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Determining Volumes by Slicing

Determine the volume of a solid by integrating a cross-section (the slicing method).

Volume and the Slicing Method

Just as area is the numerical measure of a two-dimensional region, volume is the numerical measure of a three-dimensional solid. Most of us have computed volumes of solids by using basic geometric formulas. The volume of a rectangular solid, for example, can be computed by multiplying length, width, and height: \(V=lwh.\) The formulas for the volume of a sphere \((V=\frac{4}{3}\pi {r}^{3}),\) a cone \((V=\frac{1}{3}\pi {r}^{2}h),\) and a pyramid \((V=\frac{1}{3}Ah)\) have also been introduced. Although some of these formulas were derived using geometry alone, all these formulas can be obtained by using integration.

We can also calculate the volume of a cylinder. Although most of us think of a cylinder as having a circular base, such as a soup can or a metal rod, in mathematics the word cylinder has a more general meaning. To discuss cylinders in this more general context, we first need to define some vocabulary.

We define the cross-section of a solid to be the intersection of a plane with the solid. A cylinder is defined as any solid that can be generated by translating a plane region along a line perpendicular to the region, called the axis of the cylinder. Thus, all cross-sections perpendicular to the axis of a cylinder are identical. The solid shown in is an example of a cylinder with a noncircular base. To calculate the volume of a cylinder, then, we simply multiply the area of the cross-section by the height of the cylinder: \(V=A\cdot h.\) In the case of a right circular cylinder (soup can), this becomes \(V=\pi {r}^{2}h.\)

If a solid does not have a constant cross-section (and it is not one of the other basic solids), we may not have a formula for its volume. In this case, we can use a definite integral to calculate the volume of the solid. We do this by slicing the solid into pieces, estimating the volume of each slice, and then adding those estimated volumes together. The slices should all be parallel to one another, and when we put all the slices together, we should get the whole solid. Consider, for example, the solid S shown in , extending along the \(x\text{-axis}\text{.}\)

Because the cross-sectional area is not constant, we let \(A(x)\) represent the area of the cross-section at point \(x.\) Now let \(P=\{{x}_{0},{x}_{1}\text{\ldots },{X}_{n}\}\) be a regular partition of \([a,b],\) and for \(i=1,2\text{,\ldots }n,\) let \({S}_{i}\) represent the slice of \(S\) stretching from \({x}_{i-1}\text{to}\ {x}_{i}.\) The following figure shows the sliced solid with \(n=3.\)

\[V(S)\approx \sum _{i=1}^{n}A({x}_{i}^{*})\text{\Delta }x.\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Solids of Revolution

If a region in a plane is revolved around a line in that plane, the resulting solid is called a solid of revolution, as shown in the following figure.

Solids of revolution are common in mechanical applications, such as machine parts produced by a lathe. We spend the rest of this section looking at solids of this type. The next example uses the slicing method to calculate the volume of a solid of revolution.

Example

Try it.

Use the slicing method to find the volume of the solid of revolution bounded by the graphs of \(f(x)={x}^{2}-4x+5,x=1,\ \text{and}\ x=4,\) and rotated about the \(x\text{-axis}\text{.}\)

Solution

Using the problem-solving strategy, we first sketch the graph of the quadratic function over the interval \([1,4]\) as shown in the following figure.

Next, revolve the region around the x-axis, as shown in the following figure.

Since the solid was formed by revolving the region around the \(x\text{-axis,}\) the cross-sections are circles (step 1). The area of the cross-section, then, is the area of a circle, and the radius of the circle is given by \(f(x).\) Use the formula for the area of the circle:

\[A(x)=\pi {r}^{2}=\pi {[f(x)]}^{2}=\pi {({x}^{2}-4x+5)}^{2}\ \text{(step 2)}.\]

