Examples

Revolution of a triangle

A rotation-symmetric 3D body has cross section given by the region between $y=3x+2$, $y=6-x$, $x=0$, and is rotated around the $y$-axis. Find the volume of this 3D body.

Solution

(1) Cross-section region:

Bounded above-right by $y=6-x$. Bounded below-right by $y=3x+2$. These intersect at $x=1$.

Bounded left by $x=0$.

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(2) Set up integral:

Rotated around $y$-axis, therefore use $x$ for integration variable (shells!). Formula:

$$V={\int }_a^{b}2\pi Rhdr$$

Domain is $[a,b]=[\mathrm{0,1}]$.

$R(x)=x$ because shell radius is the $x$-distance from $x=0$ to the shell position.

Height:

$$\begin{array}{c}h(x)\gg \gg (6-x)-(3x+2)\\ \\ \gg \gg 4-4x\end{array}$$

$dr$ is limit of $\mathrm{\Delta }r$ which equals $\mathrm{\Delta }x$ here, so $dr=dx$.

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(3) Evaluate integral:

$$\begin{array}{c}V={\int }_a^{b}2\pi Rhdr\gg \gg {\int }_0^{1}2\pi \cdot x(4-4x)dx\\ \\ \gg \gg {2\pi (2x^2-\frac{4x^3}{3})|}_0^1\gg \gg \frac{4\pi }{3}\end{array}$$

Revolution of a sinusoid

Consider the region given by revolving the first hump of $y=\sin (x)$ about the $y$-axis. Set up an integral that gives the volume of this region using the method of shells.

Solution

(1) Set up the integral for shells:

Integration variable: $r=x$, the distance of a shell to the $y$-axis.

Then $dr=dx$ and $h=\sin x$, the height of a shell.

Bounds: one hump is given by $x\in [0,\pi ]$. Thus:

$$V={\int }_0^{\pi }2\pi x\sin xdx$$

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(2) Perform the integral using IBP:

Choose $u=2\pi x$ and $v^{\prime }=\sin x$ since $x$ is A and $\sin x$ is T.

Then $u^{\prime }=2\pi$ and $v=-\cos x$.

Use IBP formula:

$$\begin{array}{c}\int u{v}^{\prime }dx=uv-\int u^{\prime }vdx\\ \\ \gg \gg {\int }_0^{\pi }2\pi x\sin xdx\gg \gg (2\pi x)(-\cos x){|}_0^{\pi }-{\int }_0^{\pi }2\pi (-\cos x)dx\end{array}$$

Compute first term:

$$\begin{array}{c}\gg \gg -2\pi x\cos x{|}_0^{\pi }\\ \\ \gg \gg -2\pi (\pi )(-1)--2\pi (0)(+1)\gg \gg 2{\pi }^2\end{array}$$

Compute integral term:

$$-{\int }_0^{\pi }2\pi (-\cos x)dx\gg \gg 2\pi \sin {|}_0^{\pi }\gg \gg 0$$

So the answer is $2{\pi }^2$.