maths.free › Multivariable Calculus › 5. Multiple Integration › Double Integrals in Polar Coordinates
Double Integrals in Polar Coordinates
Recognize the format of a double integral over a polar rectangular region.
Polar Rectangular Regions of Integration
When we defined the double integral for a continuous function in rectangular coordinates—say, \(g\) over a region \(R\) in the \(xy\)-plane—we divided \(R\) into subrectangles with sides parallel to the coordinate axes. These sides have either constant \(x\)-values and/or constant \(y\)-values. In polar coordinates, the shape we work with is a polar rectangle, whose sides have constant \(r\)-values and/or constant \(\theta\)-values. This means we can describe a polar rectangle as in (a), with \(R=\{(r,\theta )|a\le r\le b,\alpha \le \theta \le \beta \}.\)
In this section, we are looking to integrate over polar rectangles. Consider a function \(f(r,\theta )\) over a polar rectangle \(R.\) We divide the interval \([a,b]\) into \(m\) subintervals \([{r}_{i-1},{r}_{i}]\) of length \(\text{\Delta }r=(b-a)\text{/}m\) and divide the interval \([\alpha ,\beta ]\) into \(n\) subintervals \([{\theta }_{j-1},{\theta }_{j}]\) of width \(\text{\Delta }\theta =(\beta -\alpha )\text{/}n.\) This means that the circles \(r={r}_{i}\) and rays \(\theta ={\theta }_{j}\) for \(1\le i\le m\) and \(1\le j\le n\) divide the polar rectangle \(R\) into smaller polar subrectangles \({R}_{ij}\) ((b)).
As before, we need to find the area \(\text{\Delta }A\) of the polar subrectangle \({R}_{ij}\) and the “polar” volume of the thin box above \({R}_{ij}.\) Recall that, in a circle of radius \(r,\) the length \(s\) of an arc subtended by a central angle of \(\theta\) radians is \(s=r\theta .\) Notice that the polar rectangle \({R}_{ij}\) looks a lot like a trapezoid with parallel sides \({r}_{i-1}\text{\Delta }\theta\) and \({r}_{i}\text{\Delta }\theta\) and with a width \(\text{\Delta }r.\) Hence the area of the polar subrectangle \({R}_{ij}\) is
\[\text{\Delta }A=\frac{1}{2}\text{\Delta }r({r}_{i-1}\text{\Delta }\theta +{r}_{i}\text{\Delta }\theta ).\]Simplifying and letting \({r}_{ij}^{*}=\frac{1}{2}({r}_{i-1}+{r}_{i}),\) we have \(\text{\Delta }A={r}_{ij}^{*}\text{\Delta }r\text{\Delta }\theta .\) Therefore, the polar volume of the thin box above \({R}_{ij}\) () is
\[f({r}_{ij}^{*},{\theta }_{ij}^{*})\text{\Delta }A=f({r}_{ij}^{*},{\theta }_{ij}^{*}){r}_{ij}^{*}\text{\Delta }r\text{\Delta }\theta .\]Using the same idea for all the subrectangles and summing the volumes of the rectangular boxes, we obtain a double Riemann sum as
\[\sum _{i=1}^{m}\sum _{j=1}^{n}f({r}_{ij}^{*},{\theta }_{ij}^{*}){r}_{ij}^{*}\text{\Delta }r\text{\Delta }\theta .\]As we have seen before, we obtain a better approximation to the polar volume of the solid above the region \(R\) when we let \(m\) and \(n\) become larger. Hence, we define the polar volume as the limit of the double Riemann sum,
This becomes the expression for the double integral.
Condensed — the full section is in OpenStax Calculus Volume 3.
General Polar Regions of Integration
To evaluate the double integral of a continuous function by iterated integrals over general polar regions, we consider two types of regions, analogous to Type I and Type II as discussed for rectangular coordinates in Double Integrals over General Regions. It is more common to write polar equations as \(r=f(\theta )\) than \(\theta =f(r),\) so we describe a general polar region as \(D=\{(r,\theta )|\alpha \le \theta \le \beta ,{h}_{1}(\theta )\le r\le {h}_{2}(\theta )\}\) (see the following figure).
Example
Try it.
Evaluate the integral \(\underset{D}{∬}{r}^{2}\cdot \sin \theta \cdot r\ dr\ d\theta\) where \(D\) is the region bounded by the polar axis and the upper half of the cardioid \(r=1+\text{cos}\ \theta .\)
Solution
We can describe the region \(D\) as \(\{(r,\theta )|0\le \theta \le \pi ,0\le r\le 1+\text{cos}\ \theta \}\) as shown in the following figure.
