Question

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

Short answer

Volume of solid is$V=\frac{\mathrm{\pi }}{32}$units

Step-by-step solution

Given Information

The equations are$z=x$and$z=x^2+y^2$.

Simplification

The relationship between rectangular and cylindrical coordinates are:

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

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

Evaluation of Limits

The rectangular coordinates are $z=x$and $z=x^2+y^2$

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

The cartesian limits are $x^2+y^2\le z\le x$

$x=r\cos \theta$

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

And $r^2=x^2+y^2$

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

It implies $r^2\le z\le r\cos \theta$

Simplifying

$x^2+y^2=x$

$\Rightarrow r^2=r\cos \theta$

Simplifying role="math" localid="1652376534758" $r=0,r=\cos \theta \Rightarrow 0\le r\le \cos \theta$

The cylindrical limits are:

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

Evaluation of Volume

Required volume is given by

$V={\int }_{\theta =-\pi /2}^{\pi /2}{\int }_{r=0}^{\cos \theta }{\int }_{z=r^2}^{r\cos \theta }rdzdrd\theta$

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

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

$={\left.{\int }_{\theta =-\pi /2}^{\pi /2}\cos \theta \left(\frac{r^3}{3}\right)\right|}_{r=0}^{r=\cos \theta }d\theta -{\left.{\int }_{\theta =-\pi /2}^{\pi /2}\left(\frac{r^4}{4}\right)\right|}_{r=0}^{r=\cos \theta }d\theta$

$V={\int }_{\theta =-\pi /2}^{\pi /2}\frac{1}{3}\left({\cos }^4\theta \right)d\theta -{\int }_{\theta =-\pi /2}^{\pi /2}\frac{1}{4}\left({\cos }^4\theta \right)d\theta$

Further simplification gives

$V=\frac{2}{3}{\int }_{\theta =0}^{\pi /2}{\cos }^4\theta d\theta -\frac{1}{4}·2{\int }_{\theta =0}^{\pi /2}{\cos }^4\theta d\theta$

$V=\left(\frac{2}{3}\right)\times \frac{3}{4}\times \frac{1}{2}\times \frac{\pi }{2}-\left(\frac{1}{2}\right)\times \frac{3}{4}\times \frac{1}{2}\times \frac{\pi }{2}$

$V=\frac{4\pi }{32}-\frac{3\pi }{32}$

$V=\frac{\pi }{32}$units