# Obtain the volume of the solid which is bounded by a circular paraboloid z=x^2+y^2, cylinder x^2+y^2=4, and Coordinate plane. And the solid is in the (x>=0, y>=0, z>=0).

Obtain the volume of the solid which is bounded by a circular paraboloid $z={x}^{2}+{y}^{2}$, cylinder ${x}^{2}+{y}^{2}=4$, and Coordinate plane. And the solid is in the $\left(x\ge 0,y\ge 0,z\ge 0\right)$.
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Faiza Fuller
Cylindrical coordinates can be given as,
$x=r\mathrm{cos}\theta$
$y=r\mathrm{sin}\theta$
$z=z$
${x}^{2}+{y}^{2}={r}^{2}$
Now the r will be 2 as ${x}^{2}+y2={2}^{2}$. Limits of region can be given as,
$R\left\{\left(r,\theta ,z\right):0\le r\le 2,0\le \theta \le \frac{\pi }{2},0\le z\le {r}^{2}\right\}$
Volume of solid in cylindrical coordinate plane is given as,
$V={\oint }_{R}rdrd\theta dz$
Calculate the volume using the values,
$V={\int }_{0}^{\frac{\pi }{2}}{\int }_{0}^{2}{\int }_{0}^{{r}^{2}}rdzdrd\theta$
$={\int }_{0}^{\frac{\pi }{2}}{\int }_{0}^{2}r{\left[z\right]}_{0}^{{r}^{2}}drd\theta$
$={\int }_{0}^{\frac{\pi }{2}}{\int }_{0}^{2}{r}^{3}drd\theta$
$={\int }_{0}^{\frac{\pi }{2}}{\int }_{0}^{2}{\left[\frac{{r}^{4}}{4}\right]}_{0}^{2}d\theta$
$={\int }_{0}^{\frac{\pi }{2}}4d\theta$
$=4{\left[t\stackrel{^}{e}\right]}_{0}^{\frac{\pi }{2}}$
$=4\left[\frac{\pi }{2}-0\right]$
$=2\pi$
Volume of the given solid is $2\pi$