Hence, we have
\[\begin{array}{ll}\underset{D}{∬}{r}^{2}\cdot \sin \theta \cdot r\ dr\ d\theta & =\int _{\theta =0}^{\theta =\pi }\ \int _{r=0}^{r=1+\text{cos }\theta }{r}^{3}\cdot \sin \theta \ dr\ d\theta \\ & =\frac{1}{4}\int _{\theta =0}^{\theta =\pi }{[{r}^{4}]}_{r=0}^{r=1+\text{cos}\ \theta }\text{sin}\ \theta \ d\theta \\ & =\frac{1}{4}\int _{\theta =0}^{\theta =\pi }{(1+\text{cos}\ \theta )}^{4}\text{sin}\ \theta \ d\theta \\ & =-\frac{1}{4}{[\frac{{(1+\text{cos}\ \theta )}^{5}}{5}]}_{0}^{\pi }=\frac{8}{5}.\end{array}\]Polar Areas and Volumes
As in rectangular coordinates, if a solid \(S\) is bounded by the surface \(z=f(r,\theta ),\) as well as by the surfaces \(r=a,r=b,\theta =\alpha ,\) and \(\theta =\beta ,\) we can find the volume \(V\) of \(S\) by double integration, as
\[V=\underset{R}{∬}f(r,\theta )r\ dr\ d\theta =\int _{\theta =\alpha }^{\theta =\beta }\ \int _{r=a}^{r=b}f(r,\theta )r\ dr\ d\theta .\]If the base of the solid can be described as \(D=\{(r,\theta )|\alpha \le \theta \le \beta ,{h}_{1}(\theta )\le r\le {h}_{2}(\theta )\},\) then the double integral for the volume becomes
\[V=\underset{D}{∬}f(r,\theta )r\ dr\ d\theta =\int _{\theta =\alpha }^{\theta =\beta }\ \int _{r={h}_{1}(\theta )}^{r={h}_{2}(\theta )}f(r,\theta )r\ dr\ d\theta .\]We illustrate this idea with some examples.
Example
Try it.
Find the volume of the solid that lies under the paraboloid \(z=1-{x}^{2}-{y}^{2}\) and above the unit circle on the \(xy\)-plane (see the following figure).
Solution
By the method of double integration, we can see that the volume is the iterated integral of the form \(\underset{R}{∬}(1-{x}^{2}-{y}^{2})dA\) where \(R=\{(r,\theta )|0\le r\le 1,0\le \theta \le 2\pi \}.\)
This integration was shown before in , so the volume is \(\frac{\pi }{2}\) cubic units.
Notice in the next example that integration is not always easy with polar coordinates. Complexity of integration depends on the function and also on the region over which we need to perform the integration. If the region has a more natural expression in polar coordinates or if \(f\) has a simpler antiderivative in polar coordinates, then the change in polar coordinates is appropriate; otherwise, use rectangular coordinates.
To answer the question of how the formulas for the volumes of different standard solids such as a sphere, a cone, or a cylinder are found, we want to demonstrate an example and find the volume of an arbitrary cone.
As with rectangular coordinates, we can also use polar coordinates to find areas of certain regions using a double integral. As before, we need to understand the region whose area we want to compute. Sketching a graph and identifying the region can be helpful to realize the limits of integration. Generally, the area formula in double integration will look like
\[\text{Area}\ A=\int _{\alpha }^{\beta }\ \int _{{h}_{1}(\theta )}^{{h}_{2}(\theta )}1r\ dr\ d\theta .\]Condensed — the full section is in OpenStax Calculus Volume 3.
Key Concepts
- To apply a double integral to a situation with circular symmetry, it is often convenient to use a double integral in polar coordinates. We can apply these double integrals over a polar rectangular region or a general polar region, using an iterated integral similar to those used with rectangular double integrals.
- The area \(dA\) in polar coordinates becomes \(r\ dr\ d\theta .\)
- Use \(x=r\ \text{cos}\ \theta ,\) \(y=r\ \text{sin}\ \theta ,\) and \(dA=r\ dr\ d\theta\) to convert an integral in rectangular coordinates to an integral in polar coordinates.
- Use \({r}^{2}={x}^{2}+{y}^{2}\) and \(\theta ={\text{tan}}^{-1}(\frac{y}{x})\) to convert an integral in polar coordinates to an integral in rectangular coordinates, if needed.
- To find the volume in polar coordinates bounded above by a surface \(z=f(r,\theta )\) over a region on the \(xy\)-plane, use a double integral in polar coordinates.
Key Equations
| Double integral over a polar rectangular region \(R\) | \(\underset{R}{∬}f(r,\theta )dA=\underset{m,n\to \infty }{\text{lim}}\sum _{i=1}^{m}\sum _{j=1}^{n}f({r}_{ij}*,{\theta }_{ij}*)\Delta A=\underset{m,n\to \infty }{\text{lim}}\sum _{i=1}^{m}\sum _{j=1}^{n}f({r}_{ij}*,{\theta }_{ij}*){r}_{ij}*\Delta r\Delta \theta\) |
| Double integral over a general polar region | \(\underset{D}{∬}f(r,\theta )r\ dr\ d\theta =\int _{\theta =\alpha }^{\theta =\beta }\ \int _{r={h}_{1}(\theta )}^{r={h}_{2}(\theta )}f(r,\theta )r\ dr\ d\theta\) |
Double Integrals in Polar Coordinates
In the following exercises, express the region \(D\) in polar coordinates.
In the following exercises, the graph of the polar rectangular region \(D\) is given. Express \(D\) in polar coordinates.
In the following exercises, evaluate the double integral \(\underset{R}{∬}f(x,y)dA\) over the polar rectangular region \(D.\)
In the following exercises, the integrals have been converted to polar coordinates. Verify that the identities are true and choose the easiest way to evaluate the integrals, in rectangular or polar coordinates.
