Q. 56 - Calculus

Question

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.

Step-by-Step Solution

Answer

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

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:

$$\begin{gathered} -4\le z\le x^2-y^2 \\ -2\le x\le 2 \\ -2\le y\le 2 \end{gathered}$$

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 $\rho =k(x^2+y^2+(z+4{)}^2).$

Step 3. Solve.

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

So,

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

Let's integrate with respect to 'z'

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

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

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

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

Step 4. Solve.

By proceeding with the calculation further, 

First, we solve $I_1,$

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

Let's solve $I_2,$

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

Step 5. Solve.

Now, add the integral $I_1+I_2,$

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