Volume of a soldid of revolution

Differentials and integrals: Applications of integration

Volume of a soldid of revolution

We consider the graph of a function $y=f(x)$ that is continuous and nonnegative on an interval $[a,b]$ . The region bounded by the graph of $f$ , the $x$ -axis, and the lines $x=a$ and $x=b$ can now be revolved about the $x$ -axis and in this way a solid of revolution $S$ is created. You can just as well rotate the graph of $f$ about the $x$ -axis and bound it by means of the vertical planes through the points $(a,0,0)$ and $(b,0,0)$ .

In the left figure below, only a quarter of $S$ is sketched (with $x\ge0,y\ge 0$ and $z\ge 0$ . In the right figure below, the solid of revolution is sketched that you get by rotating the sine graph on the interval $[0,2\pi]$ about the $x$ axis. An interactive version showing the solid of revolution of the function $f(x)=x^2-3x$ on the interval $[1,3]$ is also available. You can show and rotate the graph of the function, adjust the interval by moving the points $A$ and $B$ , and change the 3D-viewpoint with the right mouse button.

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GeoGebra

Choose a number $x$ on the interval $[a,b]$ and let $V(x)$ be the volume of the portion of the solid of revolution to the left of the vertical plane through $(x,0,0)$ . Then $V(a)=0$ and $V(b)=\textrm{volume of the solid of revolution }S$ . For a small increment $\dd x$ the increment $V(x+\dd x)-V(x)$ is equal to the volume of the thin slice of the solid of revolution between the vertical planes through the points $(x,0,0)$ and $(x+\dd x,0,0)$ . For a very small (mathematicians say 'infinitesimally small') increment $\dd x$ this is almost equal to the volume of a cylindrical disk with radius $f(x)$ and thickness $\dd x$ . The volume of this cylindrical disc is equal to the product of the area of the circle with radius $f(x)$ and the thickness $\dd x$ , i.e., it is equal to $\pi\cdot \bigl(f(x)\bigr)^2\,\dd x$ . The expression $\pi\cdot \bigl(f(x)\bigr)^2\,\dd x$ is a differential and equal to the $\dd V$ differential of the volume function. So: $\dd V=\pi \cdot \bigl(f(x)\bigr)^2\,\dd x$ .

The requested volume of the solid of revolution $S$ that is obtained by revolving the area bounded by the graph of the function $f$ , the $x$ -axis and the lines $x=a$ and $x=b$ around the $x$ -axis, is equal to $\begin{aligned}\text{volume of }S&=V(b)\\[0.25cm]&=\int_a^b \pi\cdot\bigl(f(x)\bigr)^2\,\dd x\\[0.25cm]&=\pi\int_a^b\bigl(f(x)\bigr)^2\,\dd x\end{aligned}$

What is the exact volume $V(S)$ of the solid of revolution $S$ obtained by revolving the region under graph of the function $f(x)=x^2-3x$ on the interval $[1,3]$ about the $x$ -axis?

$V(S)={}$ $\frac{32}{5}\pi$

$\begin{aligned}V(S) &=\pi \int_1^3\bigl(f(x)\bigr)^2\,\dd x\\[0.25cm] &=\pi\int_{1}^{3}(x^2-3x)^2\,\dd x\\[0.25cm] &=\pi\int_1^3(x^4-6x^3+9x^2)\,\dd x\\[0.25cm] &=\pi\cdot\Bigl[\frac{1}{5}x^5-\frac{3}{2}x^4+3x^3\Bigr]_1^3\\[0.25cm] &=\pi\cdot\Bigl(\bigl(\frac{3^5}{5}-\frac{3^5}{2}+3^4\bigr)-\bigl(\frac{1}{5}-\frac{3}{2}+3\bigr)\Bigr)\\[0.25cm] &=\pi\cdot\bigl(\frac{81}{10}-\frac{17}{10}\bigr)\\[0.25cm] &=\frac{32}{5}\pi\end{aligned}$

New example

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Instead of rotation about the $x$ -axis, you can also rotate part of a graph of a function around the $y$ -axis. Also in this case, the volume of the solid of revolution can be calculated. We just need to swap the roles of $x$ and $y$ . So we should describe the curve as $x=f(y)$ instead of $y=f(x)$ . and the limits must be given in terms of $y$ , as in the figure below from $y=c$ to $y=d$ .

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The requested volume of the solid of revolution is then equal to $\pi\int_c^d f(y)^2\,\dd y$ .

The part of the parabola $y=x^2+1$ defined over the positive $x$ -axis can also be described as $x=\sqrt{y-1}$ . We can rotate the planar region $R$ bounded by the parabola, the $y$ -axis and the horizontal lines $y=2$ and $y=5$ about the $y$ -axis. See the figure below.

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Calculate the volume $V(S)$ of the so-obtained solid of revolution $S$ .

$V(S)={}$ $\frac{17}{2}\pi$

$\begin{aligned}I(L)&=\pi \int_{2}^{5}\bigl(x(y)\bigr)^2\,\dd y\\[0.25cm] &=\pi \int_{^2}^{5} \bigl(\sqrt{y-1}\bigr)^2\,\dd y\\[0.25cm] &=\pi \int_{2}^{5} (y-1)\,\dd y\\[0.25cm] &=\pi \int_{1}^{4}u\,\dd u& \blue{u=y-1, \dd u=\dd y}\\[0.25cm] &=\pi \cdot\Bigl[\frac{1}{2}u^2\Bigr]_{1}^{4}\\[0.25cm] &=\pi\cdot\Bigl(\frac{4^2}{2}-\frac{1^2}{2}\Bigr)\\[0.25cm]&=\frac{17}{2}\pi\end{aligned}$

New example

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We return to a region under the graph of a continuous nonnegative function $f$ . Instead of revolving the region $R$ under the graph about a $x$ -axis, you can also rotate such a region around the $y$ -axis.

Choose a number $x$ in the interval $[a,b]$ and let $V(x)$ be the volume of the portion of the solid of revolution to the left of the vertical plane through $(x,0,0)$ obtained by revolving about the $y$ -axis. Then $V(a)=0$ and $V(b)=\textrm{volume of the solid of revolution }S$ . For a small increment $\dd x$ the increment $V(x+\dd x)-V(x)$ is equal to the volume of the thin slice of the solid of revolution between the vertical planes through the points $(x,0,0)$ and $(x+\dd x,0,0)$ . For very small (mathematicians say 'infinitesimally small') increment $\dd x$ this is almost equal to the volume of a cylindrical shell with radius $x$ , thickness $\dd x$ and height $f(x)$ . The volume of this cylindrical shell is equal to the product of the circumference of a circle with radius $x)$ , the height $f(x)$ and the thickness $\dd x$ , i.e., it equals $2\pi x\, f(x)\,\dd x$ . This expression is a differential and equal to the $\dd V$ differential of the volume function. So: $\dd V=2\pi x\, f(x)\,\dd x$ .

The requested volume of the solid of revolution $S$ , which is obtained by revolving the region bounded by the graph of the function $f$ , the $x$ -axis and the lines $x=a$ and $x=b$ about the $y$ -axis, is equal to $\begin{aligned}\text{volume of }S &=V(b)\\[0.25cm]&=2\pi \int_a^b x\,f(x)\,\dd x\end{aligned}$

Calculate the volume of a torus obtained by rotating a circular disc with radius $r$ and center $(c,0)$ with $c>r>0$ about the $y$ -axis.

The torus with $r=1$ and $c=3$ is drawn below.

torus

$\text{volume of the torus }={}$ $2\pi^2r^2c$

The equation of a circle with radius $r$ and center $(c,0)$ is given by the equation $(x-c)^2+y^2=r^2$ The semicircle above the positive $x$ -axis is then the graph of the function $f(x)=\sqrt{r^2-(x-c)^2}$ with $c-r\le x \le c+r$ To calculate the volume of the torus, we double the volume of the solid of revolution that is obtained by revolving the area under the graph of the function $f$ about the $y$ -axis. $\begin{aligned}\text{volume of the torus } &=2\cdot 2\pi \int_{c-r}^{c+r}x\,f(x)\,\dd x\\[0.25cm] &=4\pi \int_{c-r}^{c+r}x\,\sqrt{r^2-(x-c)^2}\,\dd x\\[0.25cm] &=4\pi \int_{-r}^{r}(u+c)\sqrt{r^2-u^2}\,\dd u\\ &\phantom{abcdwxyzabcdwxyz} \blue{u=x-c, \dd u=\dd x}\\[0.25cm] &=4\pi \int_{-r}^{r}u\,\sqrt{r^2-u^2}\,\dd u + 4\pi \,c\int_{-r}^{r}\sqrt{r^2-u^2}\,\dd u\\[0.25cm] &=4\pi \,c\int_{-r}^{r}\sqrt{r^2-u^2}\,\dd u \\ &\phantom{abcdwxyzabcdwxyz} \blue{\text{because }u\,\sqrt{r^2-u^2}\text{ is an odd funtion}}\\[0.25cm] &=4\pi\,c \int_{\pi}^{0}-r^2\sin^2z\,\dd z\\ &\phantom{abcdwxyzabcdwxyz} \blue{u=r\cos{z}, \dd u=-r\sin{z}\, \dd z}\\[0.25cm] &=4\pi\,c\,r^2 \int_{0}^{\pi}\sin^2z\,\dd z\\[0.25cm] &=4\pi\,c\,r^2\cdot\frac{1}{2}\left( \int_{0}^{\pi}\sin^2z\,\dd z+\int_{0}^{\pi}\cos^2z\,\dd z\right) \\[0.25cm] &=2\pi\,c\,r^2 \int_{0}^{\pi}\bigl(\sin^2z+\cos^2z\bigr)\,\dd z \\[0.25cm] &=2\pi\,c\,r^2\int_{0}^{\pi}1\,\dd z\\[0.25cm] &=2\pi\,c\,r^2\,\pi \\[0.25cm] &=2\pi^2r^2c\end{aligned}$

New example

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Finally, we give the formulas for the volume of a solid of revolution by rotating a region under a parametric curve about the $x$ - or $y$ -axis.

When you revolve a parametric curve $\bigl(x(t),y(t)\bigr)$ on the interval $[a,b]$ about the $x$ - or $y$ -axis, then the volume $V$ of the solid of revolution obtained by revolving the region under the curve or to the left of the curve is given by the formulas below.

Rotation about the $x$ -axis

$V=\pi\int_a^b y(t)^2\cdot\frac{\dd x}{\dd t}\,\dd t$

Rotation about the $y$ -axis

$V=\pi\int_a^b x(t)^2\cdot\frac{\dd y}{\dd t}\,\dd t$

These formulas follow from the previous formulas for the volumes of solids of revolution through the following relations of differentials: $\dd x = \frac{\dd x}{\dd t}\,\dd t\quad\text{and}\quad \dd y = \frac{\dd y}{\dd t}\,\dd t$