In the following exercises, convert the integrals to polar coordinates and evaluate them.
For the following two exercises, consider a spherical ring, which is a sphere with a cylindrical hole cut so that the axis of the cylinder passes through the center of the sphere (see the following figure).
Double Integrals in Polar Coordinates
Because we choose to cover polar coordinates in the chapter on the precalculus of multivariable functions, this section jumps right into how polar coordinates work in double integrals and provide some computational advantages. If you did not cover when teaching , you will need to cover at least the subsection on polar coordinates before teaching this section. The preview activity in this section helps students recall important elements of points, graphs, and regions in polar coordinates before we apply the same slicing and iterated integral techniques as in .
This section has only two activities (plus the preview activity). For instructors who introduced polar coordinates in , it will be easy to cover this section in a single class meeting. Instead of having more activities in this section, we encourage instructors to reinforce that cylindrical coordinates are polar coordinates with \(z\) so that the section on triple integrals in cylindrical coordinates can also help students build skills in this area.
Introduction
Suppose that we would like to integrate \(f(x,y) = e^{x^2 + y^2}\) over the the unit disc \(D\) centered at the origin. Using the methods we have learned thus far tin the chapter, we could evaluate this double integral using the iterated integral \[\begin{aligned}\end{aligned}\]. This relies on treating \(D\) as vertically simple, as suggested in .
For this particular integral, we are unable to find an antiderivative of the integrand. Changing the order of integration does not help, as the region and the integrand are symmetric in \(x\) and \(y\). Furthermore, even if we could find an antiderivative, the inner limits of integration involve relatively complicated functions. However, the region of integration has rotational symmetry about the origin. This should make you think about polar coordinates, which we introduced in . In this section, we will see how we can use polar coordinates to write double integrals as iterated integrals, allowing us to both streamline the writing of the limits of integration and make evaluating the iterated integral easier.
Recall that a point \(P\) in rectangular coordinates that is described by an ordered pair \((x,y)\), where \(x\) is the displacement from \(P\) to the \(y\)-axis and \(y\) is the displacement from \(P\) to the \(x\)-axis can also be described with polar coordinates \((r,\theta)\), where \(r\) is the distance from \(P\) to the origin and \(\theta\) is the angle formed by the line segment \(\overline{OP}\) and the positive \(x\)-axis, as shown in Figure.
Trigonometry and the Pythagorean Theorem allow for straightforward conversion from rectangular to polar, and vice versa. A reminder of these formulas, which were developed in , is included below.
- From polar to rectangular
Given polar coordinates \((r,\theta)\) of a point \(P\), the rectangular coordinates \((x,y)\) of \(P\) satisfy \[\begin{aligned}x \amp = r \cos(\theta) \amp y \amp= r \sin(\theta)\end{aligned}\].
- From rectangular to polar
Given rectangular coordinates \((x,y)\) of a point \(P\), the polar coordinates \((r,\theta)\) of \(P\) satisfy \[\begin{aligned}r^2 \amp = x^2 + y^2 \amp \tan(\theta) \amp= \frac{y}{x}\end{aligned}\], assuming \(x \neq 0\).
We can draw graphs of curves in polar coordinates in a similar way to how we do in rectangular coordinates. However, when plotting in polar coordinates, we use a grid that considers changes in angles and changes in distance from the origin. In particular, the angles \(\theta\) and distances \(r\) partition the plane into small wedges as shown in Figure.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Integration in Polar Coordinates
To find a way to evaluate the double integral \[\begin{aligned}\end{aligned}\] where \(D\) is the unit disk using polar coordinates, we will find a way to write this double integral as an iterated integral in the variables \(r\) and \(\theta\). We will need to complete three steps to transform this double integral into polar coordinates:
- Convert the area element to polar coordinates.
- Describe the region of integration with inequalities that correspond to fixing one coordinate and bounding the other in a consistent way.
- \(f(x,y)\)
To write \(dA\) in polar coordinates, we start by remembering what \(dA\) represents. It comes from \(\Delta A\), which is the area of the base of a rectangular prism used to approximate a volume. In rectangular coordinates, \(\Delta A = \Delta x\, \Delta y\), which leads to \(dA = dx\, dy\). To think in polar coordinates, we need to consider the area of a polar rectangle with \(R_0\leq r\leq R_1\) and \(\theta_0 \leq \theta\leq \theta_1\). shows two polar rectangles with the same values of \(\theta_0\) and \(\theta_1\). However, the figure illustrates that a polar rectangle located farther from the origin has greater area than one located closer to the origin, even when \(\Delta\theta = \theta_1 - \theta_0\) and \(\Delta r = R_1 - R_0\) are the same for the two polar rectangles. This suggests that \(dA\) depends on not only \(dr\) and \(d\theta\) but also \(r\).
Before stating and justifying the formula for \(dA\) in polar coordinates, it's worth also thinking about what the units on \(dA\) are. They need to be units of area, so suppose that \(x\) and \(y\) are both measured in centimeters. Then \(dA = dx\, dy\) in rectangular coordinates has units square centimeters, which is an area unit. However, in polar coordinates, \(\theta\) is measured in radians. Conventionally, we treat quantities measured in radians as lacking units, so \(d\theta\, dr\) would have units of centimeters, which is a unit of length, not area.
