91Ó°ÊÓ

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

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

$$\begin{gathered} −4≤z≤x^2−y^2 \\ −2≤x≤2 \\ −2≤y≤2 \end{gathered}$$

It is given that the density at each point is proportionalto the distance of the point from the plane with equationz = −4,so$\mathrm{ÒÏ}=k(x^2+y^2+(z+4{)}^2).$

The mass of the solid is${∭}_V\mathrm{ÒÏ}dxdydz.$

So,

$={∫}_{x=−2}^2{∫}_{y=−2}^2{∫}_{z=−4}^{x^2−y^2}k(x^2+y^2+(z+4{)}^2)dxdydz$

Let's integrate with respect to'z'

$$\begin{gathered} ={k{∫}_{x=−2}^2{∫}_{y=−2}^2(x^2+y^2)\left(z\right)|}_{z=−4}^{z=x^2−y^2}dxdy \\ +{\left(\frac{{\displaystyle k}}{{\displaystyle 3}}\right){∫}_{x=−2}^2{∫}_{y=−2}^2(z+4{)}^3|}_{z=−4}^{z=x^2−y^2}dxdy \end{gathered}$$

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

$$\begin{gathered} =k{∫}_{x=−2}^2{∫}_{y=−2}^2(x^2+y^2)(x^2−y^2+4)dxdy \\ +\left(\frac{{\displaystyle k}}{{\displaystyle 3}}\right){∫}_{x=−2}^2{∫}_{y=−2}^2{(x^2−y^2+4)}^{3}dxdy \end{gathered}$$

Now, to find the integral we solve it like$I_1+I_2.$

By proceeding with the calculation further,

First, we solve $I_1,$

$$\begin{gathered} I_1=k{∫}_{x=−2}^2{∫}_{y=−2}^2(x^4+4x^2+4y^2−y^4)dxdy \\ {I}_1={k{∫}_{x=−2}^2\overline{)\left(x^{4}y+4x^{2}y+\frac{4}{3}{y}^3−\frac{y^4}{4}\right)}}_{y=−2}^{y=2}dx \\ {I}_1={k\overline{)\left(4\frac{x^5}{5}+16\frac{x^3}{3}+\frac{64}{3}x\right)}}_{x=−2}^{x=2} \\ {I}_1=k\left(\frac{256}{3}+\frac{256}{3}+\frac{256}{3}\right) \\ {I}_1=256k \end{gathered}$$

Let's solve $I_2,$

role="math" localid="1650382781558" $$\begin{gathered} I_2=\left(\frac{k}{3}\right){∫}_{x=−2}^2{∫}_{y=−2}^2{(x^2−y^2+4)}^{3}dxdy \\ {I}_2==\left(\frac{k}{3}\right){∫}_{x=−2}^2{∫}_{y=−2}^2\left(\begin{array}{l}{x}^6+12x^4−3x^4{y}^2+3x^2{y}^4−24x^2{y}^2\\ +48x^2−y^6+12y^4−48y^2+64\end{array}\right)dxdy \\ {I}_2=\left(\frac{k}{3}\right){∫}_{x=−2}^2\left(4x^6+32x^4+\frac{512}{5}{x}^2+\frac{768}{5}\right)dx \\ {I}_2=\left(\frac{k}{3}\right)\left(\left(\frac{90112}{105}\right)−\left(−\frac{90112}{105}\right)\right) \\ {I}_2=k\left(\frac{180224}{315}\right) \end{gathered}$$

Now, add the integral $I_1+I_2,$

$$\begin{gathered} =256k+\left(\frac{180224}{315}\right)k \\ =\frac{260864}{315}k \end{gathered}$$