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## Section6.2Using Definite Integrals to Find Volume

###### Motivating Questions
• How can we use a definite integral to find the volume of a three-dimensional solid?

• How can we use a definite integral to find the volume of a three-dimensional solid of revolution that results from revolving a two-dimensional region about a particular axis?

• In what circumstances do we integrate with respect to $y$ instead of integrating with respect to $x\text{?}$

• What adjustments do we need to make if we revolve about a line other than the $x$- or $y$-axis?

Just as we can use definite integrals to add the areas of rectangular slices to find the exact area that lies between two curves, we can also use integrals to find the volume of regions whose cross-sections have a particular shape.

In particular, we can determine the volume of solids whose cross-sections are all thin cylinders (or washers) by adding up the volumes of these individual slices. We first consider a familiar shape in Example6.15: a circular cone.

###### Example6.15

Consider a circular cone of radius 3 and height 5, which we view horizontally as pictured in Figure6.16. Our goal in this example is to use a definite integral to determine the volume of the cone.

1. Find a formula for the linear function $y = f(x)$ that is pictured in Figure6.16.

2. For the representative slice of thickness $\Delta x$ that is located horizontally at a location $x$ (somewhere between $x = 0$ and $x = 5$), what is the radius of the representative slice? Note that the radius depends on the value of $x\text{.}$

3. What is the volume of the representative slice you found in (b)?

4. What definite integral will sum the volumes of the thin slices across the full horizontal span of the cone? What is the exact value of this definite integral?

5. Compare the result of your work in (d) to the volume of the cone that comes from using the formula $V_{\text{cone} } = \frac{1}{3} \pi r^2 h\text{.}$

Hint
1. The line passes through the points $(0,3)$ and $(5,0)\text{.}$

3. A cylinder of radius $r$ and height $h$ will have volume $\pi r^2 h \text{,}$ so use the answer you found in part (b.) for the radius.

4. Your answer to part (c.) is the volume of one cylinder, so this will become the integrand (replacing $\Delta x$ with $dx$).

5. The two methods for finding the volume should agree.

1. $f(x)=-\frac 35 x +3 \text{.}$

2. The radius at some $x$ value will be $f(x)=-\frac 35 x +3 \text{.}$

3. \begin{equation*} V_{cylinder}= \pi (-\frac 35 x +3)^2 \Delta x \end{equation*}
4. $15 \pi$

5. The two methods for finding the volume agree: the volume formula for the cone gives $V_{cone} = \frac 13 \pi (3)^2(5) = 15\pi\text{,}$ the same as the integral method.

Solution
1. We are trying to find the formula for a linear function $y=f(x)$ which passes through the points $(0,3)$ and $(5,0)\text{.}$ The $y$-intercept is therefore 3, and the slope is $\frac{0-3}{5-0}= -\frac 35 \text{,}$ so the linear equation is $f(x)=-\frac 35 x +3 \text{.}$

2. The radius at some $x$ value will be $f(x)=-\frac 35 x +3 \text{.}$

3. A cylinder of radius $r$ and height $h$ will have volume $\pi r^2 h \text{.}$ For this cylinder, the radius is $f(x)=-\frac 35 x +3$ and the height is $\Delta x \text{,}$ so the volume of this cylinder is

\begin{equation*} V_{cylinder}= \pi (-\frac 35 x +3)^2 \Delta x \end{equation*}
4. To find the volume of the cone, we will sum up a bunch of cylinders, and take the limit as the width of the cylinders $\Delta x$ goes to 0. When we do this, the sum turns into an integral and $\Delta x$ turns into $dx\text{.}$ Thus the volume of the cone is

\begin{equation*} V_{cone}= \int_0^5 \pi (-\frac 35 x +3)^2 \, dx \\ = \pi \int_0^5 \frac{9}{25} x^2 -\frac {18}{5}x +9 \, dx \\ = \pi \left. \left( \frac{3}{25} x^3 -\frac {9}{5}x^2 +9x \right)\right |_0^5 \\ = \pi \left( \frac{3}{25} (125) -\frac {9}{5} (25) +9(5) \right) - 0 = 15 \pi \end{equation*}
5. The two methods for finding the volume agree: the volume formula for the cone gives $V_{cone} = \frac 13 \pi (3)^2(5) = 15\pi\text{,}$ the same as the integral method.

### SubsectionVolume as the Integral of Cross-Sectional Area

In the previous section, we saw that the way we find area of a 2D region is to slice it into thin rectangles, and then integrate the height of the rectangles. The same principle lets us find the volume of a 3D shape: slice it into thin cross-sections, and then integrate the area of the cross-sections. So while the formula for area is

\begin{equation*} \text{Area}= \int_a^b f(x)\, dx \end{equation*}

where $f(x)$ is the height of a rectangle at $x \text{,}$ $dx$ is the width of the rectangle, and $a$ and $b$ are the endpoints of the interval , the formula for volume is

\begin{equation*} \text{Volume}= \int_a^b A(x)\, dx \end{equation*}

where $A(x)$ is the area of the cross-section at $x \text{,}$ $dx$ is the width, and $a$ and $b$ are the endpoints of the interval. In Example6.15, we saw that the cross-sections were circles, so we used the circle area formula to get $A(x)=\pi r^2=\pi (-\frac 35 x +3)^2\text{,}$ and integrated from $x=0$ to $x=5$ since those corresponded to the top and bottom of the cone.

As we saw in the previous section, for 2D shapes it is sometimes more convenient to integrate with respect to $y$ instead of $x\text{.}$ The same is true for 3D shapes. In fact, one may want to integrate with respect to $x\text{,}$ $y\text{,}$ or $z\text{,}$ and you should try to choose the variable of integration so that all of your cross-sections are the same shape. For instance, in Example6.15, we integrated with respect to $x$ so that the cross-sections were all circles. If we had taken slices in another direction, the area of a cross-section would have been much more complicated and the integral would have been much harder!

###### Example6.17

Consider a pyramid with a 1-by-1 square base along the $xy$-plane with corners at $(0,0,0), (0,1,0), (1,1,0), (1,0,0)\text{,}$ and tapering up to a point at $(0,0,2)\text{.}$ Notice that every $z$-cross-section (i.e. slice parallel to the $xy$-plane) is a square, so we will want to integrate with respect to $z\text{.}$

1. Slicing the pyramid at a $z$-value (height) between 0 and 2 gives a square shape. Find the linear function $f(z)$ that gives the side-length of this square.

2. Since the $z$-cross-sections are squares, we will want to integrate with respect to $z\text{.}$ Therefore, the volume of the pyramid is

\begin{equation*} \text{Volume} = \int_a^b A(z) dz \end{equation*}

The endpoints $a$ and $b$ represent the $z$-values (heights) where the pyramid starts and stops. What are $a$ and $b\text{?}$

3. Combine your answers to the previous parts to set up a definite integral that represents the volume of the pyramid, then evaluate that integral to find the volume of the pyramid.

4. The volume of a square pyramid is given by the formula $V=\frac 1 3 l^2h$ where $l$ is the side length of the base and $h$ is the height. Use this formula to find the volume of the pyramid and compare to your answer to the previous part.

Hint
1. Use the fact that $f(0)=1$ and $f(2)=0$ to find the linear function.

2. $a$ and $b$ will be the $z$-values at the top and bottom of the pyramid.

3. Use $\text{Volume}=\int_a^b A(z)dz$ and the fact that the area of a square is the side length squared.

1. $f(z)=-\frac 12 z+1 \text{.}$

2. $a=0$ and $b=2 \text{.}$

3. $\frac 23$

4. $\frac 23$

Solution
1. Slicing at $z=0$ gives the base of the pyramid, which is a square of side length 1, so $f(0)=1\text{.}$ Slicing at $z=2$ gives the peak of the pyramid, which can be thought of as a square of side length 0, so $f(2)=0\text{.}$ Therefore the linear function will have intercept $b=1$ and slope $m=\frac{0-1}{2-0}=-\frac 12\text{,}$ so $f(z)=-\frac 12 z+1 \text{.}$

2. We will integrate from the bottom of the pyramid to the top, so from $z=0$ to $z=2 \text{.}$ Thus $a=0$ and $b=2 \text{.}$

3. Since the area of a square is the side length squared, then $A(z)=(-\frac 12 z+1)^2 \text{.}$ Therefore,

\begin{equation*} \text{Volume}=\int_a^b A(z)dz = \int _0^2 (-\frac 12 z+1)^2 dz \\ = \left. \frac 1{12} z^3 -\frac 1 2 z^2 +z \right |_0^2 \\ =(\frac 1 {12} (8)-\frac 1 2(4)+2) -(0)=2/3 \\ \end{equation*}
4. Since our base length is 1 and our height is 2, then the volume is $V=\frac 1 3 l^2h=\frac 1 3 (1)^2(2)=\frac 23 \text{,}$ which is the same as our answer from the integral method.

###### Example6.18

We can use the same technique to find the volume of a sphere of radius 5 centered at the origin $(0,0,0)\text{.}$

1. Since a sphere is symmetrical in all directions, it doesn't matter which direction we slice it. Let's slice perpendicular to the $x$-axis (or, in the $x$-direction) so that the integral will be with respect to $x \text{.}$ Each cross-section will be a circle of radius $r(x)\text{.}$ Use the Pythagorean theorem to find a formula for $r(x)\text{.}$

2. The volume of the sphere is

\begin{equation*} \text{Volume} = \int_a^b A(x)\, dx \end{equation*}

where $A(x)$ is the area of a circular cross-section and $a$ and $b$ represent the $x$-values where the sphere starts and stops. What are $a$ and $b\text{?}$

3. Combine your answers to the previous parts to set up a definite integral that represents the volume of the sphere, then evaluate that integral to find the volume of the sphere.

4. The volume of a sphere is given by the formula $V=\frac 4 3 \pi R^3$ where $R$ is the radius of the sphere. Use this formula to find the volume of the pyramid and compare to your answer to the previous part.

Hint
1. The Pythagorean theorem says that the side lengths of a right triangle are related by $a^2+b^2=c^2\text{.}$ For a cross-section at $x\text{,}$ the side lengths of the right triangle will be $x, r(x),$ and $5\text{.}$

2. What are the leftmost and rightmost points of the sphere along the $x$-axis?

3. Use $\text{Volume} = \int_a^b A(x)\, dx \text{,}$ and the fact that the area of a circle is $\pi r^2\text{.}$

1. $r(x)=\sqrt{25-x^2} \text{.}$

2. $a=-5$ and $b=5 \text{.}$

3. $\pi \frac {500}{3}$

4. $\pi \frac {500}{3}$

Solution
1. Slicing at an $x$-value lets us make a right triangle with hypotenuse 5 and side lengths $x$ and $r(x)\text{,}$ so the Pythagorean theorem tell us that $5^2=x^2+(r(x))^2\text{,}$ and solving for $r(x)$ gives $r(x)=\sqrt{25-x^2} \text{.}$

2. The sphere is centered at the origin and has a radius of 5, so it begins at $x=-5$ and ends at $x=5 \text{.}$ Thus $a=-5$ and $b=5 \text{.}$

3. Since the area of a circle is $\pi r^2 \text{,}$ then $A(x)=\pi (r(x))^2= \pi (25-x^2) \text{.}$ Therefore,

\begin{equation*} \text{Volume}=\int_a^b A(x)dx = \int _{-5}^5 \pi(25-x^2) dx \\ = \left. \pi(25x-\frac 1 3 x^3) \right |_{-5}^5 \\ =\pi(25(5)-\frac 1 3(125))- \pi(25(-5)-\frac 1 3(-125))=\pi \frac {500}{3} \\ \end{equation*}
4. Since our radius is 5, then the volume is $V=\frac 4 3 \pi R^3=\frac 4 3 \pi(5)^3=\pi\frac {500}3 \text{,}$ which is the same as our answer from the integral method.

###### Example6.19

Let $f(x)=x^2+6$ and $g(x)=4x+6\text{.}$ Consider a solid $S$ whose base is the finite region in the $xy$-plane bounded by $y=f(x)$ and $y=g(x)\text{,}$ and whose cross-sections parallel to the $y$-axis are squares. In this example we will find the volume of $S\text{.}$

1. Find the two $x$ values where $f(x)=x^2+6$ and $g(x)=4x+6$ intersect.

2. Consider one of the square cross-sections of $S$ at a particular $x$-value. Find a formula for the side length of this cross-section (in terms of $x$), and then find a formula for $A(x)\text{,}$ the area of this cross-section.

3. Use your answer to the previous parts to find the volume of $S\text{.}$

Hint
1. Solve $f(x)=g(x)\text{.}$

2. The side length will be the distance from the curve $y=g(x)$ to the curve $y=f(x)\text{.}$ The area of a square is its side length squared.

3. Use $\text{Volume}=\int_a^b A(x)dx\text{.}$

1. $x=0,4\text{.}$

2. The side length will be $g(x)-f(x)= 4x-x^2\text{.}$ The area of a cross-section will be $A(x)=(4x-x^2)^2\text{.}$

3. $V= 34 \frac{1}{3}$

Solution
1. We solve $f(x)=g(x)\text{,}$ and straightforward algebra gives that the solutions to $x^2+6=4x+6$ occur at $x=0,4\text{.}$

2. The side length will be $g(x)-f(x)= 4x-x^2\text{.}$ Therefore, the area of a cross-section will be $A(x)=(4x-x^2)^2\text{.}$

3. Since the object begins at $x=0$ and ends at $x=4\text{,}$ we can integrate cross-sectional area to get volume:

\begin{equation*} V=\int_a^b A(x)\, dx = \int_0^4 (4x-x^2)^2\, dx= 34 \frac{1}{3} \end{equation*}

### SubsectionThe Volume of a Solid of Revolution

A solid of revolution is a three dimensional solid that can be generated by revolving one or more curves around a fixed axis. For example, the circular cone in Figure6.16 is the solid of revolution generated by revolving the portion of the line $y = 3 - \frac{3}{5}x$ from $x = 0$ to $x = 5$ about the $x$-axis. Notice that if we slice a solid of revolution perpendicular to the axis of revolution, the resulting cross-section is a circle.

We first consider solids whose slices are thin cylinders. Recall that the volume of a cylinder is given by $V = \pi r^2 h\text{.}$

###### Example6.20

Find the volume of the solid of revolution generated when the region $R$ bounded by $y = 4-x^2$ and the $x$-axis is revolved about the $x$-axis.

Solution

First, we observe that $y = 4-x^2$ intersects the $x$-axis at the points $(-2,0)$ and $(2,0)\text{.}$ When we revolve the region $R$ about the $x$-axis, we get the three-dimensional solid pictured in Figure6.21.

We slice the solid into vertical slices of thickness $\Delta x$ between $x = -2$ and $x = 2\text{.}$ A representative slice is a cylinder of height $\Delta x$ and radius $4-x^2\text{.}$ Hence, the volume of the slice is

\begin{equation*} V_{\text{slice} } = \pi (4-x^2)^2 \Delta x\text{.} \end{equation*}

Using a definite integral to sum the volumes of the representative slices, it follows that

\begin{equation*} V = \int_{-2}^{2} \pi (4-x^2)^2 \, dx\text{.} \end{equation*}

It is straightforward to evaluate the integral and find that the volume is $V = \frac{512}{15}\pi\text{.}$

For a solid such as the one in Example6.20, where each slice is a cylindrical disk, we first find the volume of a typical slice (noting particularly how this volume depends on $x$), and then integrate over the range of $x$-values that bound the solid. Often, we will be content with simply finding the integral that represents the volume; if we desire a numeric value for the integral, we typically use a calculator or computer algebra system to find that value.

This method for finding the volume of a solid of revolution is often called the disk method.

###### The Disk Method

If $y = r(x)$ is a nonnegative continuous function on $[a,b]\text{,}$ then the volume of the solid of revolution generated by revolving the curve about the $x$-axis over this interval is given by

\begin{equation*} V = \int_a^b \pi r(x)^2 \, dx\text{.} \end{equation*}

A different type of solid can emerge when two curves are involved, as we see in the following example.

###### Example6.22

Find the volume of the solid of revolution generated when the finite region $R$ that lies between $y = 4-x^2$ and $y = x+2$ is revolved about the $x$-axis.

Solution

First, we must determine where the curves $y = 4-x^2$ and $y = x+2$ intersect. Substituting the expression for $y$ from the second equation into the first equation, we find that $x + 2 = 4-x^2\text{.}$ Rearranging, it follows that

\begin{equation*} x^2 + x - 2 = 0\text{,} \end{equation*}

and the solutions to this equation are $x = -2$ and $x = 1\text{.}$ The curves therefore cross at $(-2,0)$ and $(1,1)\text{.}$

When we revolve the region $R$ about the $x$-axis, we get the three-dimensional solid pictured at left in Figure6.23. Figure6.23At left, the solid of revolution in Example6.22. At right, a typical slice with inner radius $r(x)$ and outer radius $R(x)\text{.}$

Immediately we see a major difference between the solid in this example and the one in Example6.20: here, the three-dimensional solid of revolution isn't solid because it has open space in its center along the axis of revolution. If we slice the solid perpendicular to the axis of revolution, we observe that the resulting slice is not a solid disk, but rather a washer, as pictured at right in Figure6.23. At a given location $x$ between $x = -2$ and $x = 1\text{,}$ the small radius $r(x)$ of the inner circle is determined by the curve $y = x+2\text{,}$ so $r(x) = x+2\text{.}$ Similarly, the big radius $R(x)$ comes from the function $y = 4-x^2\text{,}$ and thus $R(x) = 4-x^2\text{.}$

To find the volume of a representative slice, we compute the volume of the outer disk and subtract the volume of the inner disk. Since

\begin{equation*} \pi R(x)^2 \Delta x - \pi r(x)^2 \Delta x = \pi [ R(x)^2 - r(x)^2] \Delta x\text{,} \end{equation*}

it follows that the volume of a typical slice is

\begin{equation*} V_{\text{slice} } = \pi [ (4-x^2)^2 - (x+2)^2 ] \Delta x\text{.} \end{equation*}

Using a definite integral to sum the volumes of the respective slices across the integral, we find that

\begin{equation*} V = \int_{-2}^1 \pi[ (4-x^2)^2 - (x+2)^2 ] \, dx\text{.} \end{equation*}

Evaluating the integral, we find that the volume of the solid of revolution is $V = \frac{108}{5}\pi\text{.}$

This method for finding the volume of a solid of revolution generated by two curves is often called the washer method.

###### The Washer Method

If $y = R(x)$ and $y = r(x)$ are nonnegative continuous functions on $[a,b]$ that satisfy $R(x) \ge r(x)$ for all $x$ in $[a,b]\text{,}$ then the volume of the solid of revolution generated by revolving the region between them about the $x$-axis over this interval is given by

\begin{equation*} V = \int_a^b \pi [R(x)^2 - r(x)^2] \, dx\text{.} \end{equation*}
###### Example6.24

In each of the following questions, draw a careful, labeled sketch of the region described, as well as the resulting solid that results from revolving the region about the stated axis. In addition, draw a representative slice and state the volume of that slice, along with a definite integral whose value is the volume of the entire solid. It is not necessary to evaluate the integrals you find.

1. The region $S$ bounded by the $x$-axis, the curve $y = \sqrt{x}\text{,}$ and the line $x = 4\text{;}$ revolve $S$ about the $x$-axis.

2. The region $S$ bounded by the $y$-axis, the curve $y = \sqrt{x}\text{,}$ and the line $y = 2\text{;}$ revolve $S$ about the $x$-axis.

3. The finite region $S$ bounded by the curves $y = \sqrt{x}$ and $y = x^3\text{;}$ revolve $S$ about the $x$-axis.

4. The finite region $S$ bounded by the curves $y = 2x^2 + 1$ and $y = x^2 + 4\text{;}$ revolve $S$ about the $x$-axis.

5. The region $S$ bounded by the $y$-axis, the curve $y = \sqrt{x}\text{,}$ and the line $y = 2\text{;}$ revolve $S$ about the $y$-axis. How is this problem different from the one posed in part (b)?

Hint
1. Use slices perpendicular to the $x$-axis.

2. Note that slices perpendicular to the $x$-axis will be washers.

3. Find the points where the two curves intersect and draw a labeled graph.

4. Think about how the solid looks like a donut; what is the outer radius?

5. Try using slices perpendicular to the $y$-axis.

1. $V = \int_0^4 \pi (\sqrt{x})^2 \, dx = \int_0^4 \pi x \, dx = 8\pi\text{.}$

2. $V = \int_0^4 \pi (4-(\sqrt{x})^2) \, dx = \int_0^4 \pi (4-x) \, dx = 8\pi\text{.}$

3. $V = \int_0^1 \pi(x - x^6) \, dx = \frac{5}{14}\pi\text{.}$

4. $V = \int_{-\sqrt{3}}^{\sqrt{3}} \pi( (x^2 + 4)^2 - (2x^2 + 1)^2) \, dx = \frac{136\sqrt{3}}{5}\pi\text{.}$

5. $V = \int_0^2 \pi y^4 \, dy = \frac{32}{5}\pi\text{.}$

Solution
1. A typical slice (perpendicular to the $x$-axis) has radius $r(x) = \sqrt{x}\text{,}$ and thus the volume of such a slice is

\begin{equation*} V_{\text{slice}} = \pi (\sqrt{x})^2 \Delta x\text{.} \end{equation*}

It follows by the disk method that the volume of the solid of revolution is

\begin{equation*} V = \int_0^4 \pi (\sqrt{x})^2 \, dx = \int_0^4 \pi x \, dx = 8\pi\text{.} \end{equation*}
2. A typical slice (perpendicular to the $x$-axis) is a washer with outer radius $R(x) = 2$ and inner radius $r(x) = \sqrt{x}\text{.}$ The volume of such a slice is

\begin{equation*} V_{\text{slice}} = \pi(R(x)^2-r(x)^2) \Delta x = \pi (2^2 - (\sqrt{x})^2) \Delta x\text{.} \end{equation*}

It follows by the washer method that the volume of the solid of revolution is

\begin{equation*} V = \int_0^4 \pi (4-(\sqrt{x})^2) \, dx = \int_0^4 \pi (4-x) \, dx = 8\pi\text{.} \end{equation*}
3. The curves $y = \sqrt{x}$ and $y = x^3$ intersect at $(0,0)$ and $(1,1)\text{;}$ revolving the region $S$ about the $x$-axis and taking slices perpendicular to the $x$-axis, we see that a typical slice is a washer with outer radius $R(x) = \sqrt{x}$ and inner radius $r(x) = x^3\text{.}$ Hence the volume of a typical slice is

\begin{equation*} V_{\text{slice}} = \pi(R(x)^2-r(x)^2) \Delta x = \pi((\sqrt{x})^2 - (x^3)^2) \Delta x = \pi(x - x^6) \Delta x\text{.} \end{equation*}

By the washer method, the volume of the solid is

\begin{equation*} V = \int_0^1 \pi(x - x^6) \, dx = \frac{5}{14}\pi\text{.} \end{equation*}
4. The curves $y = 2x^2 + 1$ and $y = x^2 + 4$ intersect at the points $(\pm \sqrt{3}, 7)\text{;}$ after we revolve the region $S$ about the $x$-axis, cross-sections perpendicular to the axis of rotation are washers with outer radius $R(x) = x^2 + 4$ and inner radius $r(x) = 2x^2 + 1\text{.}$ By the washer method, we see that

\begin{equation*} V = \int_{-\sqrt{3}}^{\sqrt{3}} \pi( (x^2 + 4)^2 - (2x^2 + 1)^2) \, dx = \frac{136\sqrt{3}}{5}\pi\text{.} \end{equation*}
5. The curve $y = \sqrt{x}$ and the line $y = 2$ intersect at $(4,2)\text{;}$ when we revolve the region $S$ about the $y$-axis, slices perpendicular to the $y$-axis are disks. Here we see that each disk's radius is a function of $y$ and the disks have thickness $\Delta y\text{,}$ so we will need to integrate with respect to $y$ and consider the radius function $r(y) = y^2$ (which comes from solving $y = \sqrt{x}$ for $x$ in terms of $y\text{.}$ Note how this is different from the problem in part (b) because of the axis we rotate about and how this forces us to integrate in terms of $y\text{.}$ A typical slice has volume

\begin{equation*} V_{\text{slice}} = \pi (y^2)^2 \Delta y\text{,} \end{equation*}

and thus by the disk method the total volume is

\begin{equation*} V = \int_0^2 \pi y^4 \, dy = \frac{32}{5}\pi\text{.} \end{equation*}

### SubsectionRevolving About the $y$-axis

When we revolve a given region about the $y$-axis, the representative slices now have thickness $\Delta y\text{,}$ which means that we must integrate with respect to $y\text{.}$

###### Example6.25

Find the volume of the solid of revolution generated when the region $R$ that lies between $y = \sqrt{x}$ and $y = x^4$ is revolved about the $y$-axis.

Solution

These two curves intersect when $x = 1\text{,}$ hence at the point $(1,1)\text{.}$ When we revolve the region $R$ about the $y$-axis, we get the three-dimensional solid pictured at left in Figure6.26. Figure6.26At left, the solid of revolution in Example6.25. At right, a typical slice with inner radius $r(y)$ and outer radius $R(y)\text{.}$

Note that the slices are cylindrical washers only if taken perpendicular to the $y$-axis. We slice the solid horizontally, starting at $y = 0$ and proceeding up to $y = 1\text{.}$ The thickness of a representative slice is $\Delta y\text{,}$ so we must express the integrand in terms of $y\text{.}$ The inner radius is determined by the curve $y = \sqrt{x}\text{,}$ so we solve for $x$ and get $x = y^2 = r(y)\text{.}$ In the same way, we solve the curve $y = x^4$ (which governs the outer radius) for $x$ in terms of $y\text{,}$ and hence $x = \sqrt{y}\text{.}$ Therefore, the volume of a typical slice is

\begin{equation*} V_{\text{slice} } = \pi [R(y)^2 - r(y)^2] = \pi[(\sqrt{y})^2 - (y^2)^2] \Delta y\text{.} \end{equation*}

We use a definite integral to sum the volumes of all the slices from $y = 0$ to $y = 1\text{.}$ The total volume is

\begin{equation*} V = \int_{y=0}^{y=1} \pi \left[ (\sqrt{y})^2 - (y^2)^2 \right] \, dy\text{.} \end{equation*}

It is straightforward to evaluate the integral and find that $V = \frac{7}{15} \pi\text{.}$

###### Example6.27

In each of the following questions, draw a careful, labeled sketch of the region described, as well as the resulting solid that results from revolving the region about the stated axis. In addition, draw a representative slice and state the volume of that slice, along with a definite integral whose value is the volume of the entire solid. It is not necessary to evaluate the integrals you find.

1. The region $S$ bounded by the $y$-axis, the curve $y = \sqrt{x}\text{,}$ and the line $y = 2\text{;}$ revolve $S$ about the $y$-axis.

2. The region $S$ bounded by the $x$-axis, the curve $y = \sqrt{x}\text{,}$ and the line $x = 4\text{;}$ revolve $S$ about the $y$-axis.

3. The finite region $S$ in the first quadrant bounded by the curves $y = 2x$ and $y = x^3\text{;}$ revolve $S$ about the $x$-axis.

4. The finite region $S$ in the first quadrant bounded by the curves $y = 2x$ and $y = x^3\text{;}$ revolve $S$ about the $y$-axis.

5. The finite region $S$ bounded by the curves $x = (y-1)^2$ and $y = x-1\text{;}$ revolve $S$ about the $y$-axis

Hint
1. Consider slices perpendicular to the $y$-axis.

2. Since we revolve around the $y$-axis, think about slices of thickness $\Delta y\text{.}$

3. Note that $y = x^3$ lies below $y = 2x$ on the interval $[0,\sqrt{2}]\text{.}$

4. The curve $y = x^3$ can be equivalently expressed as $x = \sqrt{y}\text{.}$

5. Find the intersection points by writing the second curve as $x = y+1$ and solving $(y-1)^2 = y+1\text{.}$

1. $V = \int_0^2 \pi y^4 \Delta \, dy\text{.}$

2. $V = \int_0^2 \pi (16 - y^4) \, dy\text{.}$

3. $V = int_0^{\sqrt{2}} \pi ( 4x^2 - x^6 ) \, dx\text{.}$

4. $V = \int_0^{2\sqrt{2}} \pi( y^{2/3} - y^2/4 ) \, dy\text{.}$

5. $V = \int_0^3 \pi( (y+1)^2 - (y-1)^4 ) \, dy\text{.}$

Solution
1. Observe that horizontal slices with thickness $\Delta y$ are all cylindrical disks, each with radius given by $r(y) = y^2$ since they are given by the curve $y = \sqrt{x}\text{,}$ which can be equivalently expressed as $x = y^2\text{.}$ The volume of each slice is thus

\begin{equation*} V_{\text{slice}} = \pi r(y)^2 \Delta y = \pi y^4 \Delta y\text{.} \end{equation*}

The volume of the solid is thus given by the definite integral

\begin{equation*} V = \int_0^2 \pi y^4 \Delta \, dy\text{.} \end{equation*}
2. Slicing perpendicular to the axis of rotation, we get slices of thickness $\Delta y$ that are shaped like washers. The outer radius, $R(y)\text{,}$ is given by $R(y) = 4\text{;}$ the innner radius is given by the function $y = \sqrt{x}\text{,}$ but expressed with $x$ as a function of $y\text{,}$ so $r(y) = y^2\text{.}$ It follows that a representative washer slice has volume

\begin{equation*} V_{\text{slice}} = \pi (R(y)^2 - r(y)^2) \Delta y = \pi (16 - y^4) \Delta y\text{.} \end{equation*}

Further, since $y = \sqrt{x}$ and $x = 4$ intersect at the point $(4,2)\text{,}$ we integrate from $y = 0$ to $y = 2$ and thus find that the total volume is

\begin{equation*} V = \int_0^2 \pi (16 - y^4) \, dy\text{.} \end{equation*}
3. The curves $y=x^3$ and $y=2x$ intersect where $x^3 = 2x\text{,}$ so where $x^3-2x = x(x^2 - 2) = 0\text{.}$ The two nonnegative values of $x$ that make this equation true are $x=0$ and $x=\sqrt{2}\text{,}$ and thus the points $(0,0)$ and $(\sqrt{2}, 2\sqrt{2})$ form the boundaries of the region $S$ that is being rotated about the $x$-axis. Since we rotate about the horizontal axis, we slice perpendicular to that axis with slices of thickness $\Delta x\text{.}$ A representative slice is in the shape of a washer and has volume

\begin{equation*} V_{\text{slice}} = \pi (R(x)^2 - r(x)^2) \Delta x = \pi ( (2x)^2 - (x^3)^2 ) \Delta x\text{,} \end{equation*}

and thus the total volume of the solid is

\begin{equation*} V = int_0^{\sqrt{2}} \pi ( 4x^2 - x^6 ) \, dx\text{.} \end{equation*}
4. To use the same region $S$ as in part (c), but now revolved about the $y$-axis, we shift to horizontal slices of thickness $\Delta y\text{,}$ and thus need to express the curves as functions of $y\text{:}$ the rightmost curve is $x = R(y) = \sqrt{y}\text{,}$ while the leftmost curve is $x = r(y) = \frac{1}{2}y\text{.}$ A representative slice is a washer whose volume is

\begin{equation*} V_{\text{slice}} = \pi (R(y)^2 - r(y)^2) \Delta y = \pi( (\sqrt{y})^2 - (\frac{1}{2}y)^2) \Delta y\text{.} \end{equation*}

It follows that the total volume of the solid is

\begin{equation*} V = \int_0^{2\sqrt{2}} \pi( y^{2/3} - y^2/4 ) \, dy\text{.} \end{equation*}
5. For this pair of curves, we see they intersect where $x=(y-1)^2$ and $x = (y+1)$ are both true, and hence where $(y-1)^2 = y+1\text{.}$ Expanding, moving all terms to the left, and combining like terms, we see that $y^2 - 2y + 1 = y + 1\text{,}$ so $y^2 - 3y = 0\text{.}$ It follows that $y(y-3) = 0$ and therefore $y=3$ or $y = 0\text{.}$ Plotting the region and revolving about the $y$-axis, we see that horizontal slices of thickness $\Delta y$ are shaped like washers. The outer radius is given by $R(y) = y+1\text{,}$ while the inner radius is $r(y) = (y-1)^2\text{.}$ A representative slice thus has volume

\begin{equation*} V_{\text{slice}} = \pi (R(y)^2 - r(y)^2) \Delta y = \pi( (y+1)^2 - ((y-1)^2)^2 ) \Delta y\text{.} \end{equation*}

The volume of the solid of revolution is therefore

\begin{equation*} V = \int_0^3 \pi( (y+1)^2 - (y-1)^4 ) \, dy\text{.} \end{equation*}

### SubsectionRevolving About Horizontal and Vertical Lines other than the Coordinate Axes

It is possible to revolve a region around any horizontal or vertical line. Doing so adjusts the radii of the cylinders or washers involved by a constant value. A careful, well-labeled plot of the solid of revolution will usually reveal how the different axis of revolution affects the definite integral.

###### Example6.28

Find the volume of the solid of revolution generated when the finite region $S$ that lies between $y = x^2$ and $y = x$ is revolved about the line $y = -1\text{.}$

Solution

Graphing the region between the two curves in the first quadrant between their points of intersection ($(0,0)$ and $(1,1)$) and then revolving the region about the line $y = -1\text{,}$ we see the solid shown in Figure6.29. Each slice of the solid perpendicular to the axis of revolution is a washer, and the radii of each washer are governed by the curves $y = x^2$ and $y = x\text{.}$ But we also see that there is one added change: the axis of revolution adds a fixed length to each radius. The inner radius of a typical slice, $r(x)\text{,}$ is given by $r(x) = x^2 + 1\text{,}$ while the outer radius is $R(x) = x+1\text{.}$

Therefore, the volume of a typical slice is

\begin{equation*} V_{\text{slice} } = \pi[ R(x)^2 - r(x)^2 ] \Delta x = \pi \left[ (x+1)^2 - (x^2 + 1)^2 \right] \Delta x\text{.} \end{equation*}

Finally, we integrate to find the total volume, and

\begin{equation*} V = \int_0^1 \pi \left[ (x+1)^2 - (x^2 + 1)^2 \right] \, dx = \frac{7}{15} \pi\text{.} \end{equation*}
###### Example6.30

In each of the following questions, draw a careful, labeled sketch of the region described, as well as the resulting solid that results from revolving the region about the stated axis. In addition, draw a representative slice and state the volume of that slice, along with a definite integral whose value is the volume of the entire solid. It is not necessary to evaluate the integrals you find. For each prompt, use the finite region $S$ in the first quadrant bounded by the curves $y = 2x$ and $y = x^3\text{.}$

1. Revolve $S$ about the line $y = -2\text{.}$

2. Revolve $S$ about the line $y = 4\text{.}$

3. Revolve $S$ about the line $x=-1\text{.}$

4. Revolve $S$ about the line $x = 5\text{.}$

Hint
1. Note that the two curves intersect in the first quadrant at $(0,0)$ and $(\sqrt{2}, 2\sqrt{2})\text{.}$

2. The outer radius is $R(x) = 4 - x^3\text{.}$

3. Solve the equations governing the two curves for $x$ in terms of $y\text{.}$

4. The inner radius is $r(y) = 5 - y^{1/3}\text{.}$

1. \begin{equation*} V = \int_{0}^{\sqrt{2}} \pi ( (2x+2)^2 - (x^3 + 2)^2 ) \, dx = \frac{4}{21}(21+8\sqrt{2}) \pi \approx 19.336\text{.} \end{equation*}
2. \begin{equation*} V = \int_{0}^{\sqrt{2}} \pi ( (4 - x^3)^2 - (4-2x)^2 ) \, dx = \left( 8-\frac{32\sqrt{2}}{21} \right)\pi \approx 18.3626\text{.} \end{equation*}
3. \begin{equation*} V = \int_{0}^{2\sqrt{2}} \pi( (y^{1/3} + 1)^2 - (\frac{1}{2}y + 1)^2 ) \, dy = \frac{2}{15}(15 + 8\sqrt{2}) \pi \approx 11.022\text{.} \end{equation*}
4. \begin{equation*} V = \int_{0}^{2\sqrt{2}} \pi( (5 - \frac{1}{2}y)^2 - (5 - y^{1/3})^2 ) \, dy = \frac{2}{15}(75-8\sqrt{2})\pi \approx 26.677\text{.} \end{equation*}
Solution
1. The two curves intersect in the first quadrant at $(0,0)$ and $(\sqrt{2}, 2\sqrt{2})\text{.}$ Revolving the region $S$ about the line $y = -2\text{,}$ we see that a typical slice is a washer with thickness $\Delta x\text{,}$ outer radius $R(x) = 2x + 2\text{,}$ and inner radius $r(x) = x^3 + 2\text{.}$ It follows by the washer method that

\begin{equation*} V = \int_{0}^{\sqrt{2}} \pi ( (2x+2)^2 - (x^3 + 2)^2 ) \, dx = \frac{4}{21}(21+8\sqrt{2}) \pi \approx 19.336\text{.} \end{equation*}
2. In this situation, revolving about $y = 4\text{,}$ we see that the outer radius of a typical washer is $R(x) = 4 - x^3\text{,}$ while the inner radius is $r(x) = 4 - 2x\text{,}$ and thus the volume is

\begin{equation*} V = \int_{0}^{\sqrt{2}} \pi ( (4 - x^3)^2 - (4-2x)^2 ) \, dx = \left( 8-\frac{32\sqrt{2}}{21} \right)\pi \approx 18.3626\text{.} \end{equation*}
3. Because we are revolving about $x = -1\text{,}$ when we slice perpendicular to this we get washers of thickness $\Delta y\text{,}$ and thus we need to write the two curves as functions of $y$ in order to determine the radii of the slices as functions of $y\text{,}$ we note that the line can be expressed as $x = \frac{1}{2}y$ and the cubic curve can be rewritten in the form $x = y^{1/3}\text{.}$ The outer radius of a typical slice is thus $R(y) = y^{1/3} + 1$ and the inner radius is $r(y) = \frac{1}{2}y + 1\text{.}$ It follows that the volume of the solid is

\begin{equation*} V = \int_{0}^{2\sqrt{2}} \pi( (y^{1/3} + 1)^2 - (\frac{1}{2}y + 1)^2 ) \, dy = \frac{2}{15}(15 + 8\sqrt{2}) \pi \approx 11.022\text{.} \end{equation*}
4. Here, the outer radius is $R(y) = 5 - \frac{1}{2}y$ and the inner radius is $r(y) = 5 - y^{1/3}\text{,}$ so the total volume of the solid is

\begin{equation*} V = \int_{0}^{2\sqrt{2}} \pi( (5 - \frac{1}{2}y)^2 - (5 - y^{1/3})^2 ) \, dy = \frac{2}{15}(75-8\sqrt{2})\pi \approx 26.677\text{.} \end{equation*}

### SubsectionSummary

• A definite integral may be used to represent the volume of a solid. The idea is to slice the solid into thin shapes whose volume is easy to calculate. The volume of each slice is found by taking the area of the cross-section and multiplying it by the width. Summing up the volumes of the slices and taking the limit as the width $\rightarrow 0$ results in an integral representing the volume.

• A three-dimensional solid of revolution results from revolving a two-dimensional region about a particular axis or line called the axis of revolution. We can use a definite integral to find the volume of such a solid by taking slices perpendicular to the axis of revolution. The slices will be circular disks or washers.

• If we revolve about a horizontal line and slice perpendicular to that line, then our slices are vertical and of thickness $\Delta x\text{.}$ This leads us to integrate with respect to $x\text{.}$

• If we revolve about a vertical line and slice perpendicular to that line, then our slices are horizontal and of thickness $\Delta y\text{.}$ This leads us to integrate with respect to $y\text{.}$

• If we revolve about a line other than the $x$- or $y$-axis, we need to carefully account for the shift that occurs in the radius of a typical slice. Normally, this shift involves taking a sum or difference of the function along with the constant connected to the equation for the horizontal or vertical line; a well-labeled diagram is usually the best way to decide the new expression for the radius.

### SubsectionExercises

Consider the curve $f(x) = 3 \cos(\frac{x^3}{4})$ and the portion of its graph that lies in the first quadrant between the $y$-axis and the first positive value of $x$ for which $f(x) = 0\text{.}$ Let $R$ denote the region bounded by this portion of $f\text{,}$ the $x$-axis, and the $y$-axis.

1. Set up a definite integral whose value is the exact arc length of $f$ that lies along the upper boundary of $R\text{.}$ Use technology appropriately to evaluate the integral you find.

2. Set up a definite integral whose value is the exact area of $R\text{.}$ Use technology appropriately to evaluate the integral you find.

3. Suppose that the region $R$ is revolved around the $x$-axis. Set up a definite integral whose value is the exact volume of the solid of revolution that is generated. Use technology appropriately to evaluate the integral you find.

4. Suppose instead that $R$ is revolved around the $y$-axis. If possible, set up an integral expression whose value is the exact volume of the solid of revolution and evaluate the integral using appropriate technology. If not possible, explain why.

Consider the curves given by $y = \sin(x)$ and $y = \cos(x)\text{.}$ For each of the following problems, you should include a sketch of the region/solid being considered, as well as a labeled representative slice.

1. Sketch the region $R$ bounded by the $y$-axis and the curves $y = \sin(x)$ and $y = \cos(x)$ up to the first positive value of $x$ at which they intersect. What is the exact intersection point of the curves?

2. Set up a definite integral whose value is the exact area of $R\text{.}$

3. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving $R$ about the $x$-axis.

4. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving $R$ about the $y$-axis.

5. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving $R$ about the line $y = 2\text{.}$

6. Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving $R$ about the line $x = -1\text{.}$

Consider the finite region $R$ that is bounded by the curves $y = 1+\frac{1}{2}(x-2)^2\text{,}$ $y=\frac{1}{2}x^2\text{,}$ and $x = 0\text{.}$

1. Determine a definite integral whose value is the area of the region enclosed by the two curves.

2. Find an expression involving one or more definite integrals whose value is the volume of the solid of revolution generated by revolving the region $R$ about the line $y = -1\text{.}$

3. Determine an expression involving one or more definite integrals whose value is the volume of the solid of revolution generated by revolving the region $R$ about the $y$-axis.

4. Find an expression involving one or more definite integrals whose value is the perimeter of the region $R\text{.}$

During an MRI, doctors can produce images of cross-sections of a person's brain. Suppose that during an MRI, a doctor finds that the area of a person's brain visible at a cross-section is given by $A(x)=\frac{625}{4}-\frac{x^4}{4}$ where $A(x)$ is in square centimeters and $x$ is the number of centimeters from the middle of the brain. Find the volume of the person's brain.

Consider a half-cone, meaning a shape whose base is a half-circle tapering up to a peak. Suppose that the base of the half-cone has a radius of $R\text{,}$ and that the peak is $h$ units away from the base.

1. Find a linear formula $r(x)$ for $0 \leq x \leq h$ that gives the radius of the half-circle $x$ units above the base of the half-cone. (Your answer will depend on $R$ and $h\text{.}$)

2. Use your answer to the previous part to set up and evaluate an integral that gives the volume of the half-cone. (You may want to use the fact that a half-circle has area $\frac{\pi}{2}r^2\text{.}$ Your answer will depend on $R$ and $h\text{.}$)