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Expert-verified Found in: Page 1056 ### Calculus

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 $\frac{260864}{315}k.$

See the step by step solution

## Step 1. Given Information.

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

## Step 2. Find the mass of the solid.

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 $\rho =k\left({x}^{2}+{y}^{2}+\left(z+4{\right)}^{2}\right).$

## Step 3. Solve.

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

So,

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

Let's integrate with respect to 'z'

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

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

## Step 4. Solve.

By proceeding with the calculation further,

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

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

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

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

## Step 5. Solve.

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

$=256k+\left(\frac{180224}{315}\right)k\phantom{\rule{0ex}{0ex}}=\frac{260864}{315}k$ ### Want to see more solutions like these? 