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Q. 56

Expert-verified

Calculus

Found in: Page 1056

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

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 by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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:

$- 4 \leq z \leq x^{2} - y^{2} - 2 \leq x \leq 2 - 2 \leq y \leq 2$

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 $ρ = k (x^{2} + y^{2} + (z + 4)^{2}) .$

Step 3. Solve.

The mass of the solid is $∭_{V} ρ d x d y d z .$

So,

$= \int_{x = - 2}^{2} \int_{y = - 2}^{2} \int_{z = - 4}^{x^{2} - y^{2}} k (x^{2} + y^{2} + (z + 4)^{2}) d x d y d z$

Let's integrate with respect to 'z'

$= {k \int_{x = - 2}^{2} \int_{y = - 2}^{2} (x^{2} + y^{2}) (z) |}_{z = - 4}^{z = x^{2} - y^{2}} d x d y + {(\frac{k}{3}) \int_{x = - 2}^{2} \int_{y = - 2}^{2} (z + 4)^{3} |}_{z = - 4}^{z = x^{2} - y^{2}} d x d y$

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

$= k \int_{x = - 2}^{2} \int_{y = - 2}^{2} (x^{2} + y^{2}) (x^{2} - y^{2} + 4) d x d y + (\frac{k}{3}) \int_{x = - 2}^{2} \int_{y = - 2}^{2} {(x^{2} - y^{2} + 4)}^{3} d x d y$

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 $I_{1},$

$I_{1} = k \int_{x = - 2}^{2} \int_{y = - 2}^{2} (x^{4} + 4 x^{2} + 4 y^{2} - y^{4}) d x d y I_{1} = {k \int_{x = - 2}^{2} (x^{4} y + 4 x^{2} y + \frac{4}{3} y^{3} - \frac{y^{4}}{4})}_{y = - 2}^{y = 2} d x I_{1} = {k (4 \frac{x^{5}}{5} + 16 \frac{x^{3}}{3} + \frac{64}{3} x)}_{x = - 2}^{x = 2} I_{1} = k (\frac{256}{3} + \frac{256}{3} + \frac{256}{3}) I_{1} = 256 k$

Let's solve $I_{2},$

role="math" localid="1650382781558" $I_{2} = (\frac{k}{3}) \int_{x = - 2}^{2} \int_{y = - 2}^{2} {(x^{2} - y^{2} + 4)}^{3} d x d y I_{2} = = (\frac{k}{3}) \int_{x = - 2}^{2} \int_{y = - 2}^{2} (\begin{array}{l} x^{6} + 12 x^{4} - 3 x^{4} y^{2} + 3 x^{2} y^{4} - 24 x^{2} y^{2} \\ + 48 x^{2} - y^{6} + 12 y^{4} - 48 y^{2} + 64 \end{array}) d x d y I_{2} = (\frac{k}{3}) \int_{x = - 2}^{2} (4 x^{6} + 32 x^{4} + \frac{512}{5} x^{2} + \frac{768}{5}) d x I_{2} = (\frac{k}{3}) ((\frac{90112}{105}) - (- \frac{90112}{105})) I_{2} = k (\frac{180224}{315})$

Step 5. Solve.

Now, add the integral $I_{1} + I_{2},$

$= 256 k + (\frac{180224}{315}) k = \frac{260864}{315} k$

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