Solutions - 0030-03

(a)

(1) Write down shells formula:

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

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(2) Define the cross section region:

Bounded above by $y=8-6x^2$. Bounded below by $y=2x^2$.

Bounded left by $x=0$. Bounded right by intersection at line $x=1$.

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(3) Define $R$ and $h$ and $dr$:

$$\begin{array}{cc}R(x)=x,h& =(8-6x^2)-2x^2,dr=dx\\ & \gg \gg h=8-8x^2\end{array}$$

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(4) Plug into shells formula and compute:

$$\begin{array}{cc}V& \gg \gg {\int }_a^{b}2\pi Rhdr\\ & \gg \gg {\int }_0^{1}2\pi (x)(8-8x^2)dx\\ & \gg \gg 16\pi {\int }_0^{1}x-x^{3}dx\\ & \gg \gg 16\pi {[\frac{x^2}{2}-\frac{x^4}{4}]|}_0^1\\ & \gg \gg 16\pi [\frac{1}{2}-\frac{1}{4}]\\ & \gg \gg 4\pi \end{array}$$

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(b)

(1) Write down washers formula using $dy$:

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

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(2) Rewrite bounding equations in terms of $y$:

$$\begin{array}{cc}y& =8-6x^2\gg \gg x=\sqrt{\frac{8-y}{6}}\\ & \\ y& =2x^2\gg \gg x=\sqrt{\frac{y}{2}}\end{array}$$

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(3) Determine region boundary data:

Bounded above by $y=8$. Bounded below by $y=0$. Intersection at $y=2$.

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(4) Determine $R$ in two components, with $y=2$ the dividing line:

$$R=\sqrt{\frac{y}{2}}0\le y\le 2$$ $$R=\sqrt{\frac{8-y}{6}}2\le y\le 8$$

Note that $r=0$ for both regions. These are disks.

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(5) Compute the integral:

$$\begin{array}{cc}V& \gg \gg {\int }_a^b\pi (R^2-r^2)dy\\ & \\ & \gg \gg \pi [{\int }_0^2{\left(\sqrt{\frac{y}{2}}\right)}^{2}dy+{\int }_2^8{\left(\sqrt{\frac{8-y}{6}}\right)}^{2}dy]\\ & \\ & \gg \gg \pi [{\int }_0^2\frac{y}{2}dy+{\int }_2^8\frac{8-y}{6}dy]\\ & \\ & \gg \gg \pi [{\left[\frac{y^2}{4}\right]|}_0^2+{[\frac{4}{3}y-\frac{y^2}{12}]|}_2^8]\\ & \\ & \gg \gg \pi [1+[(\frac{32}{3}-\frac{16}{3})-(\frac{8}{3}-\frac{1}{3})]]\\ & \\ & \gg \gg 4\pi \end{array}$$

(6) Why are shells preferable?

1. Only need one integral.
2. Don’t need to rewrite boundary equations in terms of $y$.