Q 61. - Calculus

Question

Find the specified quantities for the solids described below:

The mass of the region from Exercise $51$, assuming that the density at every point is proportional to the square of the point’s distance from the $xy$-plane.

Step-by-Step Solution

Answer

The mass is given by $m=\left(\frac{7k\pi }{1024}\right)$

Step 1: Given Information

The density at every point is proportional to the square of the point’s distance from the $xy$-plane.

The region bounded above by the plane is given by equation $z=x$ and bounded below by the paraboloid by equation $z=x^2+y^2$.

Step 2: Evaluation of limits

Relation between rectangular and cylindrical coordinates is given by

$r=\sqrt{x^2+y^2}, \tan \theta =\frac{y}{x}, z=z$

And

$x=r\cos \theta , y=r\sin \theta , z=z$

Rectangular coordinates are $z=x$ and $z=x^2+y^2$

Cylindrical coordinates are $z=r\cos \theta$ and $z=r^2$

Cartesian limits are $x^2+y^2\le z\le x$

$x=r\cos \theta \Rightarrow z=x \Rightarrow z=r\cos \theta$

$r^2=x^2+y^2 \Rightarrow z=x^2+y^2 \Rightarrow z=r^2$

$\Rightarrow r^2\le z\le r\cos \theta$

$x^2+y^2=x\text{ gives }{r}^2=r\cos \theta \Rightarrow r=0,r=\cos \theta \to 0\le r\le \cos \theta$

Also

$r=0\text{ gives }\cos \theta =0 \Rightarrow -\pi /2\le \theta \le \pi /2$

Cylindrical limits are

$r^2\le z\le r\cos \theta , 0\le r\le \cos \theta , -\pi /2\le \theta \le \pi /2$

Step 3: Evaluation of Mass

The density at every point is proportional to the square of the point’s distance from the $xy$ plane.

$\Rightarrow \rho =k{z}^2$

Required mass is $m={\iiint }_E\rho dxdydz$

$m={\iiint }_{E}k{z}^{2}dxdydz$

$m={\int }_{\theta =-\pi /2}^{\pi /2}{\int }_{r=0}^{\cos \theta }{\int }_{z=r^2}^{\cos \theta }\left(k{z}^2\right)rdzdrd\theta$

$m={\left.{\int }_{\theta =-\pi /2}^{\pi /2}{\int }_{r=0}^{\cos \theta }\left(kr\right)\left(\frac{z^3}{3}\right)\right|}_{z=r^2}^{z-r\cos \theta }drd\theta$

$m={\int }_{\theta =-\pi /2}^{\pi /2}{\int }_{r=0}^{\cos \theta }\left(\frac{kr}{3}\right)\left(r^3{\cos }^3\theta -r^6\right)drd\theta$

Solving further yields

$m={\left.{\int }_{\theta =-\pi /2}^{\pi /2}\left(\frac{k}{3}\right)\left(\frac{r^5}{5}{\cos }^3\theta -\frac{r^8}{8}\right)\right|}_{r=0}^{\cos \theta }d\theta$

$m={\left(\frac{k}{3}\right)}_{\theta =-\pi /2}^{\pi /2}\left(\frac{{\cos }^8\theta }{5}-\frac{{\cos }^8\theta }{8}\right)d\theta$

$m=\left(\frac{2k}{3}\right)\left(\frac{3}{40}\right){\int }_{\theta =0}^{\pi /2}{\cos }^8\theta d\theta$

Solve using integration

$m=\left(\frac{k}{20}\right){\int }_{\theta =0}^{\pi /2}{\cos }^8\theta d\theta$

$m=\left(\frac{k}{20}\right)\left(\frac{7}{8}\times \frac{5}{6}\times \frac{3}{4}\times \frac{1}{2}\times \frac{\pi }{2}\right)$

$m=\left(\frac{7k\pi }{1024}\right)$ (required mass)