The volume, then, is (step 3)

\[\begin{array}{ll}V & =\int _{a}^{b}A(x)dx \\ & ={\int }_{1}^{4}\pi {({x}^{2}-4x+5)}^{2}dx=\pi {\int }_{1}^{4}({x}^{4}-8{x}^{3}+26{x}^{2}-40x+25)dx \\ & ={\pi (\frac{{x}^{5}}{5}-2{x}^{4}+\frac{26{x}^{3}}{3}-20{x}^{2}+25x)|}_{1}^{4}=\frac{78}{5}\pi .\end{array}\]

The volume is \(78\pi \text{/}5.\)

The Disk Method

When we use the slicing method with solids of revolution, it is often called the disk method because, for solids of revolution, the slices used to over approximate the volume of the solid are disks. To see this, consider the solid of revolution generated by revolving the region between the graph of the function \(f(x)={(x-1)}^{2}+1\) and the \(x\text{-axis}\) over the interval \([-1,3]\) around the \(x\text{-axis}\text{.}\) The graph of the function and a representative disk are shown in (a) and (b). The region of revolution and the resulting solid are shown in (c) and (d).

We already used the formal Riemann sum development of the volume formula when we developed the slicing method. We know that

\[V={\int }_{a}^{b}A(x)dx.\]

The only difference with the disk method is that we know the formula for the cross-sectional area ahead of time; it is the area of a circle. This gives the following rule.

The volume of the solid we have been studying () is given by

\[\begin{array}{ll}V & ={\int }_{a}^{b}\pi {[f(x)]}^{2}dx \\ & ={\int }_{-1}^{3}\pi {[{(x-1)}^{2}+1]}^{2}dx=\pi {\int }_{-1}^{3}[{(x-1)}^{4}+2{(x-1)}^{2}+1]dx \\ & =\pi {[\frac{1}{5}{(x-1)}^{5}+\frac{2}{3}{(x-1)}^{3}+x]\ |}_{-1}^{3}=\pi [(\frac{32}{5}+\frac{16}{3}+3)-(-\frac{32}{5}-\frac{16}{3}-1)]=\frac{412\pi }{15}\ {\text{units}}^{3}.\end{array}\]

Let’s look at some examples.

Example

Try it.

Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of \(f(x)=\sqrt{x}\) and the \(x\text{-axis}\) over the interval \([1,4]\) around the \(x\text{-axis}\text{.}\)

Solution

The graphs of the function and the solid of revolution are shown in the following figure.

We have

\[\begin{array}{ll}V & ={\int }_{a}^{b}\pi {[f(x)]}^{2}dx \\ & ={\int }_{1}^{4}\pi {[\sqrt{x}]}^{2}dx=\pi {\int }_{1}^{4}x\ dx \\ & ={\frac{\pi }{2}{x}^{2}|}_{1}^{4}=\frac{15\pi }{2}.\end{array}\]

The volume is \((15\pi )\text{/}2\) units³.

The next example shows how this rule works in practice.

Condensed — the full section is in OpenStax Calculus Volume 2.

The Washer Method

Some solids of revolution have cavities in the middle; they are not solid all the way to the axis of revolution. Sometimes, this is just a result of the way the region of revolution is shaped with respect to the axis of revolution. In other cases, cavities arise when the region of revolution is defined as the region between the graphs of two functions. A third way this can happen is when an axis of revolution other than the \(x\text{-axis}\) or \(y\text{-axis}\) is selected.

When the solid of revolution has a cavity in the middle, the slices used to approximate the volume are not disks, but washers (disks with holes in the center). For example, consider the region bounded above by the graph of the function \(f(x)=\sqrt{x}\) and below by the graph of the function \(g(x)=1\) over the interval \([1,4].\) When this region is revolved around the \(x\text{-axis,}\) the result is a solid with a cavity in the middle, and the slices are washers. The graph of the function and a representative washer are shown in (a) and (b). The region of revolution and the resulting solid are shown in (c) and (d).

