MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 6 September 2014

Lecture 28

Find the volume generated by the area bounded by $y=x^2-2x$ and $y=0$ when it is rotated about the $x\text{-axis}\text{.}$ $$\begin{array}{c} 2 x y \end{array}\begin{array}{c} 2 x y \end{array}$$ $$\begin{array}{c}\text{Slice:}\begin{array}{c} R \end{array}\\ \text{Volume of a slice:}\hspace{0.17em}\pi R^{2}dx\\ \text{Add slices from}\hspace{0.17em}x=0\hspace{0.17em}\text{to}\hspace{0.17em}x=2\text{.}\end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_{x=0}^{x=2}\pi R^{2}dx}& =& {\displaystyle {\int }_{x=0}^{x=2}\pi {(-y)}^{2}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=2}\pi y^{2}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=2}\pi {(x^2-2x)}^{2}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=2}\pi (x^4-4x^3+4x^2)dx}\\ & =& {\displaystyle \pi (\frac{x^5}{5}-\frac{4x^4}{4}+\frac{4x^3}{3}){|}_{x=0}^{x=2}}\\ & =& {\displaystyle \pi (\frac{2^5}{5}-2^4+\frac{4}{3}{2}^3)-\pi (0-0+0)}\\ & =& {\displaystyle 2^3\pi (\frac{2^2}{5}-2+\frac{4}{3})}\\ & =& {\displaystyle 8\pi (\frac{4}{5}-2+\frac{4}{3})}\\ & =& {\displaystyle 8\pi (-\frac{6}{5}+\frac{4}{3})}\\ & =& {\displaystyle 8\pi (-\frac{18}{15}+\frac{20}{15})}\\ & =& {\displaystyle \frac{8\pi ·2}{15}}\\ & =& {\displaystyle \frac{16\pi }{15}\text{.}}\end{array}$$

The base of a solid is $x^2+y^2=a^2\text{.}$ Each plane section, perpendicular to the $x\text{-axis,}$ is a square, with one edge of the square in the base of the solid. Find the volume. $$\begin{array}{c} x y a a -a -a x2+y2=a2 \end{array}\begin{array}{c}\text{Slice:}\begin{array}{c} ⏟ S ⏟ dx \end{array}\end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_{x=-a}^{x=a}{S}^{2}dx}& =& {\displaystyle {\int }_{x=-a}^{x=a}{\left(2y\right)}^{2}dx}\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}4y^{2}dx}\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}4(a^2-x^2)dx}\\ & =& {\displaystyle 4(a^{2}x-\frac{x^3}{3}){|}_{x=-a}^{x=a}}\\ & =& {\displaystyle 4(a^2·a-\frac{a^3}{3})-4(a^2(-a)-\frac{{(-a)}^3}{3})}\\ & =& {\displaystyle 4(a^3-\frac{1}{3}{a}^3)-4(-a^3+\frac{1}{3}{a}^3)}\\ & =& {\displaystyle 4·\frac{2}{3}{a}^3-4(-\frac{2}{3}{a}^3)}\\ & =& {\displaystyle 4·\frac{4}{3}{a}^3}\\ & =& {\displaystyle \frac{16a^3}{3}\text{.}}\end{array}$$

Find the volume generated when the area bounded by $y=\sqrt{x},$ $y=2$ and $x=0$ is rotated about the line $y=2\text{.}$ $$\begin{array}{c} 1 4 x y y=2 y=x \end{array}\begin{array}{c}\text{Slice:}\begin{array}{c} R \end{array}\\ \text{Volume of a slice:}\hspace{0.17em}\pi R^{2}dx\\ \text{Add slices from}\hspace{0.17em}x=0\hspace{0.17em}\text{to}\hspace{0.17em}x=4\text{.}\end{array}$$ Add slices from $x=0$ to $x=4\text{.}$ $$\begin{array}{ccc}{\displaystyle {\int }_{x=0}^{x=4}\pi R^{2}dx}& =& {\displaystyle {\int }_{x=0}^{x=4}\pi {(2-y)}^{2}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=4}\pi {(2-\sqrt{x})}^{2}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=4}\pi (4-4\sqrt{x}+x)dx}\\ & =& {\displaystyle \pi (4x-\frac{2·4}{3}{x}^{\frac{3}{2}}+\frac{x^2}{2}){|}_{x=0}^{x=4}}\\ & =& {\displaystyle \pi (4·4-\frac{2}{3}·4·4^{\frac{3}{2}}+\frac{4^2}{2})-\pi (0-0+0)}\\ & =& {\displaystyle \pi (16-\frac{8}{3}·8+\frac{16}{2})}\\ & =& {\displaystyle 8\pi (2-\frac{8}{3}+1)}\\ & =& {\displaystyle 8\pi ·\frac{1}{3}}\\ & =& {\displaystyle \frac{8\pi }{3}\text{.}}\end{array}$$