Note that the unit disc \(D\) we have considered throughout this section can be considered both radially simple and angularly simple since all of the bounds of integration are constant. shows how \(D\) can be considered radially simple.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
-
Sketch the polar rectangular region \(R=\{(r,\theta )|1\le r\le 3,0\le \theta \le \pi \}.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
As we can see from , \(r=1\) and \(r=3\) are circles of radius \(1\ \text{and}\ 3\) and \(0\le \theta \le \pi\) covers the entire top half of the plane. Hence the region \(R\) looks like a semicircular band.
-
Evaluate the integral \(\underset{R}{∬}3x\ dA\) over the region \(R=\{(r,\theta )|1\le r\le 2,0\le \theta \le \pi \}.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
First we sketch a figure similar to but with outer radius \(2.\) From the figure we can see that we have
\[\begin{array}{lllll}\underset{R}{∬}3x\ dA & =\int _{\theta =0}^{\theta =\pi }\ \int _{r=1}^{r=2}3r\ \text{cos}\ \theta r\ dr\ d\theta & & & \begin{array}{l}\text{Use an iterated integral with correct limits} \\ \text{of integration.}\end{array} \\ & =\int _{\theta =0}^{\theta =\pi }\text{cos}\ \theta [{{r}^{3}|}_{r=1}^{r=2}]d\theta & & & \text{Integrate first with respect to}\ r. \\ & =\int _{\theta =0}^{\theta =\pi }7\ \text{cos}\ \theta \ d\theta ={7\ \text{sin}\ \theta |}_{\theta =0}^{\theta =\pi }=0. & & & \end{array}\] -
Sketch the region \(R=\{(r,\theta )|1\le r\le 2,-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}\},\) and evaluate \(\underset{R}{∬}x\ dA.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(\frac{14}{3}\)
-
Evaluate the integral \(\underset{R}{∬}(1-{x}^{2}-{y}^{2})dA\) where \(R\) is the unit disk on the \(xy\)-plane.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
The region \(R\) is a unit disk, so we can describe it as \(R=\{(r,\theta )|0\le r\le 1,0\le \theta \le 2\pi \}.\)
Using the conversion \(x=r\ \text{cos}\ \theta ,y=r\ \text{sin}\ \theta ,\) and \(dA=r\ dr\ d\theta ,\) we have
\[\begin{array}{ll}\underset{R}{∬}(1-{x}^{2}-{y}^{2})dA & =\int _{0}^{2\pi }\ \int _{0}^{1}(1-{r}^{2})r\ dr\ d\theta =\int _{0}^{2\pi }\ \int _{0}^{1}(r-{r}^{3})dr\ d\theta \\ & =\int _{0}^{2\pi }{[\frac{{r}^{2}}{2}-\frac{{r}^{4}}{4}]}_{0}^{1}d\theta =\int _{0}^{2\pi }\frac{1}{4}d\theta =\frac{\pi }{2}.\end{array}\] -
Evaluate the integral \(\underset{R}{∬}(x+y)dA\) where \(R=\{(x,y)|1\le {x}^{2}+{y}^{2}\le 4,x\le 0\}.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
We can see that \(R\) is an annular region that can be converted to polar coordinates and described as \(R=\{(r,\theta )|1\le r\le 2,\frac{\pi }{2}\le \theta \le \frac{3\pi }{2}\}\) (see the following graph).
Hence, using the conversion \(x=r\ \text{cos}\ \theta ,y=r\ \text{sin}\ \theta ,\) and \(dA=r\ dr\ d\theta ,\) we have
\[\begin{array}{ll}\underset{R}{∬}(x+y)dA & =\int _{\theta =\pi \text{/}2}^{\theta =3\pi \text{/}2}\ \int _{r=1}^{r=2}(r\ \text{cos}\ \theta +r\ \text{sin}\ \theta )r\ dr\ d\theta \\ & =(\int _{r=1}^{r=2}{r}^{2}dr)(\int _{\pi \text{/}2}^{3\pi \text{/}2}(\text{cos}\ \theta +\text{sin}\ \theta )d\theta ) \\ & ={[\frac{{r}^{3}}{3}]}_{1}^{2}{[\text{sin}\ \theta -\text{cos}\ \theta ]|}_{\pi \text{/}2}^{3\pi \text{/}2} \\ & =-\frac{14}{3}.\end{array}\] -
Evaluate the integral \(\underset{R}{∬}(4-{x}^{2}-{y}^{2})dA\) where \(R\) is the circle of radius \(2\) on the \(xy\)-plane.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(8\pi\)
-
Evaluate the integral \(\underset{D}{∬}{r}^{2}\cdot \sin \theta \cdot r\ dr\ d\theta\) where \(D\) is the region bounded by the polar axis and the upper half of the cardioid \(r=1+\text{cos}\ \theta .\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
We can describe the region \(D\) as \(\{(r,\theta )|0\le \theta \le \pi ,0\le r\le 1+\text{cos}\ \theta \}\) as shown in the following figure.