The cross-sectional area, then, is the area of the outer circle less the area of the inner circle. In this case,

\[A(x)=\pi {(\sqrt{x})}^{2}-\pi {(1)}^{2}=\pi (x-1).\]

Then the volume of the solid is

\[\begin{array}{ll}V & ={\int }_{a}^{b}A(x)dx \\ & ={\int }_{1}^{4}\pi (x-1)dx={\pi [\frac{{x}^{2}}{2}-x]\ |}_{1}^{4}=\frac{9}{2}\pi \ {\text{units}}^{3}.\end{array}\]

Generalizing this process gives the washer method.

Example

Try it.

Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of \(f(x)=x\) and below by the graph of \(g(x)=1\text{/}x\) over the interval \([1,4]\) around the \(x\text{-axis}\text{.}\)

Solution

The graphs of the functions and the solid of revolution are shown in the following figure.

We have

\[\begin{array}{ll}V & ={\int }_{a}^{b}\pi [{(f(x))}^{2}-{(g(x))}^{2}]dx \\ & =\pi {\int }_{1}^{4}[{x}^{2}-{(\frac{1}{x})}^{2}]dx\ \ ={\pi [\frac{{x}^{3}}{3}+\frac{1}{x}]\ |}_{1}^{4}=\frac{81\pi }{4}\ {\text{units}}^{3}.\end{array}\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Key Concepts

Definite integrals can be used to find the volumes of solids. Using the slicing method, we can find a volume by integrating the cross-sectional area.
For solids of revolution, the volume slices are often disks and the cross-sections are circles. The method of disks involves applying the method of slicing in the particular case in which the cross-sections are circles, and using the formula for the area of a circle.
If a solid of revolution has a cavity in the center, the volume slices are washers. With the method of washers, the area of the inner circle is subtracted from the area of the outer circle before integrating.

Key Equations

Disk Method along the x-axis	\(V={\int }_{a}^{b}\pi {[f(x)]}^{2}dx\)
Disk Method along the y-axis	\(V={\int }_{c}^{d}\pi {[g(y)]}^{2}dy\)
Washer Method	\(V={\int }_{a}^{b}\pi [{(f(x))}^{2}-{(g(x))}^{2}]dx\)

Determining Volumes by Slicing

For the following exercises, draw a typical slice and find the volume using the slicing method for the given volume.

For the following exercises, draw an outline of the solid and find the volume using the slicing method.

For the following exercises, draw the region bounded by the curves. Then, use the disk method to find the volume when the region is rotated around the x-axis.

For the following exercises, draw the region bounded by the curves. Then, find the volume when the region is rotated around the y-axis.

For the following exercises, draw the region bounded by the curves. Then, find the volume when the region is rotated around the x-axis.

For the following exercises, draw the region bounded by the curves. Then, use the washer method to find the volume when the region is revolved around the y-axis.

For the following exercises, find the volume of the solid described.

Volume and the Slicing Method

\[V(S)\approx \sum _{i=1}^{n}A({x}_{i}^{*})\text{\Delta }x.\]

Condensed — the full section is in OpenStax Calculus Volume 1.

Solids of Revolution

If a region in a plane is revolved around a line in that plane, the resulting solid is called a solid of revolution, as shown in the following figure.

Example

Try it.

Use the slicing method to find the volume of the solid of revolution bounded by the graphs of \(f(x)={x}^{2}-4x+5,x=1,\ \text{and}\ x=4,\) and rotated about the \(x\text{-axis}\text{.}\)

Solution

Using the problem-solving strategy, we first sketch the graph of the quadratic function over the interval \([1,4]\) as shown in the following figure.

Next, revolve the region around the x-axis, as shown in the following figure.

\[A(x)=\pi {r}^{2}=\pi {[f(x)]}^{2}=\pi {({x}^{2}-4x+5)}^{2}\ \text{(step 2)}.\]

The volume, then, is (step 3)

The volume is \(78\pi \text{/}5.\)

The Disk Method

We already used the formal Riemann sum development of the volume formula when we developed the slicing method. We know that

\[V={\int }_{a}^{b}A(x)dx.\]

The only difference with the disk method is that we know the formula for the cross-sectional area ahead of time; it is the area of a circle. This gives the following rule.

