Let $y=f(x)$ and $y=g(x)$ be continuous functions on a closed interval $[a,b]$ such that $f(x)\geq g(x)$ on $[a,b]$ as shown in Figure 1.

Figure 1

If the region bounded by $f(x)$ and $g(x)$ on $[a,b]$ is revolved about the $x$-axis, the volume of the resulting solid is obtained by \begin{align*}V&=\int_a^b\pi [f(x)]^2dx-\int_a^b\pi [g(x)]^2dx\\&=\int_a^b\pi ([f(x)]^2-[g(x)]^2)dx\end{align*} This is essentially the disk method but it is rather called the washer method because each cross section is a washer due to a hollow core within the solid.

Example. The region bounded by  the curve $y=x^2+1$ and the line $y=-x+3$ is revolved about the $x$-axis. Find the volume of the resulting solid.

Solution. Figure 2 shows the bounded region

Figure 1

and Figure 2 shows the solid of revolution.

Figure 2

The volume $V$ is given by $$V=\int_{-2}^1\pi[(-x+3)^2-(x^2+1)^2]dx=\frac{117\pi}{5}$$

Example. The region bounded by the parabola $y=x^2$ and the line $y=2x$ in the first quadrant is revolved about the $y$-axis to generate a solid. Find the volume of the solid.

Solution. Figure 3 shows the bounded region

Figure 3

and Figure 4 shows the solid of revolution.

Figure 4

The volume of the solid is given by $$V=\int_0^4\pi\left[(\sqrt{y})^2-\left(\frac{y}{2}\right)^2\right]dy=\frac{8\pi}{3}$$

Suppose that the region bounded by $y=x^4+1$, $x$-axis, $x=1$ and $x=2$, as shown in Figure 1, is revolved about the $y$-axis and we want to find the volume of the resulting solid.

Figure 1

You will find it difficult to apply the disk method or the washer method for this though it’s doable. Here we want to devise another method that can come in handy for this kind of problems. First let us take a look at Figure 2.

Figure 2

Imagine that each rectangle is rotated about the $y$-axis. Then we would obtain layers of cylindrical shells and the sum of the volumes of these cylindrical shells would approximate the volume of the solid in question. If we use infinitesimally thin rectangles, the sum of the volumes of cylindrical shells would be exactly the volume of the solid. The radius of the cylindrical shell at $x$ is $x$ since the axis of rotation is the $y$-axis ($x=0$), its height is $x^4+1$, and its infinitesimal thickness is $dx$. So the volume of the cylindrical shell at $x$ is $2\pi x(x^4+1)dx$. Therefore the volume of the solid is $$V=\int_1^2 2\pi x(x^4+1)dx=24\pi$$

Example. Find the volume of the solid formed by rotating the region enclosed by $$x=0,\ x=1,\ y=0,\ y=9+x^3$$ about the $y$-axis.

Solution.

Figure 3

The enclosed region is shown in Figure 3. Using the cylindrical shell method, the volume is \begin{align*}V&=\int_0^1 2\pi x(9+x^3)dx\\&=2\pi\int_0^1(9x+x^4)dx\\&=2\pi\left[\frac{9}{2}x^2+\frac{x^5}{5}\right]_0^1\\&=\frac{47}{5}\pi\end{align*}

Remark. The volume can be also found by the disk/washer method but its solution is more complicated than the one by cylindrical shell method. In order to use the disk/washer method, we need to break the enclosed region into two parts as shown in Figure 4.

Figure 4

When rotated about the $y$-axis, the region in red will give rise to a solid cylinder with the radius 1 and the height 9. Its volume is $9\pi$. For the solid formed by rotating the region in blue, using the washer method, its volume is given by \begin{align*}\int_9^{10}\pi [1^2-((y-9)^{\frac{1}{3}})^2]dy&=\pi\int_9^{10}[1-(y-9)^{\frac{2}{3}}]dy\\&=\pi\left[y-\frac{3}{5}(y-9)^{\frac{5}{3}}\right]_9^{10}\\&=\frac{2}{5}\pi\end{align*} Therefore, the volume $V$ of the solid in question is given by $$V=9\pi+\frac{2}{5}\pi=\frac{47}{5}\pi$$

More generally we have

Cylindrical Shell Method

Suppose that the region under the curve $y=f(x)$, $a\leq x\leq b$ is revolved about the line $x=L$. Then the volume $V$ of the resulting solid is obtained by $$V=\left\{\begin{array}{ccc}\int_a^b 2\pi (x-L)f(x)dx & \mbox{if} & L\leq a\\\int_a^b 2\pi (L-x)f(x)dx & \mbox{if} & L\geq b\end{array}\right.$$

Example. The region enclosed by the $x$-axis and the parabola $y=3x-x^2$ is revolved about the vertical line $x=-1$. Find the volume of the resulting solid.

Solution. The region (in red) and the axis of rotation (in blue) are shown in Figure 5.

Figure 5

By cylindrical shell method, the volume $V$ is given by $$V=\int_0^3 2\pi(x+1)(3x-x^2)dx=\frac{45\pi}{2}$$

Example. The region bounded by the curve $y=\sqrt{x}$, the $x$-axis and the line $x=4$ is revolved about the $y$-axis. Find the volume of the resulting solid.

Solution. Figure 6 shows the region.

Figure 6

Since the axis of rotation is $x=0$, by cylindrical shell method the volume $V$ is given by $$V=\int_0^42\pi x\sqrt{x}dx=\frac{128\pi}{5}$$

Note that while cylindrical shell method is simpler, it is also easy to find the volume using washer method. $$V=\int_0^2\pi[4^2-(y^2)^2]dy=\int_0^2(16-y^4)dy=\frac{128\pi}{5}$$

Example. The same region in the previous example is revolved about the $x$-axis this time. Find the volume of the resulting solid using cylindrical shell method.

Solution. The solid and a cylindrical shell are shown in Figure 7.

Figure 7

The shell radius is $y$ and the shell height $4-y^2$. Hence the volume is $$V=\int_0^2 2\pi y(4-y^2)dy=8\pi$$

Note that it would actually be easier to use disk method for this problem. Using disk method the volume $V$ is given by $$V=\int_0^4\pi(\sqrt{x})^2dx=\int_0^4\pi xdx=8\pi$$

Let $y=f(x)$ and $y=g(x)$ be two continuous functions on a closed interval $[a,b]$ such that $f(x)\geq g(x)$ on $[a,b]$ as seen in Figure 1.

Figure 1

The area $A$ of the region between the curves $f(x)$ and $g(x)$ is \begin{align*}A&=\int_a^b f(x)dx-\int_a^b g(x)dx\\&=\int_a^b[f(x)-g(x)]dx\end{align*}

Example. Find the area of the region enclosed by the parabola $y=2-x^2$ and the line $y=-x$.

Solution. First we need to find the $x$-coordinates of the points at which the two curves meet. Set $2-x^2=-x$. Then the quadratic equation $x^2-x-2=0$ has two solutions $x=-1,2$. From Figure 2

Figure 2

we see that the area is \begin{align*}A&=\int_{-1}^2[(2-x^2)-(-x)]dx\\&=\int_{-1}^2(2+x-x^2)dx\\&=\frac{9}{2}\end{align*}

Example. Find the area of the region in the first quadrant that is bounded above by $y=\sqrt{x}$ and below by the $x$-axis and the line $y=x-2$.

Solution. To find the $x$-coordinates of the points at which $y=\sqrt{x}$ and $y=x-2$ meet, set $\sqrt{x}=x-2$. Squaring this equation, we obtain the quadratic equation $x^2-5x+4=0$ whose solutions are $x=-1$ and $x=4$. Since we are considering only the first quadrant $x=-1$ is not our concern. The bounded region is shown in Figure 3.

Figure 3

Note that on $[0,2]$ the part of the region is bounded by $y=\sqrt{x}$ and the $x$-axis and on $[2,4]$ the part of the reagion is bounded by $y=\sqrt{x}$ and $y=x-2$. Hence, the area is \begin{align*}A&=\int_0^2\sqrt{x}dx+\int_2^4[\sqrt{x}-(x-2)]dx\\&=\frac{10}{3}\end{align*}

If $x=f(y)$ and $x=g(y)$ are two continuous functions on a closed interval $[c,d]$ such that $f(y)\geq g(y)$ as seen in Figure 4,

Figure 4

then the area $A$ bounded by $f(y)$ and $g(y)$ is given by $$A=\int_c^d[f(y)-g(y)]dy$$

Example. Redo the previous example by integrating with respect to $y$.

Solution. The same region in Figure 3 can be considered as the region bounded by $x=y^2$ ($y\geq 0$ and $x=y+2$. Hence, $$A=\int_0^2[y+2-y^2]dy=\frac{10}{3}$$

Often the fastest way to find an area may be to combine calculus with geometry. The same region in Figure 3 can be viewed as in Figure 5.

Figure 5

The area of the right triangle in blue is $\frac{1}{2}\cdot 2\cdot 2=2$ hence, the area of the region is $$A=\int_0^4\sqrt{x}dx-2=\frac{10}{3}$$

Disk Method

Consider the solid obtained by rotating a region bounded by the function $y=f(x)$ and the $x$-axis, $a\leq x\leq b$.

Figure 1

The cross section of the resulting sold at $x$ is a disk of radius $f(x)$ as seen in Figure 1 and so its area is $A(x)=\pi [f(x)]^2$. Hence the volume of the solid of revolution is \begin{equation}\label{eq:diskmethod}V=\pi\int_a^b [f(x)]^2dx\end{equation}

Example. The region between the curve $y=\sqrt{x}$, $0\leq x\leq 4$ and the $x$-axis is revolved about the $x$-axis. Find the volume of the resulting solid.

Solution. Figure 2 shows the region

Figure 2

and Figure 3 shows the resulting solid.

Figure 3

The volume $V$ is given by $$V=\pi\int_0^4[\sqrt{x}]^2dx=\pi\int_0^4 xdx=8\pi$$

Example. Find the volume of the solid generated by revolving the region bounded by $y=\sqrt{x}$ and the lines $y=1$, $x=4$ about the line $y=1$.

Solution. Figure 4 shows the region

Figure 4

and Figure 5 shows the resulting solid.

Figure 5

Note that the radius of the cross section at $x$ is $\sqrt{x}-1$ and so the volume is given by $$V=\pi\int_1^4[\sqrt{x}-1]^2dx=\frac{7}{6}\pi$$

Example. Find the volume of the solid obtained by revolving the region between the $y$-axis and the curve $x=\frac{2}{y}$, $1\leq y\leq 4$ about the $y$-axis.

Solution. Figure 6 shows the region

Figure 6

and Figure 7 shows the resulting solid.

Figure 7

The area of the cross section at $y$ is $A(y)=\pi\left(\frac{2}{y}\right)^2$. Hence the volume is $$V=\int_1^4 A(y)dy=\pi\int_1^4\left(\frac{2}{y}\right)^2dy=3\pi$$

Consider a solid each of which parallel cross sections has its area represented by the function $A(x)$, $a\leq x\leq b$. The volume element is then given by $dV=A(x)dx$. Hence the volume of the solid is obtained by the integral \begin{equation}\label{eq:volslice}V=\int_{x=a}^{x=b}dV=\int_a^b A(x)dx\end{equation}

Example. Find the volume of a pyramid whose base is a square with side $L$ and whose height is $h$.

Solution. See Figure 1.

Figure 1

We find the ratio $x$ to $h$ $\frac{x}{h}=\frac{\frac{s}{2}}{{\frac{L}{2}}}=\frac{s}{L}$. So we have $s=\frac{L}{h}x$. (Another observation is the line segment $\overline{OP}$ has slope $\frac{\frac{L}{2}}{h}=\frac{L}{2h}$ so the equation of line through $\overline{OP}$ is $y=\frac{L}{2h}x$ and for $y=\frac{s}{2}$ we get $s=\frac{L}{h}x$.) The area $A(x)$ of the cross section at $x$ is $A(x)=s^2=\frac{L^2}{h^2}x^2$. Hence $$V=\int_0^h A(x)dx=\frac{L^2}{h^2}\int_0^h x^2dx=\frac{1}{3}L^2h$$

Example. Show that the volume of a right circular cone whose circular base has radius $r$ and whose height is $h$ is $V=\frac{1}{3}\pi r^2h$.

Solution. Left as an exercise.

Example. Show that the volume of a sphere of radius $r$ is $V=\frac{4}{3}\pi r^3$.

Solution. See Figure 2.

Figure 2

$y=\sqrt{r^2-x^2}$ so $A(x)=\pi y^2=\pi (r^2-x^2)$. Hence the volume $V$ is \begin{align*}V&=\int_{-r}^r A(x)dx\\&=\int_{-r}^r \pi (r^2-x^2)dx\\&=2\pi\int_0^r (r^2-x^2)dx\\&=2\pi\left[r^2x-\frac{x^3}{3}\right]_0^r\\&=\frac{4}{3}\end{align*}

Example. Consider a solid with a circular base of radius 1 whose parallel cross secxtions perpendicular to the base are equilateral triangles. Find the volume of the solid.

Solution. See Figure 3.

Figure 3

From the picture we see that $$A(x)=\frac{1}{2}(2y)(\sqrt{3}y)=\sqrt{3}y^2=\sqrt{3}(1-x^2)$$ Hence the volume $V$ is given by $$V=\int_{-1}^1 A(x)dx=\sqrt{3}\int_{-1}^1(1-x^2)dx=2\sqrt{3}\int_0^1(1-x^2)dx=\frac{4\sqrt{3}}{3}$$

Example. A Curved wedge is cut from a cylinder of radius 3 by two planes. One plane is perpendicular to the axis of the cylinder. The other plane is crossing the first plane at a $45^\circ$ angle at the center of the cylinder.

Solution. See Figure 4.

Figure 4

The cross section at $x$ is the rectangle in blue. It’s area is $$A(x)=\mbox{length}\times\mbox{width}=2\sqrt{9-x^2}\times x=2x\sqrt{9-x^2}$$ Hence the volume $V$ is given by \begin{align*}V&=\int_0^3 2x\sqrt{9-x^2}dx\\&=-\int_9^0\sqrt{u}du\ (u=9-x^2)\\&=\int_0^9\sqrt{u}du\\&=\frac{2}{3}[u^{\frac{3}{2}}]_0^9\\&=18\end{align*}