Hence, we have
\[\begin{array}{ll}\underset{D}{∬}{r}^{2}\cdot \sin \theta \cdot r\ dr\ d\theta & =\int _{\theta =0}^{\theta =\pi }\ \int _{r=0}^{r=1+\text{cos }\theta }{r}^{3}\cdot \sin \theta \ dr\ d\theta \\ & =\frac{1}{4}\int _{\theta =0}^{\theta =\pi }{[{r}^{4}]}_{r=0}^{r=1+\text{cos}\ \theta }\text{sin}\ \theta \ d\theta \\ & =\frac{1}{4}\int _{\theta =0}^{\theta =\pi }{(1+\text{cos}\ \theta )}^{4}\text{sin}\ \theta \ d\theta \\ & =-\frac{1}{4}{[\frac{{(1+\text{cos}\ \theta )}^{5}}{5}]}_{0}^{\pi }=\frac{8}{5}.\end{array}\] -
Evaluate the integral
\[\underset{D}{∬}{r}^{2}{\text{sin}}^{2}\left(2\theta \right)r\ dr\ d\theta \ \text{where}\ D=\{(r,\theta )|-\frac{\pi }{4}\le \theta \le \frac{\pi }{4},0\le r\le 2\sqrt{\text{cos}\ 2\theta }\}.\]ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(\pi \text{/}4\)
-
Find the volume of the solid that lies under the paraboloid \(z=1-{x}^{2}-{y}^{2}\) and above the unit circle on the \(xy\)-plane (see the following figure).
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
By the method of double integration, we can see that the volume is the iterated integral of the form \(\underset{R}{∬}(1-{x}^{2}-{y}^{2})dA\) where \(R=\{(r,\theta )|0\le r\le 1,0\le \theta \le 2\pi \}.\)
This integration was shown before in , so the volume is \(\frac{\pi }{2}\) cubic units.
-
Find the volume of the solid that lies under the paraboloid \(z=4-{x}^{2}-{y}^{2}\) and above the disk \({(x-1)}^{2}+{y}^{2}=1\) on the \(xy\)-plane. See the paraboloid in intersecting the cylinder \({(x-1)}^{2}+{y}^{2}=1\) above the \(xy\)-plane.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
First change the disk \({(x-1)}^{2}+{y}^{2}=1\) to polar coordinates. Expanding the square term, we have \({x}^{2}-2x+1+{y}^{2}=1.\) Then simplify to get \({x}^{2}+{y}^{2}=2x,\) which in polar coordinates becomes \({r}^{2}=2r\ \text{cos}\ \theta\) and then either \(r=0\) or \(r=2\ \text{cos}\ \theta .\) Similarly, the equation of the paraboloid changes to \(z=4-{r}^{2}.\) Therefore we can describe the disk \({(x-1)}^{2}+{y}^{2}=1\) on the \(xy\)-plane as the region
\[D=\{(r,\theta )|0\le \theta \le \pi ,0\le r\le 2\ \text{cos}\ \theta \}.\]Hence the volume of the solid below the paraboloid \(z=4-{x}^{2}-{y}^{2}\) and above \(r=2\ \text{cos}\ \theta\) is
\[\begin{array}{ll}V & =\underset{D}{∬}f(r,\theta )r\ dr\ d\theta =\int _{\theta =-\frac{\pi }{2}}^{\theta =\frac{\pi }{2}}\ \int _{r=0}^{r=2\ \text{cos}\ \theta }(4-{r}^{2})r\ dr\ d\theta \\ & =\int _{\theta =-\frac{\pi }{2}}^{\theta =\frac{\pi }{2}}[4\frac{{r}^{2}}{2}-{\frac{{r}^{4}}{4}|}_{0}^{2\ \text{cos}\ \theta }]d\theta \\ & =\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}[8\ {\text{cos}}^{2}\theta -4\ {\text{cos}}^{4}\theta ]d\theta ={[\frac{5}{2}\theta +\frac{5}{2}\text{sin}\ \text{2}\theta -\frac{1}{8}\text{sin}\ \text{4}\theta ]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}=\frac{5}{2}\pi .\end{array}\] -
Find the volume of the region that lies under the paraboloid \(z={x}^{2}+{y}^{2}\) and above the triangle enclosed by the lines \(y=x,x=0,\) and \(x+y=2\) in the \(xy\)-plane ().
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
First examine the region over which we need to set up the double integral and the accompanying paraboloid.