The volume of the solid we have been studying () is given by

Let’s look at some examples.

Example

Try it.

Solution

The graphs of the function and the solid of revolution are shown in the following figure.

We have

\[\begin{array}{ll}V & ={\int }_{a}^{b}\pi {[f(x)]}^{2}dx \\ & ={\int }_{1}^{4}\pi {[\sqrt{x}]}^{2}dx=\pi {\int }_{1}^{4}x\ dx \\ & ={\frac{\pi }{2}{x}^{2}|}_{1}^{4}=\frac{15\pi }{2}.\end{array}\]

The volume is \((15\pi )\text{/}2\) units³.

The next example shows how this rule works in practice.

Condensed — the full section is in OpenStax Calculus Volume 1.

The Washer Method

The cross-sectional area, then, is the area of the outer circle less the area of the inner circle. In this case,

\[A(x)=\pi {(\sqrt{x})}^{2}-\pi {(1)}^{2}=\pi (x-1).\]

Then the volume of the solid is

\[\begin{array}{ll}V & ={\int }_{a}^{b}A(x)dx \\ & ={\int }_{1}^{4}\pi (x-1)dx={\pi [\frac{{x}^{2}}{2}-x]\ |}_{1}^{4}=\frac{9}{2}\pi \ {\text{units}}^{3}.\end{array}\]

Generalizing this process gives the washer method.

Example

Try it.

Solution

The graphs of the functions and the solid of revolution are shown in the following figure.

We have

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

Definite integrals can be used to find the volumes of solids. Using the slicing method, we can find a volume by integrating the cross-sectional area.
For solids of revolution, the volume slices are often disks and the cross-sections are circles. The method of disks involves applying the method of slicing in the particular case in which the cross-sections are circles, and using the formula for the area of a circle.
If a solid of revolution has a cavity in the center, the volume slices are washers. With the method of washers, the area of the inner circle is subtracted from the area of the outer circle before integrating.

Key Equations

Disk Method along the x-axis	\(V={\int }_{a}^{b}\pi {[f(x)]}^{2}dx\)
Disk Method along the y-axis	\(V={\int }_{c}^{d}\pi {[g(y)]}^{2}dy\)
Washer Method	\(V={\int }_{a}^{b}\pi [{(f(x))}^{2}-{(g(x))}^{2}]dx\)

Determining Volumes by Slicing

For the following exercises, draw a typical slice and find the volume using the slicing method for the given volume.

For the following exercises, draw an outline of the solid and find the volume using the slicing method.

For the following exercises, draw the region bounded by the curves. Then, use the disk method to find the volume when the region is rotated around the x-axis.

For the following exercises, draw the region bounded by the curves. Then, find the volume when the region is rotated around the y-axis.

For the following exercises, draw the region bounded by the curves. Then, find the volume when the region is rotated around the x-axis.

For the following exercises, draw the region bounded by the curves. Then, use the washer method to find the volume when the region is revolved around the y-axis.

For the following exercises, find the volume of the solid described.

Practice (40)

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

We know from geometry that the formula for the volume of a pyramid is \(V=\frac{1}{3}Ah.\) If the pyramid has a square base, this becomes \(V=\frac{1}{3}{a}^{2}h,\) where \(a\) denotes the length of one side of the base. We are going to use the slicing method to derive this formula.

