Find the volume of the region above the Cone $ z=\sqrt{x^2+y^2} $ and below the Paraboloid $ z=2-x^2-y^2 $ .
The options are (i) $\frac{13 \pi}{6}$, (ii) $\frac{7 \pi}{3}$, (iii) $\frac{13 \pi}{3}$, (iv) $4 \pi$
Answers:
The intersections of the above two surface is given by
$\sqrt{x^2+y^2}=2-x^2-y^2$
or, $ z=2-z^2 $,
or, $z=1$, since $z \geq 0$
Let , $x=r \cos \theta$ and $y=r \sin \theta$.
Then the volume is
$ V=\int_{0}^{2 \pi} \int_{r=0}^{1} \int_{z=r}^{2-r^2} r dz dr d \theta = \frac{5 \pi}{6} $ ,
which does not any of the given options . Am I right ? Any help is there ? Does it matter '' above '' cone and '' below '' paraboloid ?
Converting to cylindrical coordinates, the paraboloid is given by the equation $z = 2-r^2$ and the cone is given by the equation $z=r$. These curves intersect when $$2-r^2 = r \implies r=1\, \text{ or } r=-2$$ Since $r\geq0$, $r=1$. The volume is computed to be \begin{align} V &= \int_0^{2\pi}\int_0^1\int_r^{2-r^2}r\,dz\,dr\,d\theta \\ &=\int_0^{2\pi}\int_0^1 (2r-r^3-r^2)\,dr\,d\theta \\ &=\int_0^{2\pi}\frac{5}{12}\,d\theta \\ &=\frac{5\pi}{6} \end{align} It appears there must be an error in the options.