The region \(D\) is \(\{(x,y)|0\le x\le 1,x\le y\le 2-x\}.\) Converting the lines \(y=x,x=0,\) and \(x+y=2\) in the \(xy\)-plane to functions of \(r\) and \(\theta ,\) we have \(\theta =\pi \text{/}4,\) \(\theta =\pi \text{/}2,\) and \(r=2\text{/}(\text{cos}\ \theta +\text{sin}\ \theta ),\) respectively. Graphing the region on the \(xy\)-plane, we see that it looks like \(D=\{(r,\theta )|\pi \text{/}4\le \theta \le \pi \text{/}2,0\le r\le 2\text{/}(\text{cos}\ \theta +\text{sin}\ \theta )\}.\) Now converting the equation of the surface gives \(z={x}^{2}+{y}^{2}={r}^{2}.\) Therefore, the volume of the solid is given by the double integral
\[\begin{array}{ll}V & =\underset{D}{∬}f(r,\theta )r\ dr\ d\theta =\int _{\theta =\pi \text{/}4}^{\theta =\pi \text{/}2}\ \int _{r=0}^{r=2\text{/}(\text{cos}\ \theta +\text{sin}\ \theta )}{r}^{2}r\ dr\ d\theta ={\int _{\pi \text{/}4}^{\pi \text{/}2}[\frac{{r}^{4}}{4}]}_{0}^{2\text{/}(\text{cos}\ \theta +\text{sin}\ \theta )}d\theta \\ & =\frac{1}{4}{\int _{\pi \text{/}4}^{\pi \text{/}2}(\frac{2}{\text{cos}\ \theta +\text{sin}\ \theta })}^{4}d\theta =\frac{16}{4}{\int _{\pi \text{/}4}^{\pi \text{/}2}(\frac{1}{\text{cos}\ \theta +\text{sin}\ \theta })}^{4}d\theta =4{\int _{\pi \text{/}4}^{\pi \text{/}2}(\frac{1}{\text{cos}\ \theta +\text{sin}\ \theta })}^{4}d\theta .\end{array}\]As you can see, this integral is very complicated. So, we can instead evaluate this double integral in rectangular coordinates as
\[V=\int _{0}^{1}\ \int _{x}^{2-x}({x}^{2}+{y}^{2})dy\ dx.\]Evaluating gives
\[\begin{array}{ll}V & =\int _{0}^{1}\ \int _{x}^{2-x}({x}^{2}+{y}^{2})dy\ dx={\int _{0}^{1}[{x}^{2}y+\frac{{y}^{3}}{3}]|}_{x}^{2-x}dx \\ & =\int _{0}^{1}\frac{8}{3}-4x+4{x}^{2}-\frac{8{x}^{3}}{3}dx \\ & ={[\frac{8x}{3}-2{x}^{2}+\frac{4{x}^{3}}{3}-\frac{2{x}^{4}}{3}]|}_{0}^{1}=\frac{4}{3}.\end{array}\] -
Use polar coordinates to find the volume inside the cone \(z=2-\sqrt{{x}^{2}+{y}^{2}}\) and above the \(xy\text{-plane}\text{.}\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
The region \(D\) for the integration is the base of the cone, which appears to be a circle on the \(xy\text{-plane}\) (see the following figure).
We find an equation of the circle by setting \(z=0\text{:}\)
\[\begin{array}{lll}0 & = & 2-\sqrt{{x}^{2}+{y}^{2}} \\ 2 & = & \sqrt{{x}^{2}+{y}^{2}} \\ {x}^{2}+{y}^{2} & = & 4.\end{array}\]This means the radius of the circle is \(2,\) so for the integration we have \(0\le \theta \le 2\pi\) and \(0\le r\le 2.\) Substituting \(x=r\ \text{cos}\ \theta\) and \(y=r\ \text{sin}\ \theta\) in the equation \(z=2-\sqrt{{x}^{2}+{y}^{2}}\) we have \(z=2-r.\) Therefore, the volume of the cone is
\(\int _{\theta =0}^{\theta =2\pi }\ \int _{r=0}^{r=2}(2-r)r\ dr\ d\theta =2\pi \frac{4}{3}=\frac{8\pi }{3}\) cubic units.
-
Use polar coordinates to find an iterated integral for finding the volume of the solid enclosed by the paraboloids \(z={x}^{2}+{y}^{2}\) and \(z=16-{x}^{2}-{y}^{2}.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(V=\int _{0}^{2\pi }\ \int _{0}^{2\sqrt{2}}(16-2{r}^{2})r\ dr\ d\theta =64\pi\) cubic units
-
Evaluate the area bounded by the curve \(r=\text{cos}\ 4\theta .\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
Sketching the graph of the function \(r=\text{cos}\ 4\theta\) reveals that it is a polar rose with eight petals (see the following figure).
Using symmetry, we can see that we need to find the area of one petal and then multiply it by \(8.\) Notice that the values of \(\theta\) for which the graph passes through the origin are the zeros of the function \(\text{cos}\ 4\theta ,\) and these are odd multiples of \(\pi \text{/}8.\) Thus, one of the petals corresponds to the values of \(\theta\) in the interval \([\text{-}\pi \text{/}8,\pi \text{/}8].\) Therefore, the area bounded by the curve \(r=\text{cos}\ 4\theta\) is
\[\begin{array}{ll}A & =8\int _{\theta =\text{-}\pi \text{/}8}^{\theta =\pi \text{/}8}\ \int _{r=0}^{r=\text{cos}\ 4\theta }1r\ dr\ d\theta \\ & =8\int _{\text{-}\pi \text{/}8}^{\pi \text{/}8}[\frac{1}{2}{{r}^{2}|}_{0}^{\text{cos}\ 4\theta }]d\theta =8\int _{\text{-}\pi \text{/}8}^{\pi \text{/}8}\frac{1}{2}{\text{cos}}^{2}4\theta \ d\theta =8[\frac{1}{4}\theta +{\frac{1}{16}\text{sin}\ 4\theta \ \text{cos}\ 4\theta |}_{\text{-}\pi \text{/}8}^{\pi \text{/}8}]=8[\frac{\pi }{16}]=\frac{\pi }{2}.\end{array}\] -
Find the area enclosed by the circle \(r=3\ \text{cos}\ \theta\) and the cardioid \(r=1+\text{cos}\ \theta .\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
First and foremost, sketch the graphs of the region ().