Bonisa impendulo

We want to apply the slicing method to a pyramid with a square base. To set up the integral, consider the pyramid shown in , oriented along the \(x\text{-axis}\text{.}\)
We first want to determine the shape of a cross-section of the pyramid. We know the base is a square, so the cross-sections are squares as well (step 1). Now we want to determine a formula for the area of one of these cross-sectional squares. Looking at (b), and using a proportion, since these are similar triangles, we have
\[\frac{s}{a}=\frac{x}{h}\ \text{or}\ s=\frac{ax}{h}.\]
Therefore, the area of one of the cross-sectional squares is
\[A(x)={s}^{2}={(\frac{ax}{h})}^{2}(\text{step}\ 2).\]
Then we find the volume of the pyramid by integrating from \(0\ \text{to}\ h\) (step \(3)\text{:}\)
\[\begin{array}{ll}V & =\int _{0}^{h}A(x)dx \\ & =\int _{0}^{h}{(\frac{ax}{h})}^{2}dx=\frac{{a}^{2}}{{h}^{2}}\int _{0}^{h}{x}^{2}dx \\ & ={[\frac{{a}^{2}}{{h}^{2}}(\frac{1}{3}{x}^{3})]\ |}_{0}^{h}=\frac{1}{3}{a}^{2}h.\end{array}\]
This is the formula we were looking for.
Use the slicing method to derive the formula \(V=\frac{1}{3}\pi {r}^{2}h\) for the volume of a circular cone.
Use the slicing method to find the volume of the solid of revolution bounded by the graphs of \(f(x)={x}^{2}-4x+5,x=1,\ \text{and}\ x=4,\) and rotated about the \(x\text{-axis}\text{.}\)

Bonisa impendulo

Using the problem-solving strategy, we first sketch the graph of the quadratic function over the interval \([1,4]\) as shown in the following figure.
Next, revolve the region around the x-axis, as shown in the following figure.
Since the solid was formed by revolving the region around the \(x\text{-axis,}\) the cross-sections are circles (step 1). The area of the cross-section, then, is the area of a circle, and the radius of the circle is given by \(f(x).\) Use the formula for the area of the circle:
\[A(x)=\pi {r}^{2}=\pi {[f(x)]}^{2}=\pi {({x}^{2}-4x+5)}^{2}\ \text{(step 2)}.\]
The volume, then, is (step 3)
\[\begin{array}{ll}V & =\int _{a}^{b}A(x)dx \\ & ={\int }_{1}^{4}\pi {({x}^{2}-4x+5)}^{2}dx=\pi {\int }_{1}^{4}({x}^{4}-8{x}^{3}+26{x}^{2}-40x+25)dx \\ & ={\pi (\frac{{x}^{5}}{5}-2{x}^{4}+\frac{26{x}^{3}}{3}-20{x}^{2}+25x)|}_{1}^{4}=\frac{78}{5}\pi .\end{array}\]
The volume is \(78\pi \text{/}5.\)
Use the method of slicing to find the volume of the solid of revolution formed by revolving the region between the graph of the function \(f(x)=1\text{/}x\) and the \(x\text{-axis}\) over the interval \([1,2]\) around the \(x\text{-axis}\text{.}\) See the following figure.

Bonisa impendulo

\(\frac{\pi }{2}\)
Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of \(f(x)=\sqrt{x}\) and the \(x\text{-axis}\) over the interval \([1,4]\) around the \(x\text{-axis}\text{.}\)

Bonisa impendulo

The graphs of the function and the solid of revolution are shown in the following figure.
We have
\[\begin{array}{ll}V & ={\int }_{a}^{b}\pi {[f(x)]}^{2}dx \\ & ={\int }_{1}^{4}\pi {[\sqrt{x}]}^{2}dx=\pi {\int }_{1}^{4}x\ dx \\ & ={\frac{\pi }{2}{x}^{2}|}_{1}^{4}=\frac{15\pi }{2}.\end{array}\]
The volume is \((15\pi )\text{/}2\) units³.
Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of \(f(x)=\sqrt{4-x}\) and the \(x\text{-axis}\) over the interval \([0,\ 4]\) around the \(x\text{-axis}\text{.}\)

Bonisa impendulo

\(8\pi\) units³
Let \(R\) be the region bounded by the graph of \(g(y)=\sqrt{4-y}\) and the \(y\text{-axis}\) over the \(y\text{-axis}\) interval \([0,4].\) Use the disk method to find the volume of the solid of revolution generated by rotating \(R\) around the \(y\text{-axis}\text{.}\)

