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Q. 56

Expert-verifiedFound in: Page 1056

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Find the masses of the solids described in Exercises 53–56.

The solid bounded above by the hyperboloid with equation $z={x}^{2}-{y}^{2}$ and bounded below by the square with vertices *(2, 2, −4), (2, −2, −4), (−2, −2, −4),* and *(−2, 2, −4)* if the density at each point is proportional to the distance of the point from the plane with equation* z = −4.*

The mass of the solid is $\frac{260864}{315}k.$

The given equation of hyperboloid is $z={x}^{2}-{y}^{2}$.

To find the mass, let's find the limits:

$-4\le z\le {x}^{2}-{y}^{2}\phantom{\rule{0ex}{0ex}}-2\le x\le 2\phantom{\rule{0ex}{0ex}}-2\le y\le 2$

It is given that the density at each point is proportional to the distance of the point from the plane with equation* z = −4, *so $\rho =k({x}^{2}+{y}^{2}+(z+4{)}^{2}).$

The mass of the solid is ${\iiint}_{V}\rho dxdydz.$

So,

$={\int}_{x=-2}^{2}{\int}_{y=-2}^{2}{\int}_{z=-4}^{{x}^{2}-{y}^{2}}k({x}^{2}+{y}^{2}+(z+4{)}^{2})dxdydz$

Let's integrate with respect to *'z'*

$={k{\int}_{x=-2}^{2}{\int}_{y=-2}^{2}({x}^{2}+{y}^{2})\left(z\right)|}_{z=-4}^{z={x}^{2}-{y}^{2}}dxdy\phantom{\rule{0ex}{0ex}}+{\left(\frac{k}{3}\right){\int}_{x=-2}^{2}{\int}_{y=-2}^{2}(z+4{)}^{3}|}_{z=-4}^{z={x}^{2}-{y}^{2}}dxdy$

Now, let's integrate with respect to *'y'*

$=k{\int}_{x=-2}^{2}{\int}_{y=-2}^{2}({x}^{2}+{y}^{2})({x}^{2}-{y}^{2}+4)dxdy\phantom{\rule{0ex}{0ex}}+\left(\frac{k}{3}\right){\int}_{x=-2}^{2}{\int}_{y=-2}^{2}{({x}^{2}-{y}^{2}+4)}^{3}dxdy$

Now, to find the integral we solve it like ${I}_{1}+{I}_{2}.$

By proceeding with the calculation further,

First, we solve ${I}_{1},$

${I}_{1}=k{\int}_{x=-2}^{2}{\int}_{y=-2}^{2}({x}^{4}+4{x}^{2}+4{y}^{2}-{y}^{4})dxdy\phantom{\rule{0ex}{0ex}}{I}_{1}={k{\int}_{x=-2}^{2}\overline{)\left({x}^{4}y+4{x}^{2}y+\frac{4}{3}{y}^{3}-\frac{{y}^{4}}{4}\right)}}_{y=-2}^{y=2}dx\phantom{\rule{0ex}{0ex}}{I}_{1}={k\overline{)\left(4\frac{{x}^{5}}{5}+16\frac{{x}^{3}}{3}+\frac{64}{3}x\right)}}_{x=-2}^{x=2}\phantom{\rule{0ex}{0ex}}{I}_{1}=k\left(\frac{256}{3}+\frac{256}{3}+\frac{256}{3}\right)\phantom{\rule{0ex}{0ex}}{I}_{1}=256k$

Let's solve ${I}_{2},$

role="math" localid="1650382781558" ${I}_{2}=\left(\frac{k}{3}\right){\int}_{x=-2}^{2}{\int}_{y=-2}^{2}{({x}^{2}-{y}^{2}+4)}^{3}dxdy\phantom{\rule{0ex}{0ex}}{I}_{2}==\left(\frac{k}{3}\right){\int}_{x=-2}^{2}{\int}_{y=-2}^{2}\left(\begin{array}{l}{x}^{6}+12{x}^{4}-3{x}^{4}{y}^{2}+3{x}^{2}{y}^{4}-24{x}^{2}{y}^{2}\\ +48{x}^{2}-{y}^{6}+12{y}^{4}-48{y}^{2}+64\end{array}\right)dxdy\phantom{\rule{0ex}{0ex}}{I}_{2}=\left(\frac{k}{3}\right){\int}_{x=-2}^{2}\left(4{x}^{6}+32{x}^{4}+\frac{512}{5}{x}^{2}+\frac{768}{5}\right)dx\phantom{\rule{0ex}{0ex}}{I}_{2}=\left(\frac{k}{3}\right)\left(\left(\frac{90112}{105}\right)-\left(-\frac{90112}{105}\right)\right)\phantom{\rule{0ex}{0ex}}{I}_{2}=k\left(\frac{180224}{315}\right)$

Now, add the integral ${I}_{1}+{I}_{2},$

$=256k+\left(\frac{180224}{315}\right)k\phantom{\rule{0ex}{0ex}}=\frac{260864}{315}k$

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