We can from see the symmetry of the graph that we need to find the points of intersection. Setting the two equations equal to each other gives
\[3\ \text{cos}\ \theta =1+\text{cos}\ \theta .\]One of the points of intersection is \(\theta =\pi \text{/}3.\) The area above the polar axis consists of two parts, with one part defined by the cardioid from \(\theta =0\) to \(\theta =\pi \text{/}3\) and the other part defined by the circle from \(\theta =\pi \text{/}3\) to \(\theta =\pi \text{/}2.\) By symmetry, the total area is twice the area above the polar axis. Thus, we have
\[A=2[\int _{\theta =0}^{\theta =\pi \text{/}3}\ \int _{r=0}^{r=1+\text{cos}\ \theta }1r\ dr\ d\theta +\int _{\theta =\pi \text{/}3}^{\theta =\pi \text{/}2}\ \int _{r=0}^{r=3\ \text{cos}\ \theta }1r\ dr\ d\theta ].\]Evaluating each piece separately, we find that the area is
\[A=2(\frac{1}{4}\pi +\frac{9}{16}\sqrt{3}+\frac{3}{8}\pi -\frac{9}{16}\sqrt{3})=2(\frac{5}{8}\pi )=\frac{5}{4}\pi \ \text{square units}\text{.}\] -
Find the area enclosed inside the cardioid \(r=3-3\ \text{sin}\ \theta\) and outside the cardioid \(r=1+\text{sin}\ \theta .\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(A=2\int _{\text{-}\pi \text{/}2}^{\pi \text{/}6}\ \int _{1+\text{sin}\ \theta }^{3-3\ \text{sin}\ \theta }r\ dr\ d\theta =8\pi +9\sqrt{3}\)
-
Evaluate the integral \(\underset{{ℝ}^{2}}{∬}{e}^{-10({x}^{2}+{y}^{2})}dx\ dy.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
This is an improper integral because we are integrating over an unbounded region \({ℝ}^{2}.\) In polar coordinates, the entire plane \({ℝ}^{2}\) can be seen as \(0\le \theta \le 2\pi ,\) \(0\le r<\infty .\)
Using the changes of variables from rectangular coordinates to polar coordinates, we have
\[\begin{array}{ll}\underset{{ℝ}^{2}}{∬}{e}^{-10({x}^{2}+{y}^{2})}dx\ dy & =\int _{\theta =0}^{\theta =2\pi }\ \int _{r=0}^{r=\infty }{e}^{-10{r}^{2}}r\ dr\ d\theta =\int _{\theta =0}^{\theta =2\pi }(\underset{a\to \infty }{\text{lim}}\int _{r=0}^{r=a}{e}^{-10{r}^{2}}r\ dr)d\theta \\ & =(\int _{\theta =0}^{\theta =2\pi }d\theta )(\underset{a\to \infty }{\text{lim}}\int _{r=0}^{r=a}{e}^{-10{r}^{2}}r\ dr) \\ & =2\pi (\underset{a\to \infty }{\text{lim}}\int _{r=0}^{r=a}{e}^{-10{r}^{2}}r\ dr) \\ & =2\pi \underset{a\to \infty }{\text{lim}}(-\frac{1}{20})({{e}^{-10{r}^{2}}|}_{0}^{a}) \\ & =2\pi (-\frac{1}{20})\underset{a\to \infty }{\text{lim}}({e}^{-10{a}^{2}}-1) \\ & =\frac{\pi }{10}.\end{array}\] -
Evaluate the integral \(\underset{{ℝ}^{2}}{∬}{e}^{-4({x}^{2}+{y}^{2})}dx\ dy.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(\frac{\pi }{4}\)
-
\(D\) is the region of the disk of radius \(2\) centered at the origin that lies in the first quadrant.
-
\(D\) is the region between the circles of radius \(4\) and radius \(5\) centered at the origin that lies in the second quadrant.