Bonisa impendulo

shows the function and a representative disk that can be used to estimate the volume. Notice that since we are revolving the function around the \(y\text{-axis,}\) the disks are horizontal, rather than vertical.
The region to be revolved and the full solid of revolution are depicted in the following figure.
To find the volume, we integrate with respect to \(y.\) We obtain
\[\begin{array}{ll}V & ={\int }_{c}^{d}\pi {[g(y)]}^{2}dy \\ & ={\int }_{0}^{4}\pi {[\sqrt{4-y}]}^{2}dy=\pi {\int }_{0}^{4}(4-y)dy \\ & ={\pi [4y-\frac{{y}^{2}}{2}]\ |}_{0}^{4}=8\pi .\end{array}\]
The volume is \(8\pi\) units³.
Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of \(g(y)=y\) and the \(y\text{-axis}\) over the interval \([1,4]\) around the \(y\text{-axis}\text{.}\)

Bonisa impendulo

\(21\pi\) units³
Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of \(f(x)=x\) and below by the graph of \(g(x)=1\text{/}x\) over the interval \([1,4]\) around the \(x\text{-axis}\text{.}\)

Bonisa impendulo

The graphs of the functions and the solid of revolution are shown in the following figure.
We have
\[\begin{array}{ll}V & ={\int }_{a}^{b}\pi [{(f(x))}^{2}-{(g(x))}^{2}]dx \\ & =\pi {\int }_{1}^{4}[{x}^{2}-{(\frac{1}{x})}^{2}]dx\ \ ={\pi [\frac{{x}^{3}}{3}+\frac{1}{x}]\ |}_{1}^{4}=\frac{81\pi }{4}\ {\text{units}}^{3}.\end{array}\]
Find the volume of a solid of revolution formed by revolving the region bounded by the graphs of \(f(x)=\sqrt{x}\) and \(g(x)=1\text{/}x\) over the interval \([1,3]\) around the \(x\text{-axis}\text{.}\)

Bonisa impendulo

\(\frac{10\pi }{3}\) units³
Find the volume of a solid of revolution formed by revolving the region bounded above by \(f(x)=4-x\) and below by the \(x\text{-axis}\) over the interval \([0,4]\) around the line \(y=-2.\)

Bonisa impendulo

The graph of the region and the solid of revolution are shown in the following figure.
We can’t apply the volume formula to this problem directly because the axis of revolution is not one of the coordinate axes. However, we still know that the area of the cross-section is the area of the outer circle less the area of the inner circle. Looking at the graph of the function, we see the radius of the outer circle is given by \(f(x)+2,\) which simplifies to
\[f(x)+2=(4-x)+2=6-x.\]
The radius of the inner circle is \(g(x)=2.\) Therefore, we have
\[\begin{array}{ll}V & ={\int }_{0}^{4}\pi [{(6-x)}^{2}-{(2)}^{2}]dx \\ & =\pi {\int }_{0}^{4}({x}^{2}-12x+32)dx\ \ ={\pi [\frac{{x}^{3}}{3}-6{x}^{2}+32x]\ |}_{0}^{4}=\frac{160\pi }{3}\ {\text{units}}^{3}.\end{array}\]
Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of \(f(x)=x+2\) and below by the \(x\text{-axis}\) over the interval \([0,3]\) around the line \(y=-1.\)

Bonisa impendulo

\(60\pi\) units³
Derive the formula for the volume of a sphere using the slicing method.
Use the slicing method to derive the formula for the volume of a cone.
Use the slicing method to derive the formula for the volume of a tetrahedron with side length \(a.\)
Use the disk method to derive the formula for the volume of a trapezoidal cylinder.
Explain when you would use the disk method versus the washer method. When are they interchangeable?
A pyramid with height 6 units and square base of side 2 units, as pictured here.

Bonisa impendulo

8 units³
A pyramid with height 4 units and a rectangular base with length 2 units and width 3 units, as pictured here.
A tetrahedron with a base side of 4 units, as seen here.