Find the volume generated when the area bounded by $y=\text{sin}\hspace{0.17em}x,$ $0\le x\le \pi ,$ and $y=0$ is rotated about the $y\text{-axis.}$ $$\begin{array}{c} π 2 π x 1 y y=sin x \end{array}\begin{array}{c}\text{Slice:}\begin{array}{c} } dy Ri R0 \end{array}\\ \text{Volume of slice:}\hspace{0.17em}(\pi R_0^2-\pi R_i^2)dy\\ \text{Add slices from}\hspace{0.17em}y=0\hspace{0.17em}\text{to}\hspace{0.17em}y=1\text{.}\end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_{y=0}^{y=1}(\pi R_0^2-\pi R_i^2)dy}& =& {\displaystyle {\int }_{y=0}^{y=1}\pi (x_{\text{right}}^2-x_{\text{left}}^2)dy}\\ & =& {\displaystyle {\int }_{y=0}^{y=1}\pi ({(\pi -x_{\text{left}})}^2-x_{\text{left}}^2)dy}\\ & =& {\displaystyle {\int }_{y=0}^{y=1}\pi ({\pi }^2-2\pi x+x^2-x^2)dy}\\ & =& {\displaystyle {\int }_{y=0}^{y=1}\pi ({\pi }^2-2\pi x)\frac{dy}{dx}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=\frac{\pi }{2}}\pi ({\pi }^2-2\pi x)\text{cos}\hspace{0.17em}x\hspace{0.17em}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=\frac{\pi }{2}}({\pi }^3\text{cos}\hspace{0.17em}x-2{\pi }^{2}x\text{cos}\hspace{0.17em}x)dx}\\ & =& {\displaystyle {\pi }^3\text{sin}\hspace{0.17em}x-2{\pi }^2(x\text{sin}\hspace{0.17em}x+\text{cos}\hspace{0.17em}x){|}_{x=0}^{x=\frac{\pi }{2}}}\\ & =& {\displaystyle {\pi }^3\text{sin}\hspace{0.17em}\frac{\pi }{2}-2{\pi }^2(\frac{\pi }{2}\text{sin}\hspace{0.17em}\frac{\pi }{2}+\text{cos}\hspace{0.17em}\frac{\pi }{2})-({\pi }^3\text{sin}\hspace{0.17em}0-2{\pi }^2(0+\text{cos}\hspace{0.17em}0))}\\ & =& {\displaystyle {\pi }^3-\frac{2{\pi }^2·\pi }{2}+2{\pi }^2}\\ & =& {\displaystyle 2{\pi }^2\text{.}}\end{array}$$

Find the volume of a bagel produced by rotating the circle $x^2+y^2=a^2$ about the line $y=b\text{.}$ $$\begin{array}{c} - a a x a y y=b x2+y2=a2 \end{array}\begin{array}{c}\text{Slice:}\begin{array}{c} } H R dx ⏟ \end{array}\\ \text{Volume of slice:}\hspace{0.17em}2\pi RHdx\\ \text{Add slices from}\hspace{0.17em}x=a\hspace{0.17em}\text{to}\hspace{0.17em}x=a\text{.}\end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_{x=-a}^{x=a}2\pi RHdx}& =& {\displaystyle {\int }_{x=-a}^{x=a}2\pi (b-x)2ydx}\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}2\pi (b-x)2\sqrt{a^2-x^2}dx}\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}(4\pi b\sqrt{a^2-x^2}-4\pi x\sqrt{a^2-x^2})dx}\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}4\pi b\sqrt{a^2-x^2}dx-\frac{4\pi }{-2}{\int }_{x=-a}^{x=a}-2x\sqrt{a^2-x^2}dx}\\ & =& {\displaystyle 4\pi b\left(\text{area of a semicircle of radius}\hspace{0.17em}a\right)+2\pi \left({(a^2-x^2)}^{\frac{3}{2}}\right){|}_{x=-a}^{x=a}}\\ & =& {\displaystyle 4\pi b\frac{\pi a^2}{2}+2\pi \left(0^{\frac{3}{2}}\right)-2\pi \left(0^{\frac{3}{2}}\right)}\\ & =& {\displaystyle \frac{4{\pi }^2{a}^{2}b}{2}}\\ & =& {\displaystyle 2{\pi }^2{a}^{2}b\text{.}}\end{array}$$

Notes and References

These are a typed copy of Lecture 28 from a series of handwritten lecture notes for the class MATH 221 given on November 13, 2000.

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