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Volume

Many three-dimensional solids can be generated by revolving a curve about the $x$-axis or $y$-axis. For example, if we revolve the semi-circle given by $f(x)=\sqrt{r^2-x^2}$ about the $x$-axis, we obtain a sphere of radius $r$. We can derive the familiar formula for the volume of this sphere.

Finding the Volume of a Sphere

Consider a cross-section of the sphere as shown. It is a circle with radius $f(x)$ and area $\pi [f(x)]^2$. Informally speaking, if we "slice" the sphere vertically into discs, each disc having infinitesimal thickness $dx$, the volume of each disc is approximately $\pi [f(x)]^2\, dx$. If we "add up" the volumes of the discs, we will get the volume of the sphere: \begin{eqnarray*} V&=&\int^r_{-r} \pi [f(x)]^2\, dx\\ &=& \int^r_{-r} \pi (r^2-x^2)\, dx\\ &=& \left.\pi \left(r^2x-\frac{x^3}{3}\right)\right|^r_{-r}\\ &=& \pi \left(\frac{2}{3}r^3\right)-\pi \left(-\frac{2}{3}r^3\right)\\ &=& \frac{4}{3}\pi r^3,\quad{\small\textrm{as expected.}} \end{eqnarray*} This is called the Method of Discs. In general, suppose $y=f(x)$ is nonnegative and continuous on $[a,b]$. If the region bounded above by the graph of $f$, below by the $x$-axis, and on the sides by $x=a$ and $x=b$ is revolved about the $x$-axis, the volume $V$ of the generated solid is given by \[V=\int^a_b \pi [f(x)]^2\, dx.\] We can also obtain solids by revolving curves about the $y$-axis.

Revolving a Region about the $y$-axis

If we revolve the region enclosed by $y=x^2$ and $y=2x$, $0\leq x\leq 2$, about the $y$-axis, we generate a three-dimensional solid.

Let's find the volume of this solid. If we "slice" the solid horizontally, each slice is a "washer." The outer radius is $\sqrt{y}$ (since $y=x^2 \rightarrow x=\sqrt{y}$), the inner radius is $y/2$ (since $y=2x \rightarrow x=y/2$), and the thickness is $dy$. The volume of each washer is therefore \[ [\pi (\sqrt{y})^2-\pi (y/2)^2]\, dy.\] Then the volume of the entire solid is given by \begin{eqnarray*} \int^4_0 [\pi (\sqrt{y})^2-\pi (y/2)^2]\, dy&=&\int^4_0 \pi \left[y-\frac{y^2}{4}\right]\, dy\\ &=& \left.\pi \left[\frac{y^2}{2}-\frac{y^3}{12}\right]\right|^4_0\\ &=& \pi \left( 8-\frac{16}{3}\right)-\pi \left(0-0\right)\\ &=&\frac{8\pi}{3}. \end{eqnarray*} This generalization of the Method of Discs is called the Method of Washers. As we have seen, these methods may be used when a region is revolved about either axis.

  Suppose $y=f(x)$ and $y=g(x)$ are Suppose $x=F(y)$ and $x=G(y)$ are
  continuous and nonnegative on $[a,b]$ continuous and nonnegative on $[c,d]$
  with $g(x)\leq f(x)$ for all $x\in [a,b].$ with $G(y)\leq F(y)$ for all $y\in [c,d]$.
  If the region bounded above by $f$, If the region bounded on the right
  below by $g$, and on the sides by by $F$, on the left by $G$, and on the
  $x=a$ and $x=b$ is revolved about top and bottom by $y=d$ and $y=c$
  the $x$ axis, the volume of the is revolved about the $y$ axis, the
  solid generated is volume of the solid generated is
  $V=\int^b_a \pi ([f(x)]^2-[g(x)]^2)\, dx$. $V=\int^d_c \pi ([F(y)]^2-[G(y)]^2)\, dy$.


We could have taken a different approach in the previous example:

Another Method

Look again at the volume of the solid generated by revolving the region enclosed by $y=2x$, $y=x^2$, $0\leq x\leq 2$ about the $y$-axis. This time, we will view the solid as being composed of a collection of concentric cylindrical shells of radius $x$, height $2x-x^2$, and infinitesimal thickness $dx$. The volume of each shell is approximately given by the lateral surface area ($=2\pi\cdot {\small\textrm{radius}}\cdot {\small\textrm{height}}$) multiplied by the thickness: \[2\pi x[2x-x^2]\, dx.\] "Adding up" the volumes of the cylindrical shells, \begin{eqnarray*} V&=& \int^2_0 2\pi x[2x-x^2]\, dx\\ &=& \int^2_0 2\pi [2x^2-x^3]\, dx\\ &=& \left.\left(\frac{4}{3}\pi x^3-\frac{1}{2}\pi x^4\right)\right|^2_0\\ &=& \left(\frac{32}{3}\pi-8\pi\right)-\left(0-0\right)\\ &=& \frac{8\pi}{3},\quad{\small\textrm{as found earlier.}} \end{eqnarray*} This is called the Method of Cylindrical Shells. Suppose $f(x)$, $g(x)$, $F(y)$, $G(y)$ satisfy all the requirements given earlier. Then, for a region revolved about the $y$-axis, \[V=\int_a^b 2\pi xf(x)\, dx \qquad{\small\textrm{or}}\qquad V=\int_a^b 2\pi x[f(x)-g(x)]\, dx.\] For a region revolved about the $x$-axis, \[V=\int_c^d 2\pi yF(y)\, dy \qquad{\small\textrm{or}}\qquad V=\int_c^d 2\pi y[F(y)-G(y)]\, dy.\]

Notes

  • In the disc and washer methods, you integrate with respect to the same variable as the axis about which you revolved the region.

  • In the method of cylindrical shells, you integrate with respect to the other variable.

Computing volumes using these methods takes some practice. With experience, you will be better able to visualize the solids and determine which method to apply.


Key Concepts

Method of Washers: \begin{eqnarray*} V = \int^b_a \pi ([f(x)]^2 - [g(x)]^2)\, dx & \qquad{\small\textrm{or}}\qquad & V = \int^d_c \pi ([F(y)]^2 - [G(y)]^2)\, dy. \\ \end{eqnarray*}

Method of Cylindrical Shells: \begin{eqnarray*} V = \int^b_a 2\pi xf(x)\, dx & \qquad{\small\textrm{or}}\qquad & V = \int^b_a 2\pi x[f(x)-g(x)]\, dx. \\ V = \int^d_c 2\pi yF(y)\, dy & \qquad{\small\textrm{or}}\qquad & V = \int^d_c 2\pi y[F(y)-G(y)]\, dy. \\ \end{eqnarray*}