Bonisa impendulo

\(\frac{32}{3\sqrt{2}}\) units³
A pyramid with height 5 units, and an isosceles triangular base with lengths of 6 units and 8 units, as seen here.
A cone of radius \(r\) and height \(h\) has a smaller cone of radius \(r\text{/}2\) and height \(h\text{/}2\) removed from the top, as seen here. The resulting solid is called a frustum.

Bonisa impendulo

\(\frac{7}{24}\pi {r}^{2}h\) units³
The base is a circle of radius \(a.\) The slices perpendicular to the base are squares.
The base is a triangle with vertices \((0,0),(1,0),\) and \((0,1).\) Slices perpendicular to the x-axis are semicircles.

Bonisa impendulo

\(\frac{\pi }{24}\) units³
The base is the region under the parabola \(y=1-{x}^{2}\) in the first quadrant. Slices perpendicular to the xy-plane and parallel to the y-axis are squares.
The base is the region under the parabola \(y=1-{x}^{2}\) and above the \(x\text{-axis}\text{.}\) Slices perpendicular to the \(y\text{-axis}\) are squares.

Bonisa impendulo

\(2\) units³
The base is the region enclosed by \(y={x}^{2}\) and \(y=9.\) Slices perpendicular to the x-axis are right isosceles triangles. The intersection of one of these slices and the base is the leg of the triangle.
The base is the area between \(y=x\) and \(y={x}^{2}.\) Slices perpendicular to the x-axis are semicircles.

Bonisa impendulo

\(\frac{\pi }{240}\) units³
\(x+y=8,x=0,\ \text{and}\ y=0\)
\(y=2{x}^{2},x=0,x=4,\ \text{and}\ y=0\)

Bonisa impendulo

\(\frac{4096\pi }{5}\) units³
\(y={e}^{x}+1,x=0,x=1,\ \text{and}\ y=0\)
\(y={x}^{4},x=0,\ \text{and}\ y=1\text{for}\ x\ge 0\)

Bonisa impendulo

\(\frac{8\pi }{9}\) units³
\(y=\sqrt{x},x=0,x=4,\ \text{and}\ y=0\)
\(y=\text{sin}\ x,y=\text{cos}\ x,\ \text{and}\ x=0\)

Bonisa impendulo

\(\frac{\pi }{2}\) units³
\(y=\frac{1}{x},x=2,\ \text{and}\ y=3\)
\({x}^{2}-{y}^{2}=9\ \text{and}\ x+y=9,y=0\ \text{and}\ x=0\)

Bonisa impendulo

\(207\pi\) units³
\(y=4-\frac{1}{2}x,x=0,\ \text{and}\ y=0\)
\(y=2{x}^{3},x=0,x=1,\ \text{and}\ y=0\)

Bonisa impendulo

\(\frac{4\pi }{5}\) units³
\(y=3{x}^{2},x=0,\ \text{and}\ y=3\)
\(y=\sqrt{4-{x}^{2}},y=0,\ \text{and}\ x=0\)

Bonisa impendulo

\(\frac{16\pi }{3}\) units³

Symbols used here

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

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

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

\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…

imaginary unit
i² = −1.

\approx

approximately equal
Equal to the precision shown, not exactly.

\leq,\ \geq

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

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

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Determining Volumes by Slicing

Determine the volume of a solid by integrating a cross-section (the slicing method).
Find the volume of a solid of revolution using the disk method.
Find the volume of a solid of revolution with a cavity using the washer method.
Examine the solid and determine the shape of a cross-section of the solid. It is often helpful to draw a picture if one is not provided.
Determine a formula for the area of the cross-section.
Integrate the area formula over the appropriate interval to get the volume.
Definite integrals can be used to find the volumes of solids. Using the slicing method, we can find a volume by integrating the cross-sectional area.
For solids of revolution, the volume slices are often disks and the cross-sections are circles. The method of disks involves applying the method of slicing in the particular case in which the cross-sections are circles, and using the formula for the area of a circle.

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

Zama wena

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Okuningi Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests