Question

PLEASE HELP!! WILL MARK BRAINLIEST!!!

Answer 1

In Cartesian coordinates, the volume is

$\displaystyle \iiint_R \mathrm dV$

where R is the set

$R = \left\{(x,y,z) \, : \, -4\le x\le4 \text{ and } -\sqrt{16-x^2} \le y \le \sqrt{16-x^2} \text{ and }\right. \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~\left.-\sqrt{16-x^2-y^2} \le z \le \sqrt{16-x^2-y^2} \right\}$

Converting to spherical coordinates, we take

$\begin{cases}x=\rho\cos(\theta)\sin(\phi) \\ y=\rho\sin(\theta)\sin(\phi) \\ z = \rho\cos(\phi)\end{cases}$

and we can compute the Jacobian to find

$dV = dx\,dy\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi$

and the set R transforms to

$R' = \left\{(\rho,\theta,\phi) \,:\, 0\le \rho\le4 \text{ and }0\le\theta\le2\pi \text{ and } 0\le\phi\le\pi\right\}$

The new integral and hence the volume is then

$\displaystyle \iiint_R \mathrm dV = \iiint_{R'} \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi \\\\ = \boxed{\int_0^\pi \int_0^{2\pi} \int_0^4 \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi}$