1. ## spherical coordinates

Use spherical coordinates to show that $\int\int\int_{E} z dV = \frac{15\pi}{16}$, where $E$ lies between the spheres $x^2 + y^2 + z^2 = 1$ and $x^2 + y^2 + z^2 = 4$ in the first octant (i.e. $x, y, z \geq 0$).

2. Originally Posted by wik_chick88
Use spherical coordinates to show that $\int\int\int_{E} z dV = \frac{15\pi}{16}$, where $E$ lies between the spheres $x^2 + y^2 + z^2 = 1$ and $x^2 + y^2 + z^2 = 4$ in the first octant (i.e. $x, y, z \geq 0$).
$z=\rho\cos\phi$
$\rho^2 = x^2+y^2+z^2$
$dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$

So the bounds are:

$\rho: 1\to 2$
$\phi: 0\to\pi/2$
$\theta: 0\to\pi/2$

This makes the integrand:

$\rho\cos\phi*\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = \rho^3\sin(\phi)\cos(\phi)\,d\rho\,d\phi\,d\theta$

The integral is:

$\int_0^{\pi/2}\int_0^{\pi/2}\int_1^2 \rho^3\sin(\phi)\cos(\phi)\,d\rho\,d\phi\,d\theta$

Here's the evaluation of the integral, if you need it:
Spoiler:

Evaluating the innermost integral gives you $\left(\frac{\rho^4}{4}\sin(\phi)\cos(\phi)\right)\ bigg|_1^2 = \frac{15}{4}\sin(\phi)\cos(\phi)$

Noticing that $\sin(\phi)\cos(\phi)=\frac{1}{2}\sin(2\phi)$, our integral is now:

$\frac{15}{8}\int_0^{\pi/2}\int_0^{\pi/2}\sin(2\phi)\,d\phi\,d\theta$

Evaluating the next integral, we have $\left(-\frac{1}{2}\cos(2\phi)\right)\bigg|_0^{\pi/2} = -\frac{1}{2}(-1-1) = 1$

Now the integral is $\frac{15}{8}\int_0^{\pi/2}\,d\theta = \left(\frac{15}{8}\theta\right)\bigg|_0^{\pi/2} = \frac{15\pi}{16}$

So the integral is $\frac{15\pi}{16}.$