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(D=\{(r,\theta )|4\le r\le 5,\frac{\pi }{2}\le \theta \le \pi \}\)
-
\(D\) is the region bounded by the \(y\)-axis and \(x=\sqrt{1-{y}^{2}}.\)
-
\(D\) is the region bounded by the \(x\)-axis and \(y=\sqrt{2-{x}^{2}}.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(D=\{(r,\theta )|0\le r\le \sqrt{2},0\le \theta \le \pi \}\)
-
\(D=\{(x,y)|{x}^{2}+{y}^{2}\le 4x\}\)
-
\(D=\{(x,y)|{x}^{2}+{y}^{2}\le 4y\}\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(D=\{(r,\theta )|0\le r\le 4\ \text{sin}\ \theta ,0\le \theta \le \pi \}\)
-
In the following graph, the region \(D\) is situated below \(y=x\) and is bounded by \(x=1,x=5,\) and \(y=0.\)
-
In the following graph, the region \(D\) is bounded by \(y=x\) and \(y={x}^{2}.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(D=\{(r,\theta )|0\le r\le \text{tan}\ \theta \ \text{sec}\ \theta ,0\le \theta \le \frac{\pi }{4}\}\)
-
\(f(x,y)={x}^{2}+{y}^{2},D=\{(r,\theta )|3\le r\le 5,0\le \theta \le 2\pi \}\)
-
\(f(x,y)=x+y,\ \ D=\{(r,\theta )|3\le r\le 5,0\le \theta \le 2\pi \}\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(0\)
-
\(f(x,y)={x}^{2}+xy,D=\{(r,\theta )|1\le r\le 2,\pi \le \theta \le 2\pi \}\)
-
\(f(x,y)={x}^{4}+{y}^{4},D=\{(r,\theta )|1\le r\le 2,\frac{3\pi }{2}\le \theta \le 2\pi \}\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(\frac{63\pi }{16}\)
-
\(f(x,y)=\sqrt[3]{{x}^{2}+{y}^{2}},\) where \(D=\{(r,\theta )|0\le r\le 1,\frac{\pi }{2}\le \theta \le \pi \}.\)
-
\(f(x,y)={x}^{4}+2{x}^{2}{y}^{2}+{y}^{4},\) where \(D=\{(r,\theta )|3\le r\le 4,\frac{\pi }{3}\le \theta \le \frac{2\pi }{3}\}.\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(\frac{3367\pi }{18}\)
-
\(f(x,y)=\text{sin}(\text{arctan}\ \frac{y}{x}),\) where \(D=\{(r,\theta )|1\le r\le 2,\frac{\pi }{6}\le \theta \le \frac{\pi }{3}\}\)
-
\(f(x,y)=\text{arctan}(\frac{y}{x}),\) where \(D=\{(r,\theta )|2\le r\le 3,\frac{\pi }{4}\le \theta \le \frac{\pi }{3}\}\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(\frac{35{\pi }^{2}}{576}\)
-
\(\underset{D}{∬}{e}^{{x}^{2}+{y}^{2}}[1+2\ \text{arctan}(\frac{y}{x})]dA\text{,}\ D=\{(r,\theta )|1\le r\le 2,\frac{\pi }{6}\le \theta \le \frac{\pi }{3}\}\)
-
\(\underset{D}{∬}({e}^{{x}^{2}+{y}^{2}}+{x}^{4}+2{x}^{2}{y}^{2}+{y}^{4})\text{arctan}(\frac{y}{x})dA\text{,}\ D=\{(r,\theta )|1\le r\le 2,\frac{\pi }{4}\le \theta \le \frac{\pi }{3}\}\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(\frac{7{\pi }^{2}}{576}\left(21-{e}^{2}+{e}^{4}\right)\)
-
\(\int _{1}^{2}\ \int _{0}^{x}({x}^{2}+{y}^{2})dy\ dx=\int _{0}^{\frac{\pi }{4}}\ \int _{\text{sec}\ \theta }^{2\ \text{sec}\ \theta }{r}^{3}dr\ d\theta\)
-
\(\int _{2}^{3}\ \int _{0}^{x}\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}dy\ dx=\int _{0}^{\pi \text{/}4}\ \int _{2\ \text{sec}\ \theta }^{3\text{sec}\ \theta }r\ \text{cos}\ \theta \ dr\ d\theta\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(\frac{5}{2}\text{ln}(1+\sqrt{2})\)
-
\(\int _{0}^{1}\ \int _{{x}^{2}}^{x}\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}dy\ dx=\int _{0}^{\pi \text{/}4}\ \int _{0}^{\text{tan}\ \theta \ \text{sec}\ \theta }dr\ d\theta\)
-
\(\int _{0}^{1}\ \int _{{x}^{2}}^{x}\frac{y}{\sqrt{{x}^{2}+{y}^{2}}}dy\ dx=\int _{0}^{\pi \text{/}4}\ \int _{0}^{\text{tan}\ \theta \ \text{sec}\ \theta }r\ \text{sin}\ \theta \ dr\ d\theta\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
\(\frac{1}{6}(2-\sqrt{2})\)
Symbols used here
Add a_k for k = 1 up to n.
Antiderivative (indefinite) or signed area from a to b (definite).
The non-negative number whose square (n-th power) is x.
Not a number: "grows without bound" in limits and intervals.
Ratio of a circle's circumference to its diameter, 3.14159…
The usual name for an angle.
Ratios of sides in a right triangle; coordinates on the unit circle.
i² = −1.
Inequalities that allow equality; < and > exclude it.
Least upper bound, greatest lower bound.
Derivative with respect to x, holding the other variables fixed.
Vector of partial derivatives; points uphill.
Integral over a region of the plane; integral around a closed curve.
A quantity with magnitude and direction; a column of numbers.
Σ u_i v_i; the length of v, √(v·v).
How to: Double Integrals in Polar Coordinates
- Recognize the format of a double integral over a polar rectangular region.
- Evaluate a double integral in polar coordinates by using an iterated integral.
- Recognize the format of a double integral over a general polar region.
- Use double integrals in polar coordinates to calculate areas and volumes.
- To apply a double integral to a situation with circular symmetry, it is often convenient to use a double integral in polar coordinates. We can apply these double integrals over a polar rectangular region or a general polar region, using an iterated integral similar to those used with rectangular double integrals.
- The area
- Use
- Use
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.
ನಿಮ್ಮದೇ ಆದದ್ದನ್ನು ಪ್ರಯತ್ನಿಸಿ
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.
ಇನ್ನಷ್ಟು Multivariable Calculus
Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems