## Section3.3Volume of Revolution: Disk Method

We have seen how to compute certain areas by using integration; we will now look into how some volumes may also be computed by evaluating an integral. Generally, the volumes that we can compute this way have cross-sections that are easy to describe. For example, circular cross-sections are easy to describe as their area just depends on the radius, and so they are one of the central topics in this section. However, we first discuss the general idea of calculating the volume of a solid by slicing up the solid.

### Subsection3.3.1Computing Volumes with Cross-sections

Let us first formalize what is meant by a cross-section.

###### Definition3.21. Cross-section.

A cross-section of a solid is the region obtained by intersecting the solid with a plane.

Examples of cross-sections are the circular region above the right cylinder in Figure 3.(a), the star above the star-prism in Figure 3.(b), and the square we see in the pyramid on the left side of Figure 3.11. For now, we are only interested in solids, whose volumes are generated through cross-sections that are easy to describe. For example, the right cylinder in Figure 3.(a) is generated by translating a circular region along the $x$-axis for a certain length $h\text{.}$ Every cross-section of the right cylinder must therefore be circular, when cutting the right cylinder anywhere along length $h$ that is perpendicular to the $x$-axis.

In a similar manner, many other solids can be generated and understood as shown with the translated star in Figure 3.(b).

The volume of both the right cylinder and the translated star can be thought of as

\begin{equation*} V= (\text{ area of cross-section } ) \cdot (\text{ length } )=A\cdot h\text{.} \end{equation*}

Let us now turn towards the calculation of such volumes by working through two examples. The right pyramid with square base shown in Figure 3.11 has cross-sections that must be squares if we cut the pyramid parallel to its base. The following example makes use of these cross-sections to calculate the volume of the pyramid for a certain height.

###### Example3.22. Volume of a Pyramid.

Find the volume of a pyramid that is 20 metres tall with a square base 20 metres on a side.

Solution

As with most of our applications of integration, we begin by asking how we might approximate the volume. Since we can easily compute the volume of a rectangular prism (that is, a “box”), we will use some boxes to approximate the volume of the pyramid, as shown in Figure 3.11: Suppose we cut up the pyramid into $n$ slices. On the left is a 3D view that shows cross-sections cut parallel to the base of the pyramid and replaced with rectangular boxes that are used to approximate the volume. On the right is a 2D view that now shows a cross-section perpendicular to the base of the pyramid so that we can identify the width and height of a box.

Having to use width and height means that we have two variables. However, by overlaying a Cartesian coordinate system with the origin at the midpoint of the base on to the 2D view of Figure 3.11 as shown below, we can relate these two variables to each other.

Since the cross-sectional view is placed symmetrically about the $y$-axis, we see that a height of 20 is achieved at the midpoint of the base. Thus at $x=0\text{,}$ $y=20\text{,}$ at $x=10\text{,}$ $y=0\text{,}$ and we have a slope of $m = -2\text{.}$ So

\begin{align*} y\amp =-2x+b\\ 20\amp =-2(0)+b\\ 20\amp =b\text{.} \end{align*}

Therefore, $y=20-2x\text{,}$ and in the terms of $x$ we have that $x=10-y/2\text{.}$ We could have also used similar triangles here to derive the relationship between $x$ and $y\text{.}$ From the right diagram in Figure 3.11, we see that each box has volume of the form

\begin{equation*} (2x_i)(2x_i)\Delta y\text{.} \end{equation*}

Then the total volume is approximately

\begin{equation*} \sum_{i=0}^{n-1} (2x_i)(2x_i)\Delta y = \sum_{i=0}^{n-1} 4(10-\frac{y_i}{2})^2\Delta y \end{equation*}

and when we apply the limit $\Delta y \to 0$ we get the volume as the value of a definite integral as defined in Section 1.4:

\begin{equation*} \begin{split} V \amp = \lim_{\Delta y \to 0} \sum_{i=0}^{n-1} 4(10-\frac{y_i}{2})^2\Delta y = \int_0^{20} 4(10-\frac{y}{2})^2\,dy \\[1ex] \amp =\int_0^{20} (20-y)^2\,dy \\[1ex] \amp = \left.-{(20-y)^3\over3}\right|_0^{20}\\[1ex] \amp = -{0^3\over3}-\left(-{20^3\over3}\right)={8000\over3}. \end{split} \end{equation*}

As you may know, the volume of a pyramid is given by the formula

\begin{equation*} (1/3)(\hbox{height})(\hbox{area of base})\text{.} \end{equation*}

Using this formula we calculate

\begin{equation*} (1/3)(20)(400) = \frac{8000}{3}\text{,} \end{equation*}

###### Example3.23. Volume of an Object.

The base of a solid is the region between $\ds f(x)=x^2-1$ and $\ds g(x)=-x^2+1$ as shown to the right of Figure 3.12, and its cross-sections perpendicular to the $x$-axis are equilateral triangles, as indicated in Figure 3.12 to the left.

The solid has been truncated to show a triangular cross-section above $x=1/2\text{.}$

Find the volume of the solid.

Solution

We begin by drawing the equilateral triangle above any $x_i$ and identify its base and height as shown below to the left.

The equilateral triangle has base

\begin{equation*} g(x_i)-f(x_i) = (1-x_i^2)-(x_i^2-1) = 2(1-x_i^2)\text{,} \end{equation*}

and height

\begin{equation*} \sqrt{3}g(x_i) = \sqrt{3}(1-x_i^2)\text{.} \end{equation*}

So the area of the cross-section is

\begin{equation*} {1\over2}(\hbox{base})(\hbox{height})= (1-x_i^2)\sqrt3(1-x_i^2)\text{.} \end{equation*}

The diagram above to the right indicates the position of an arbitrary thin equilateral triangle in the given region. Therefore, the volume of this thin equilateral triangle is given by

\begin{equation*} {1\over2}(\hbox{base})(\hbox{height})(\hbox{thickness})=(1-x_i^2)\sqrt3(1-x_i^2)\Delta x\text{.} \end{equation*}

If we have sliced our solid into $n$ thin equilateral triangles, then the volume can be approximated with the sum

\begin{equation*} \sum_{i=0}^{n-1}(1-x_i^2)\sqrt{3}(1-x_i^2)\Delta x = \sum_{i=0}^{n-1}\sqrt{3}(1-x_i^2)^2\Delta x\text{.} \end{equation*}

Similar to the previous example, when we apply the limit $\Delta x \to 0\text{,}$ the total volume is

\begin{equation*} V = \lim_{\Delta x \to 0} \sum_{i=0}^{n-1}\sqrt{3}(1-x_i^2)^2\Delta x = \int_{-1}^1 \sqrt3(1-x^2)^2\,dx={16\over15}\sqrt3\text{.} \end{equation*}

Notice that the limits of integration, namely -1 and 1, are the left and right bounding values of $x\text{,}$ because we are slicing the solid perpendicular to the $x$-axis from left to right.

### Subsection3.3.2Disk Method: Integration w.r.t. $x$

One easy way to get “nice” cross-sections is by rotating a plane figure around a line, also called the axis of rotation, and therefore such a solid is also referred to as a solid of revolution. For example, in Figure 3.13 we see a plane region under a curve and between two vertical lines $x=a$ and $x=b\text{,}$ which creates a solid when the region is rotated about the $x$-axis, and naturally, a typical cross-section perpendicular to the $x$-axis must be circular as shown.

Of course a real “slice” of this figure will not be cylindrical in nature, but we can approximate the volume of the slice by a cylinder or so-called disk with circular top and bottom and straight sides parallel to the axis of rotation; the volume of this disk will have the form $\ds \pi r^2\Delta x\text{,}$ where $r$ is the radius of the disk and $\Delta x$ is the thickness of the disk. As long as we can write $r$ in terms of $x$ we can compute the volume by an integral. Often, the radius $r$ is given by the height of the function, i.e. $r=f(x_i)$ and so we compute the volume in a similar manner as in Section 3.3.1:

Suppose there are $n$ disks on the interval $[a,b]\text{,}$ then the volume of the solid of revolution is approximated by

\begin{equation*} \sum_{i=0}^{n-1} \pi \left[f(x_i)\right]^2\Delta x\text{,} \end{equation*}

and when we apply the limit $\Delta x \to 0\text{,}$ the volume computes to the value of a definite integral

\begin{equation*} V = \lim_{\Delta x\to 0} \sum_{i=0}^{n-1} \pi \left[f(x_i)\right]^2\Delta x = \int_a^b \pi \left[f(x)\right]^2\,dx\text{.} \end{equation*}

Because the volume of the solid of revolution is calculated using disks, this type of computation is often referred to as the Disk Method.

We capture our results in the following theorem.

Note: Either of the variations 1 and 2 described below will most likely create disks with holes in the solid of revolution, and so we would have to apply the so-called Washer Method that is described in the next section.
1. We could rotate the area of any region around an axis of rotation, including the area of a region bounded above by a function $y=f(x)$ and below by a function $y=g(x)$ on an interval $x \in [a,b]\text{.}$

2. The axis of rotation can be any axis parallel to the $x$-axis for this method to work.

We now provide one further example of the Disk Method.

###### Example3.25. Volume of a Right Circular Cone.

Find the volume of a right circular cone with

1. base radius 10 and height 20.

2. base radius $r$ and height $h\text{.}$

(A right circular cone is one with a circular base and with the tip of the cone directly over the centre of the base.)

Solution
1. We can view this cone as produced by the rotation of the line $y=x/2$ rotated about the $x$-axis, as indicated below.

At a particular point on the $x$-axis, say $\ds x_i\text{,}$ the radius of the resulting cone is the $y$-coordinate of the corresponding point on the line, namely $\ds y_i=x_i/2\text{.}$ Thus the total volume is approximately

\begin{equation*} \sum_{i=0}^{n-1} \pi (x_i/2)^2\,dx \end{equation*}

and the exact volume is

\begin{equation*} \int_0^{20} \pi \frac{x^2}{4}\,dx= \frac{\pi}{4}\frac{x^3}{3}\bigg\vert_0^{20} = \frac{\pi}{4}\frac{20^3}{3}=\frac{2000 \pi}{3}\text{.} \end{equation*}
2. Note that we can instead do the calculation with a generic height and radius:

\begin{equation*} \int_0^{h} \pi{r^2\over h^2}x^2\,dx ={\pi r^2\over h^2}{h^3\over3}={\pi r^2h\over3}\text{,} \end{equation*}

giving us the usual formula for the volume of a cone.

### Subsection3.3.3Washer Method: Integration w.r.t $x$

Suppose the axis of revolution is not part of the boundary of an area as shown below in two different scenarios:

When either of the above area is rotated about its axis of rotation, then the solid of revolution that is created has a hole on the inside — like a distorted donut. If we now slice the solid perpendicular to the axis of rotation, then the cross-section shows a disk with a hole in it as indicated below. Such a disk looks like a “washer” and so the method that employs these disks for finding the volume of the solid of revolution is referred to as the Washer Method. The following example demonstrates how to find a volume that is created in this fashion.

###### Example3.26. Volume of an Object with a Hole.

Find the volume of the object generated when the area between $\ds y=x^2$ and $y=x$ is rotated around the $x$-axis.

Solution

We begin by graphing the area between $y=x^2$ and $y=x$ and note that the two curves intersect at the point $(1,1)$ as shown below to the left.

We now rotate this around around the $x$-axis as shown above to the right. We notice that the solid has a hole in the middle and we now consider two methods for calculating the volume.

Method 1: Subtraction of Volumes

We can think of the volume of the solid of revolution as the subtraction of two volumes: the outer volume is that of the solid of revolution created by rotating the line $y=x$ around the $x$-axis (see left graph in the figure below) — namely the volume of a cone, and the inner volume is that of the solid of revolution created by rotating the parabola $y=x^2$ around the $x$-axis (see right graph in the figure below) — namely the volume of the hornlike shape.

We have already computed the volume of a cone; in this case it is $\pi/3\text{.}$ At a particular value of $x\text{,}$ say $\ds x_i\text{,}$ the cross-section of the horn is a circle with radius $\ds x_i^2\text{,}$ so the volume of the horn is

\begin{equation*} \int_0^1 \pi(x^2)^2\,dx=\int_0^1 \pi x^4\,dx=\pi{1\over 5}\text{,} \end{equation*}

so the desired volume is $\pi/3-\pi/5=2\pi/15\text{.}$

Method 2: Washer Method

As with the area between curves, there is an alternate approach that computes the desired volume “all at once” by approximating the volume of the actual solid. We can approximate the volume of a slice of the solid with a washer-shaped volume as shown below.

The volume of such a washer is the area of the face times the thickness. The thickness, as usual, is $\Delta x\text{,}$ while the area of the face is the area of the outer circle minus the area of the inner circle, say $\ds \pi R^2-\pi r^2\text{.}$ In the present example, at a particular $\ds x_i\text{,}$ the radius $R$ is $\ds x_i$ and $r$ is $\ds x_i^2\text{.}$ Hence, the whole volume is

\begin{equation*} \int_0^1 \pi x^2-\pi x^4\,dx= \left.\pi\left({x^3\over3}-{x^5\over5}\right)\right|_0^1= \pi\left({1\over3}-{1\over5}\right)={2\pi\over15}\text{.} \end{equation*}

Of course, what we have done here is exactly the same calculation as before.

We now formalize the Washer Method employed in the above example.

Note: The most common mistake in the Washer Method is to write $(f(x)-g(x))^2$ rather than $f(x)^2-g(x)^2\text{.}$

We now present one more example that uses the Washer Method.

###### Example3.28. Washer Method.

Find the volume of the object generated when the area between $g(x)=x^2-x$ and $f(x)=x$ is rotated about the line $y=3\text{.}$

Solution

The area between the two curves is graphed below to the left, noting the intersection points $(0,0)$ and $(2,2)\text{:}$

From the graph, we see that the inner radius must be $r = 3-f(x) = 3-x\text{,}$ and the outer radius must be $R=3-g(x) = 3-x^2+x\text{.}$ Therefore, the volume of the object is

\begin{equation*} \begin{split} V \amp = \int_0^2 \pi\left(\left[3-x^2+x\right]^2-\left[3-x\right]^2\right)\,dx\\ \amp = \int_0^2 \pi \left(x^4 - 2 x^3 - 6 x^2 + 12 x\right)\,dx \\ \amp = \pi \left[\frac{x^5}{5} - \frac{x^4}{2} - 2 x^3 + 6 x^2\right]_0^2 \\ \amp = \frac{32 \pi}{5}. \end{split} \end{equation*}

### Subsection3.3.4Disk and Washer Methods: Integration w.r.t. $y$

We have already seen in Section 3.1 that sometimes a curve is described as a function of $y\text{,}$ namely $x=g(y)\text{,}$ and so the area of the region under the curve $g$ over an interval $[c,d]$ as shown to the left of Figure 3.14 can be rotated about the $y$-axis to generate a solid of revolution as indicated to the right in Figure 3.14.

We are readily convinced that the volume of such a solid of revolution can be calculated in a similar manner as those discussed earlier, which is summarized in the following theorem.

Note: Either of the variations 1 and 2 described below will most likely create disks with holes in the solid of revolution, and so we would have to apply the Washer Method.
1. Again, we could rotate the area of any region around an axis of rotation, including the area of a region bounded to the right by a function $x=f(y)$ and to the left by a function $x=g(y)$ on an interval $y \in [c,d]\text{.}$

2. The axis of rotation can be any axis parallel to the $y$-axis for this method to work.

We now provide an example of the Disk Method, where we integrate with respect to $y\text{.}$

###### Example3.30. Disk Method.

Find the volume of the object generated when the area between the curve $f(x)=x^2$ and the line $y=1$ in the first quadrant is rotated about the $y$-axis.

Solution

The area between $y=f(x)$ and $y=1$ is shown below to the right. We notice that the two curves intersect at $(1,1)\text{,}$ and that this area is contained between the two curves and the $y$-axis.

We now solve for $x$ as a function of $y\text{:}$

\begin{equation*} y = x^2 \implies x = \pm \sqrt{y}\text{,} \end{equation*}

and since we want the region in the first quadrant, we take $x=\sqrt{y}\text{.}$ The desired volume is found by integrating

\begin{equation*} \begin{split} V \amp = \pi\int_0^1 \left(\sqrt{y}\right)^2\,dy \\[1ex] \amp = \pi\int_0^1 y\,dy \\[1ex] \amp = \frac{\pi y^2}{2}\bigg\vert_0^1 = \frac{\pi}{2}. \end{split} \end{equation*}

Similar to the Washer Method when integrating with respect to $x\text{,}$ we can also define the Washer Method when we integrate with respect to $y\text{:}$

### Subsection3.3.5Summary

There are many different scenarios in which Disk and Washer Methods can be employed, which are not discussed here; however, we provide a general guideline.

###### Guideline for Disk and Washer Methods.

The following steps outline how to employ the Disk or Washer Method.

1. Graph the bounded region.

2. Construct an arbitrary cross-section perpendicular to the axis of rotation.

4. Determine the thickness of the disk or washer.

5. Set up the definite integral by making sure you are computing the volume of the constructed cross-section.

##### Exercises for Section 3.3.

Use the method from Section 3.3.1 to find each volume.

1. The base of a tetrahedron (a triangular pyramid) of height $h$ is an equilateral triangle of side $s\text{.}$ Its cross-sections perpendicular to an altitude are equilateral triangles. Express its volume $V$ as an integral, and find a formula for $V$ in terms of $h$ and $s\text{.}$ Verify that your answer is $(1/3)(\hbox{area of base})(\hbox{height})\text{.}$

$(1/3)(\hbox{area of base})(\hbox{height})$
Solution

We make a diagram below of the base of the tetrahedron:

Therefore, the vertical cross-sections at $x_i$ are equilateral triangles with side length $\sqrt{3}x_i\text{.}$ Therefore, the area of this cross-section is

\begin{equation*} A(x_i) = \frac{\sqrt{3}}{4} \bigl(3 x_i^2\bigr) \end{equation*}

for $0 \leq x_i \leq \frac{s}{2}\text{.}$ Note that at $x_i = s/2\text{,}$ we must have:

\begin{equation*} h = \frac{\sqrt{3}}{2}\left(\frac{\sqrt{3}s}{4}\right) = \frac{3s}{4}\text{,} \end{equation*}

which gives the relationship between $h$ and $s\text{.}$ We now compute the volume of the solid:

\begin{equation*} V = 2\int_0^{s/2} A(x) \,dx = 2\int_0^{s/2} \frac{\sqrt{3}}{4} \bigl(3 x^2\bigr)\,dx = \sqrt{3} \frac{s^3}{16}\text{.} \end{equation*}

We now check that this is equivalent to $\frac{1}{3}\bigl(\text{ area base } \bigr)h\text{:}$

\begin{equation*} \frac{1}{3}\bigl(\text{ area base } \bigr)h = \frac{1}{3} \left(\frac{\sqrt{3}}{4} s^2\right) h= \sqrt{3}\frac{s^3}{16}\text{,} \end{equation*}

as we found above.

2. The base of a solid is the region between $f(x)=\cos x$ and $g(x)=-\cos x\text{,}$ $-\pi/2\le x\le\pi/2\text{,}$ and its cross-sections perpendicular to the $x$-axis are squares. Find the volume of the solid.

$2\pi$
Solution

We draw a diagram below of the base of the solid:

Therefore, the cross-section at $x_i$ must be a square with side length $g(x_i)-f(x_i)\text{.}$ And so the area of this cross-section is

\begin{equation*} A(x) = \bigl(g(x_i)-f(x_i)\bigr)^2 = 4\cos^2(x_i) \end{equation*}

for $0 \leq x_i \leq \frac{\pi}{2}\text{.}$ We now compute the volume of the solid by integrating over these cross-sections:

\begin{equation*} V = 8\int_0^{\pi/2} \cos^2(x)\,dx = 2\pi\text{.} \end{equation*}

Find the volume of the solid generated by revolving the shaded region about the given axis.

1. About the $x$-axis:

$4\pi/3$
Solution
We first write $y=2-2x\text{.}$ Now integrate:
\begin{equation*} \begin{split} V \amp= \int_0^1 \pi \left[f(x)\right]^2 \,dx \\ \amp=\pi \int_0^1 \left[2-2x\right]^2\,dx \end{split} \end{equation*}
Let $u=2-2x\text{,}$ with $du = -2x\,dx\text{.}$ Then:
\begin{equation*} \begin{split} V \amp=\pi \int_0^1 \left[2-2x\right]^2\,dx \\ \amp= \pi \int_2^0 \frac{u^2}{2} \,-du\\ \amp= \frac{\pi}{2} \int_0^2 u^2 \,du\\ \amp= \frac{\pi}{6}u^3 \big\vert_0^2 \\ \amp= \frac{4\pi}{3}. \end{split} \end{equation*}
2. About the $y$-axis:

$50\pi/3$
Solution
We integrate with respect to $y\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^2 \pi \left[\frac{5y}{2}\right]^2\,dy \\ \amp= \frac{25\pi}{4}\int_0^2 y^2\,dy \\ \amp= \frac{25\pi}{12} y^3 \big\vert_0^2\\ \amp= \frac{50\pi}{3}. \end{split} \end{equation*}
3. About the $y$-axis:

$\pi/2$
Solution
We integrate with respect to $y\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^1 ]pi \left[\sqrt{y}\right]^2\,dy \\ \amp= \pi \int_0^1 y\,dy \\ \amp= \frac{\pi}{2} y^2 \big\vert_0^1\\ \amp= \frac{\pi}{2}. \end{split} \end{equation*}
4. About the $x$-axis:

$\pi^2/32$
Solution
We interate with respect to $x\text{:}$
\begin{equation*} \begin{split} V \amp= \int_{\pi/2}^{\pi/4} \pi\left[\sin x \cos x\right]^2 \,dx \\ \amp= \pi \int_{\pi/2}^{\pi/4} \sin^2 x \cos^2x \,dx \\ \amp= \pi \int_{\pi/2}^{\pi/4} \frac{1-\cos^2(2x)}{4} \,dx \\ \amp= \frac{\pi}{4} \int_{\pi/2}^{\pi/4} \left(1- \frac{1+\cos(4x)}{2}\right)\,dx\\ \amp= -\frac{\pi}{32} \left[\sin(4x)-4x\right]_{\pi/4}^{\pi/2}\\ \amp= \frac{\pi^2}{32}. \end{split} \end{equation*}

Find the volume of the solid generated by revolving the given bounded region about the $x$-axis.

1. $y=x^2, \ y=1, \ x=2, \ y = 0$

$6\pi/5$
Solution
We begin by plotting the area bounded by the given curves:
We split the region into two: one where the top curve is given by $y=x^2\text{,}$ and the second where the region is bounded by $y=1\text{.}$ We now use the disk method, and integrate with repect to $x\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^1 \pi \left[x^2\right]^2\,dx + \int_1^2 \pi \left[1\right]^2\,dx \\ \amp= \pi \int_0^1 x^4\,dx + \pi\int_1^2 \,dx \\ \amp= \frac{\pi x^5}{5}\big\vert_0^1 + \pi x \big\vert_1^2\\ \amp= \frac{\pi}{5} + \pi = \frac{6\pi}{5}. \end{split} \end{equation*}
2. $y=x^3, \ y=0, \ x=1$

$\pi/7$
Solution
We first plot the area bounded by the given curves:
We now use the Disk method, and integrate with respect to $x\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^1 \pi \left[x^3\right]^2\,dx \\ \amp= \pi \int_0^1 x^6 \,dx \\ \amp= \left[\frac{\pi x^7}{7}\right]_0^1\\ \amp= \frac{\pi}{7}. \end{split} \end{equation*}
3. $y=\sqrt{4-x^2}, \ y=0$

$32\pi/3$
Solution
We begin by plotting the area bounded by the given curves:
We now use the Disk method, and integrate with respect to $x\text{:}$
\begin{equation*} \begin{split} V \amp= \int_{-2}^2 \pi \left[\sqrt{4-x^2}\right]^2\,dx \\ \amp= \pi \int_{-2}^2 4-x^2\,dx \\ \amp= \pi \left[4x - \frac{x^3}{3}\right]_{-2}^2\\ \amp= \frac{32\pi}{3}. \end{split} \end{equation*}
4. $y=\sqrt{\sin x}, \ 0\leq x\leq \pi/2$

$\pi$
Solution
We begin by plotting the area bounded by the given curves:
We now use the Disk method, and integrate with respect to $x\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^{\pi/2} \pi \left[\sqrt{\sin x}\right]^2 \,dx \\ \amp= \pi \int_0^{\pi/2} \sin x \,dx \\ \amp= -\pi \cos x\big\vert_0^{\pi/2}\\ \amp= \pi. \end{split} \end{equation*}
5. $y=e^{-x}, \ y=0, \ x=2$

1.5420
Solution

We begin by plotting the area bounded by the given curves:

We wish to calculate the volume of the solid generated by rotating this area about the $x$-axis. We therefore use the disk method, and integrate with respect to $x\text{:}$

\begin{equation*} V = \int_0^2 \pi (e^{-x})^2 \,dx = \pi \int_0^2 e^{-2x}\,dx = -\frac{\pi}{2}e^{-2x}\bigg\vert_0^2 = -\frac{\pi}{2}\left(e^{-4}-1\right)\text{.} \end{equation*}
6. $x+y=2, \ y=0, \ x=0$

$8\pi/3$
Solution
We begin by plotting the area bounded by the given curves:
We now use the Disk method, and integrate with respect to $x\text{,}$ and use the substituion $u=2-x$ with $du = -dx\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^2 \pi\left[2-x\right]^2\,dx\\ \amp= -\pi \int_2^0 u^2 \,du\\ \amp= \pi \int_0^2 u^2 \,du\\ \amp= \frac{\pi u^3}{3} \bigg\vert_0^2\\ \amp= \frac{8\pi}{3}. \end{split} \end{equation*}
7. $y=x-x^2, \ y=0, \ x=0$

$\pi/30$
Solution
We begin by plotting the area bounded by the curves:
Therefore, we use the Disk method and integrate with respect to $x\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^1 \pi \left[x-x^2\right]^2 \,dx\\ \amp= \pi \int_0^1 x^4-2x^3+x^2 \,dx \\ \amp= \pi \left[\frac{x^5}{5}-\frac{2x^4}{4} + \frac{x^3}{3}\right]_0^1\\ \amp= \pi \left[\frac{1}{5} - \frac{1}{2} + \frac{1}{3} \right]\\ \amp= \frac{\pi}{30}. \end{split} \end{equation*}
8. $\ds y=\sqrt{\sin x}, \ x=0$

$2\pi$
Solution
We notice that $y=\sqrt(\sin(x)) = 0$ at $x=\pi\text{.}$ We plot the region below:
We now use the Disk method, and integrate with respect to $x\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^{\pi} \pi \left[\sqrt{\sin x}\right]^2 \,dx \\ \amp= \pi \int_0^{\pi} \sin x \,dx \\ \amp= -\pi \cos x\big\vert_0^{\pi}\\ \amp= 2 \pi. \end{split} \end{equation*}

Find the volume of the solid generated by revolving the given bounded region about the $y$-axis.

1. $x=y^2, \ x=0, \ y=-1, \ y=1$

$2\pi/5$
Solution
Due to symmetry, the area bounded by the given curves will be twice the green shaded area below:
We therefore use the disk method, and integrate with respect to $y\text{:}$
\begin{equation*} \begin{split} V \amp= 2\int_0^1 \pi \left[y^2\right]^2 \,dy \\ \amp= 2\pi \int_0^1 y^4\,dy \\ \amp= \frac{2\pi y^5}{5} \big\vert_0^1\\ \amp= \frac{2\pi}{5}. \end{split} \end{equation*}
2. $x=y^{3/2}, \ x=0, \ y=4$

$64\pi$
Solution
We begin by plotting the area bounded by the given curves:
We now use the Disk method, and integrate with respect to $y\text{:}$
\begin{equation*} \begin{split} V\amp= \int_0^4 \pi \left[y^{3/2}\right]^2\,dy \\ \amp= \pi \int_0^4 y^3 \,dy \\ \amp= \pi \frac{y^4}{4}\big\vert_0^4 \\ \amp= 64\pi. \end{split} \end{equation*}
3. $x=\sqrt{\sin(2y)}, \ 0\leq y\leq \pi/2, \ x=0$

$\pi$
Solution

The area contained between $x=0$ and the curve $x=\sqrt{\sin(2y)}$ for $0\leq y\leq \frac{\pi}{2}$ is shown below.

Rotating this area around the $y$-axis results in a solid with volume

\begin{equation*} \begin{split} V\amp=\int_0^{\frac{\pi}{2}} \pi \left(\sqrt{\sin(2y)}\right)^2\,dy\\ \amp = \pi\int_0^{\frac{\pi}{2}} \sin(2y)\,dy\\ a\mp = -\frac{\pi}{2}\cos(2y)\bigg\vert_0^{\frac{\pi}{2}}\\ \amp = -\frac{\pi}{2} (-1-1) = \pi.\end{split} \end{equation*}
4. $x=\sqrt{\cos(2y)},\ 0\leq y\leq \pi/2, \ x=0$

$\pi/2$

Find the volume of the solid generated by revolving the shaded region about the given axis.

1. The $x$-axis:

\begin{equation*} \end{equation*}

$4\pi \left(\pi-2\right)$
Solution
We notice that the region is bounded on top by the curve $y=2\text{,}$ and on the bottom by the curve $y=\sqrt{\cos x}\text{.}$ Therefore, we use the Washer method and integrate with respect to $x\text{.}$ By symmetry, we have:
\begin{equation*} \begin{split} V \amp= 2\int_{0}^{\pi/2} \pi \left[2^2 - \left(2\sqrt(\cos x)\right)^2 \right]\,dx\\ \amp= 2\pi \int_{0}^{\pi/2} 4-4\cos x \,dx\\ \amp= 8 \pi \left[x - \sin x\right]_0^{\pi/2}\\ \amp= 4\pi \left(\pi-2\right). \end{split} \end{equation*}

2. The $y$-axis:

\begin{equation*} \end{equation*}

$\frac{\pi}{4}\left(2\pi-1\right)$
Solution
We notice that the region is bounded on the left by the curve $x=\sin y$ and on the right by the curve $x=1\text{.}$ We therefore use the Washer method and integrate with respect to $y\text{:}$
\begin{equation*} \begin{split} V \amp = \int _0^{\pi/2} \pi \left[1 - \sin^2 y\right]\,dy \\ \amp= \pi \int_0^{\pi/2} 1 - \frac{1}{2}\left(1-\cos(2y)\right)\,dy \\ \amp= \frac{\pi}{4}\left(2\pi-1\right). \end{split} \end{equation*}

Find the volume of the solid generated by revolving the given bounded region about the $x$-axis.

1. $y=x, \ y=2, \ x=0$

$16\pi/3$
Solution
We first plot the area bounded by the given curves:
We notice that the area is bounded above by the curve $y=2$ for $x\in[0,2]\text{,}$ and bounded below by the curve $y=x$ for $x\in[0,2]\text{.}$ Therefore, we use the Washer method and integrate with respect to $x\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^2 \pi \left[2^2-x^2\right]\,dx\\ \amp= \pi \int_0^2 \left[4-x^2\right]\,dx\\ \amp= \pi\left[4x-\frac{x^3}{3}\right]_0^2\\ \amp=\frac{16\pi}{3}. \end{split} \end{equation*}
2. $y=3\sqrt{x}, \ y=3, \ x=0$

$9\pi/2$
Solution
We first plot the area bounded by the given curves:
We notice that the region is bounded above by the curve $y=3$ for $x\in[0,1]$ and bounded below by the curve $y=3\sqrt{x}$ for $x\in[0,1]\text{.}$ Therefore, we use the Washer method and integrate with respect to $x\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^1 \pi \left[3^2-\bigl(3\sqrt{x}\bigr)^2\right]\,dx\\ \amp= \pi \int_0^1 \left[9-9x\right]\,dx\\ \amp= \pi\left[9x-\frac{9x^2}{2}\right]_0^1\\ \amp=\frac{9\pi}{2}. \end{split} \end{equation*}
3. $y=x^2+1,\ y+x=3$

$117\pi/5$
Solution

The points of intersection of the curves $y=x^2+1$ and $y+x=3$ are calculated to be

\begin{equation*} \begin{gathered} x^2+1=3-x \\ x^2+x-2 = 0 \\ (x-1)(x+2) = 0 \\ \implies x=1,-2. \end{gathered} \end{equation*}

Let $f(x)=x^2+1$ and $g(x)=3-x\text{.}$ We now plot the area contained between the two curves:

To find the volume of the solid which is generated when this area is rotated around the $x$-axis, we use the washer method, and integrate with respect to $x\text{.}$ We find

\begin{equation*} V = \int_{-2}^1 \pi\left[(3-x)^2 - (x^2+1)^2\right]\,dx = \pi \left[-\frac{x^5}{5} - \frac{x^3}{3} - 3x^2 + 8x\right]_{-2}^1 = \frac{117\pi}{5}\text{.} \end{equation*}
4. $y=9-x^2, \ y=3-x$

$125/3 (6\pi-1)$
Solution
We first compute the intersection point(s) of the two curves:
\begin{equation*} \begin{gathered} x^2+1=3-x \\ x^2-x-6 = 0 \\ (x-3)(x+2) = 0 \\ \implies x=3,-2. \end{gathered} \end{equation*}
We now plot the area contained between the two curves:
The region is bounded above by the curve $y=9-x^2$ for $x\in[-2,3]\text{,}$ and bounded below by the curve $y=3-x$ for $x\in[-2,3]\text{.}$ We therefore use the Washer method and integrate with respect to $x\text{:}$
\begin{equation*} \begin{split} V \amp= \int_{-2}^3 \pi \left[(9-x^2)^2 - (3-x)^2\right)\,dx \\ \amp= \pi \int_{-2}^3 \left[x^4-19x^2+6x+72\right]\,dx\\ \amp = \pi \left[\frac{x^5}{5}-19\frac{x^3}{3} + 3x^2 + 72x\right]_{-2}^3\\ \amp= \frac{125}{3}\bigl(6\pi-1\bigr) \end{split} \end{equation*}

The equation $\ds x^2/9+y^2/4=1$ describes an ellipse. Find the volume of the solid obtained by rotating the ellipse around the $x$-axis and also around the $y$-axis. These solids are called ellipsoids; one is vaguely rugby-ball shaped, one is sort of flying-saucer shaped, or perhaps squished-beach-ball-shaped.

$16\pi\text{,}$ $24\pi$
Solution
For the first solid, we consider the following region:
Notice that for this region, we have $y = 2\sqrt{1-\frac{x^2}{9}}\text{.}$ Therefore, we use the Disk method and integrate with respect to $x\text{:}$
\begin{equation*} \begin{split} V \amp= \int_{-3}^3 \pi \left[2\sqrt{1-\frac{x^2}{9}}\right]^2\,dx \\ \amp= 4\pi \int_{-3}^3 \left(1-\frac{x^2}{9}\right)\,dx\\ \amp= 4\pi \left[x - \frac{x^3}{9(3)}\right]_{-3}^3\\ \amp= 16 \pi. \end{split} \end{equation*}
Next, we consider the following region:
Now we write $x = 3\sqrt{1-\frac{y^2}{4}}$ and integrate with respect to $y\text{:}$
\begin{equation*} \begin{split} V \amp= \int_{-2}^2 \pi \left[3\sqrt{1-\frac{y^2}{4}}\right]^2\,dy \\ \amp= 9\pi \int_{-2}^2 \left(1-\frac{y^2}{4}\right)\,dx\\ \amp= 9\pi \left[x - \frac{y^3}{4(3)}\right]_{-2}^2\\ \amp= 24 \pi. \end{split} \end{equation*}

Use integration to compute the volume of a sphere of radius $r\text{.}$ You should of course get the well-known formula $\ds 4\pi r^3/3\text{.}$

Solution
Consider the region the curve $y^2+x^2=r^2$ such that $y \geq 0\text{:}$
Notice that the shaded region is bounded above by the curve $y = \sqrt{r^2-x^2}$ for $x \in[-r,r]$ and below by the $x$-axis. Therefore, to find the volume of the sphere of radius $r\text{,}$ we use the Disk method and integrate with respect to $x\text{:}$
\begin{equation*} \begin{split} V \amp= \int_{-r}^r \pi \left[\sqrt{r^2-x^2}\right]^2\,dx\\ \amp= \pi \int_{-r}^r \left(r^2-x^2\right)\,dx\\ \amp= \pi \left[r^2 x - \frac{x^3}{3}\right]_{-r}^r \\ \amp= \pi \left[\left(r^3-\frac{r^3}{3}\right)-\left(-r^3+\frac{r^3}{3}\right)\right]\\ \amp= \pi \left(2r^3-\frac{2r^3}{3}\right)\\ \amp= \frac{4\pi r^3}{3}, \end{split} \end{equation*}
which is exactly the formula for the volume of a sphere of radius $r\text{.}$

A hemispheric bowl of radius $r$ contains water to a depth $h\text{.}$ Find the volume of water in the bowl.

$\ds \frac{2}{3}\pi h r^2$
(Note: this is the part of the ellipse $\frac{y^2}{h^2} + \frac{x^2}{r^2} = 1$ for which $y \geq 0$ and $0\leq x\leq r$). Therefore, to find the volume of the solid, we use the Disk method and integrate with respect to $y\text{:}$