \(D\) be the region bounded by the curve \(y=2-{x}^{2}\) and below the equations \(y=x,y=\text{-}x,\) Explain why \(D\) is neither of Type I nor II.
Let \(D\) be the region bounded above by the curve of the equation \(y=4-{x}^{2}\) and below by \(y=\text{cos}x\) and the \(x\)-axis. Explain why \(D\) is neither of Type I nor II.

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The region \(D\) is not of Type I: it does not lie between two vertical lines and the graphs of two continuous functions \({g}_{1}(x)\) and \({g}_{2}(x).\) The region \(D\) is not of Type II: it does not lie between two horizontal lines and the graphs of two continuous functions \({h}_{1}(y)\) and \({h}_{2}(y).\)
\(f(x,y)=2x+5y\) and \(D=\{(x,y)|0\le x\le 1,{x}^{3}\le y\le {x}^{3}+1\}\)
\(f(x,y)=1\) and \(D=\{(x,y)|0\le x\le \frac{\pi }{2},\text{sin}\ x\le y\le 1+\text{sin}\ x\}\)

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\(\frac{\pi }{2}\)
\(f(x,y)=2\) and \(D=\{(x,y)|0\le y\le 1,y-1\le x\le \text{arccos}\ y\}\)

Symbols used here

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\iint,\ \oint

double / contour integral
Integral over a region of the plane; integral around a closed curve.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\frac{\partial f}{\partial x}

partial derivative
Derivative with respect to x, holding the other variables fixed.

\nabla f

gradient (nabla, del)
Vector of partial derivatives; points uphill.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|

dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: Double Integrals over General Regions

Recognize when a function of two variables is integrable over a general region.
Evaluate a double integral by computing an iterated integral over a region bounded by two vertical lines and two functions of
Simplify the calculation of an iterated integral by changing the order of integration.
Use double integrals to calculate the volume of a region between two surfaces or the area of a plane region.
Solve problems involving double improper integrals.
integrating first with respect to
integrating first with respect to
One way to look at it is by first integrating

Questions people ask

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

Vyskúšajte si vlastné

Parts of this page are adapted from Boelkins et al., Active Calculus Multivariable (CC BY-SA 4.0), OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Viac v kategórii Multivariable Calculus

Functions of several variables Partial derivatives and the gradient Optimisation in several variables Double and triple integrals Vector fields, line integrals and the big theorems