Q. 54 - Calculus

Question

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

The solid bounded above by the plane with equation 2x + 3y − z = 2 and bounded below by the triangle with vertices (1, 0, 0), (4, 0, 0), and (0, 2, 0) if the density at each point is proportional to the distance of the point from the

xy-plane.

Step-by-Step Solution

Answer

The mass of the solid is $19k.$

Step 1. Given Information.

The given equation of the plane is $2x+3y-z=2.$

Step 2. Find the mass of the solid.

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

$$\begin{gathered} 0\le z\le 2x+3y-2 \\ \frac{2-y}{2}\le x\le 4-2y \\ 0\le y\le 2 \end{gathered}$$

It is given that the density at each point is proportional to the distance of the point from the xy-plane, so $\rho =kz.$

Step 3. Solve.

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

So,

$={\int }_{y=0}^2{\int }_{x=\frac{\left(2-y\right)}{2}}^{4-2y}{\int }_{z=0}^{2x+3y-2}kz dxdydz.$

Let's integrate with respect to 'z'

$$\begin{gathered} ={k{\int }_{y=0}^2{\int }_{x=\frac{(2-y)}{2}}^{4-2y}\overline{)\left(\frac{z^2}{2}\right)}}_{z=0}^{2x+3y-2}dxdy \\ =\left(\frac{k}{2}\right){\int }_{y=0}^2{\int }_{x=\frac{(2-y)}{2}}^{4-2y}(2x+3y-2{)}^{2}dxdy \end{gathered}$$

Step 4. Solve.

By proceeding with the calculation further,

$\begin{array}{rl}=& \left(\frac{k}{2}\right){\int }_{y=0}^2{\int }_{x=\frac{(2-y)}{2}}^{4-2y}4x^2+(3y-2{)}^2+4x(3y-2)dxdy\\ =& \left(\frac{k}{2}\right){\int }_{y=0}^2{\int }_{x=\frac{(2-y)}{2}}^{2(2-y)}4x^2+9y^2+4-12y+12xy-8xdxdy\\ =& \left(\frac{k}{2}\right){\int }_{y=0}^2{\overline{)\left(\frac{4x^3}{3}+9y^{2}x+4x-12xy+6x^{2}y-4x^2\right)}}_{x=\frac{(2-y)}{2}}^{2(2-y)}dy\\ =& \left(\frac{k}{2}\right){\int }_{y=0}^2\left(-\frac{14}{3}{y}^3+12y^2-24y+\frac{112}{3}\right)-\left(-\frac{19}{6}{y}^3+9y^2-6y+\frac{4}{3}\right)dy\end{array}$

Now,

$\begin{array}{rl}=& \left(\frac{k}{2}\right){\int }_{y=0}^2\left(-\frac{3}{2}{y}^3+3y^2-18y+36\right)dy\\ =& {\left(\frac{k}{2}\right)\overline{)\left(-\frac{3}{2}\frac{y^4}{4}+y^3-9y^2+36y\right)}|}_{y=0}^2\\ =& \left(\frac{k}{2}\right)\left(-\frac{3}{2}\times \frac{2^4}{4}+2^3-9(2{)}^2+36(2)\right)\\ =& \left(\frac{k}{2}\right)(-6+8-36+72)\\ =& 19k\end{array}$