Q. 53 - Calculus

Question

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

The first-octant solid bounded by the coordinate planes and the plane 3x + 4y + 6z = 12 if the density at each point is proportional to the distance of the point from the xz-plane.

Step-by-Step Solution

Answer

The mass of the solid is $3k.$

Step 1. Given Information.

The given plane is $3x+4y+6z=12.$

Step 2. Find the mass of the solid.

The given plane is $3x+4y+6z=12$ we can also write it as $\frac{x}{4}+\frac{y}{3}+\frac{z}{2}=1.$

Now, let's find the limits in the first octant, so the limits of z is $z=0 \text{and }z=2\left(1-\frac{x}{4}-\frac{y}{3}\right).$

The limits of y in the xy plane is $y=0 \text{and }y=3\left(1-\frac{x}{4}\right)$ and the limits of x is $x=0 \text{and }x=4.$

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

Step 3. Solve.

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

So,

$={\int }_{x=0}^4{\int }_{y=0}^{3\left(1-\frac{x}{4}\right)}{\int }_{z=0}^{2\left(1-\frac{x}{4}-\frac{y}{3}\right)}ky dxdydz.$

Let's integrate with respect to 'z'

$\begin{array}{rl}=& k{\int }_{x=0}^4{\int }_{y=0}^{3(1-\frac{x}{4})}\left(z\right){|}_{z=0}^{z=2\left(1-\frac{x}{4}-\frac{y}{3}\right)}ydxdy\\ =& k{\int }_{x=0}^4{\int }_{y=0}^{3(1-\frac{x}{4})}2(1-\frac{x}{4})y-2\left(\frac{y}{3}\right)ydxdy\end{array}$

Step 4. Solve.

Now, integrate with respect to 'y'

$$\begin{gathered} =k{\int }_{x=0}^4{\int }_{y=0}^{3\left(1-\frac{x}{4}\right)}2\left(1-\frac{x}{4}\right)ydxdy-k\left(\frac{2}{3}\right){\int }_{x=0}^4{\int }_{y=0}^{3\left(1-\frac{x}{4}\right)}{y}^{2}dxdy \\ =k{\int }_{x=0}^4\left(1-\frac{x}{4}\right){\overline{){\left(y^2\right)}_{y=0}^{3\left(1-\frac{x}{4}\right)}}dx-k\left(\frac{2}{3}\right){\int }_{x=0}^4\overline{)\left(\frac{y^3}{3}\right)}}_{y=0}^{3\left(1-\frac{x}{4}\right)}dx \\ =9k{\int }_{x=0}^4{\left(1-\frac{x}{4}\right)}^{3}dx-k\left(\frac{2\times 27}{9}\right){\int }_{x=0}^4{\left(1-\frac{x}{4}\right)}^{3}dx \\ =3k{\int }_{x=0}^4{\left(1-\frac{x}{4}\right)}^{3}dx \end{gathered}$$

Integrate with respect to 'x'

$$\begin{gathered} =3k\underset{x=0}{\overset{4}{\int }}{\left(1-\frac{{\displaystyle x}}{{\displaystyle 4}}\right)}^3 \\ =\left(3k\right)\left(-4\right)\underset{x=0}{\overset{4}{\int }}{\left(1-\frac{{\displaystyle x}}{{\displaystyle 4}}\right)}^3\left(-\frac{1}{4}\right)dx \\ =\left(-12k\right){\overline{)\left(\frac{{\left(1-\frac{{\displaystyle x}}{{\displaystyle 4}}\right)}^3}{4}\right)}}_{x=0}^{x=4} \\ =\left(-12k\right)\left(0-\left(\frac{1}{4}\right)\right) \\ =3k \end{gathered}$$