Q. 55 - Calculus

Question

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

The solid bounded above by the paraboloid with equation $z=8-x^2-y^2$ and bounded below by the rectangle $\mathfrak{R}=\left\{\right(x,y,0\left)\right|1\le x\le 2 \mathrm{and} 0\le y\le 2\}$in the xy-plane if the density at each point is proportional to the square of the distance of the point from the origin.

Step-by-Step Solution

Answer

The mass of the solid is $\frac{31412k}{315}.$

Step 1. Given Information.

The given equation of paraboloid is $z=8-x^2-y^2.$

Step 2. Find the mass of the solid.

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

$$\begin{gathered} 0\le z\le 8-x^2-y^2 \\ 1\le x\le 2 \\ 0\le y\le 2 \end{gathered}$$

It is given that the density at each point is proportional to the square of the  distance of the point from the origin, so $\rho =k(x^2+y^2+z^2).$

Step 3. Solve.

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

So,

$={\int }_{x=1}^2{\int }_{y=0}^2{\int }_{z=0}^{8-x^2-y^2}k(x^2+y^2+z^2)dxdydz.$

Let's integrate with respect to 'z'

$\begin{array}{rl}=k& {{\int }_{x=1}^2{\int }_{y=0}^2(x^2+y^2)\overline{)\left(z\right)}}_{z=0}^{8-x^2-y^2}dxdy+{\left(\frac{k}{3}\right){\int }_{x=1}^2{\int }_{y=0}^2\left(z^3\right)|}_{z=0}^{8-x^2-y^2}dxdy\\ =k& {\int }_{x=1}^2{\int }_{y=0}^2(x^2+y^2)(8-x^2-y^2)dxdy+\left(\frac{k}{3}\right){\int }_{x=1}^2{\int }_{y=0}^2{(8-x^2-y^2)}^{3}dxdy\end{array}$

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 $I_1,$

$$\begin{gathered} I_1=k{\int }_{x=1}^2{\int }_{y=0}^2(x^2+y^2)(8-x^2-y^2)dxdy \\ {I}_1=k{\int }_{x=1}^2{\int }_{y=0}^2(-x^4-2x^2{y}^2+8x^2-y^4+8y^2)dxdy \end{gathered}$$

Integrate with respect to 'y'

$$\begin{gathered} ={k{\int }_{x=1}^2\overline{)\left[-x^{4}y-2x^2\left(\frac{y^3}{3}\right)+8x^{2}y-\left(\frac{y^5}{5}\right)+8\left(\frac{y^3}{3}\right)\right]}}_{y=0}^{2}dx \\ =k{\int }_{x=1}^2-2x^4-x^2\left(\frac{16}{3}\right)+16x^2-\left(\frac{32}{5}\right)+\left(\frac{64}{3}\right)dx \end{gathered}$$

Integrate with respect to 'x'

$$\begin{gathered} ={k\overline{)\left(-2\frac{x^5}{5}-\frac{16}{3}\times \left(\frac{x^3}{3}\right)+16\left(\frac{x^3}{3}\right)-\left(\frac{32}{5}\right)x+\left(\frac{64}{3}\right)x\right)}}_{x=1}^2 \\ =k\left(\left(\frac{2048}{45}\right)-\left(\frac{814}{45}\right)\right) \\ =k\left(\frac{1234}{45}\right) \end{gathered}$$

Step 5. Solve.

Let's solve $I_2,$

$$\begin{gathered} I_2=\left(\frac{k}{3}\right){\int }_{x=1}^2{\int }_{y=0}^2{(8-x^2-y^2)}^{3}dxdy \\ {I}_2=\left(\frac{k}{3}\right){\int }_{x=1}^2{\int }_{y=0}^2(-x^6-3x^4{y}^2+24x^4-3x^2{y}^4+ \\ 48x^2{y}^2-192x^2-y^6+24y^4-192y^2+512)dxdy \end{gathered}$$

Integrate with respect to 'y'

$$\begin{gathered} =\left(\frac{k}{3}\right){\int }_{x=1}^2(-x^{6}y-x^4{y}^3+24x^{4}y-3x^2\left(\frac{y^5}{5}\right)+ \\ 16x^2{y}^3-192x^{2}y-\left(\frac{y^7}{7}\right)+24\left(\frac{y^5}{5}\right)-64y^3{\overline{)+512y)}}_{y=0}^{y=2} \\ =\left(\frac{k}{3}\right)\left({\int }_{x=1}^2-2x^6+40x^4-x^2\left(\frac{1376}{5}\right)+\left(\frac{22656}{35}\right)\right)dx \\ =\left(\frac{k}{3}\right)\left(\left(\frac{16384}{21}\right)-\left(\frac{59146}{105}\right)\right) \\ =\frac{22774k}{315} \end{gathered}$$

Step 6. Solve.

Now, add the integral $I_1+I_2,$

$\begin{array}{rl}=& \left(\frac{1234k}{45}\right)+\left(\frac{22774k}{315}\right)\\ =& \frac{31412k}{315}\end{array}$