Volume of Revolution

Tags

Calculus

Cegep/2

Word count

587 words

Reading time

4 minutes

Application of the definite integral

For one curve:

Integration perpendicular to the axis of rotation:

Let $r$ be the radius function, then

$$V=\pi {\int }_a^b{r}^2(x)\mathrm{d}x$$

Integration parallel to the axis of rotation:

Let $r$ and $h$ be the radius and height functions, then

$$V=2\pi {\int }_a^{b}r(x)h(x)\mathrm{d}x$$

For two curves:

Let $R$ and $r$ be the outer and inner radius functions, then

$$V=\pi {\int }_a^b(R^2(x)-r^2(x))\mathrm{d}x$$

Examples

Find the formula for the volume of a sphere.

$$\begin{array}{rl}V& =\pi {\int }_{-r}^r{f}^2(x)\mathrm{d}x\\ & =2\pi {\int }_0^r{f}^2(x)\mathrm{d}x\\ & =2\pi {\int }_0^r|r^2-x^2|\mathrm{d}x\\ & =2\pi {|r^3-\frac{x^3}{3}|}_0^r\\ & =2\pi ((r^3-\frac{r^3}{3})-(r^3-\frac{0^3}{3}))\\ & =\end{array}$$

Find the volume of the solid generated by rotating the region bounded by:

$y=x^2$ and $y=3x$. about 1. the x axis, 2. the y axis.

The two lines intersect at $(0,0)$ and $(3,9)$.

x axis:

$$\begin{array}{rl}V& =\pi {\int }_0^3(9x^2-x^4)\mathrm{d}x\\ & =\pi {[3x^3-\frac{x^5}{5}]}_0^3\\ & =\pi (3\cdot 3^3-\frac{3^5}{5})\\ & =32.4\pi \end{array}$$

y axis:

$$\begin{array}{rl}V& =\pi {\int }_0^9(y-\frac{y^2}{9})\mathrm{d}y\\ & =\pi {[\frac{y^2}{2}-\frac{y^3}{27}]}_0^9\\ & =\pi (\frac{9^2}{2}-\frac{9^3}{27})\\ & =13.5\pi \end{array}$$$$\begin{array}{rl}V& =\pi {\int }_0^4((\sqrt{y}+1-6{)}^2-(1-6{)}^2)\mathrm{d}y\\ & =\pi {\int }_0^4(y-10\sqrt{y})\mathrm{d}y\\ & =\pi {[\frac{y^2}{2}-\frac{20y^{\frac{3}{2}}}{3}]}_0^4\\ & =\pi (\frac{4^2}{2}-\frac{20\cdot 4^{\frac{3}{2}}}{3})\\ & =-\frac{136}{3}\pi \end{array}$$

$x=y^2-4$, the x-axis, the y-axis, and $y=1$.

$$\begin{array}{rl}V& =\pi {\int }_0^1{f}^2(y)\mathrm{d}y\\ & =\pi {\int }_0^1(0-y^2+4{)}^2\mathrm{d}y\\ & =\pi {[\frac{y^5}{5}-\frac{8y^3}{3}+16y]}_0^1\\ & =\pi (\frac{1^5}{5}-\frac{8\cdot 1^3}{3}+16\cdot 1)\\ & =\frac{203}{15}\pi \end{array}$$

$y=e^{-x}$, $y=1$, $x=2$ about $y=2$.

$$\begin{array}{rl}V& =\pi {\int }_0^2(f^2(x)-g^2(x))\mathrm{d}x\\ & =\pi {\int }_0^2(1-(e^{-x}{)}^2)\mathrm{d}x\\ & =\pi {[x+\frac{e^{-2x}}{2}]}_0^2\\ & =\pi (2+\frac{e^{-2\cdot 2}}{2}-0-\frac{e^{-2\cdot 0}}{2})\\ & =1.5091\dots \pi \end{array}$$

$x^2+y^2-2x=0$ about the y axis.

$$\begin{array}{rl}V& =2\pi {\int }_a^{b}r(x)h(x)\mathrm{d}x\\ & =2\pi {\int }_0^{2}x(2\sqrt{2x-x^2})\mathrm{d}x\\ & =4\pi {\int }_0^{2}x\sqrt{2x-x^2}\mathrm{d}x\\ & =4\pi {\int }_0^{2}x\sqrt{-(x^2-2x+1)+1}\mathrm{d}x\end{array}$$

Find the volume of a right circular cone with height $h$ and base radius $r$.

$$\begin{array}{rl}V& =2\pi {\int }_0^{r}r(x)h(x)\mathrm{d}x\\ & =2\pi {\int }_0^{r}xh(1-\frac{x}{r})\mathrm{d}x\\ & =2\pi h{\int }_0^r(x-\frac{x^2}{r})\mathrm{d}x\\ & =2\pi h{[\frac{x^2}{2}-\frac{x^3}{3r}]}_0^r\\ & =\frac{\pi h{r}^2}{3}\end{array}$$

Contributors

Ajitani Hifumi

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