Finding the limits of a triple integral.

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Evaluate:

$$\iiint_V (x^2+y^2)\:\mathrm{d}x\:\mathrm{d}y\:\mathrm{d}z,$$

where $V$ is the region in the positive octant bounded by the sphere $\|\vec{r}\|=a$.

I am unsure how to get the limits of integration here.

I believe I need to use cylindrical coordinates

$x = r\cos \theta$

$y = r\sin \theta$

$z=z$

$$\iiint_V f(x,y,z)\:\mathrm{d}x\:\mathrm{d}y\:\mathrm{d}z = \iiint_V f(r\cos \theta, r\sin \theta, z)r \:\mathrm{d}z\:\mathrm{d}r\:\mathrm{d}\theta = \iiint_V r^3 \:\mathrm{d}z\:\mathrm{d}r\:\mathrm{d}\theta,$$

But because the bound is only $\|\vec{r}\|=a$.. i'm unsure what to do now.

Or am i completely wrong :L?

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The boundary is a piece of a sphere so use spherical coordinates:

$$dV = r^2 \, \sin{\theta} \, dr \, d\theta \, d\phi$$

$$x^2+y^2=r^2 \sin^2{\theta}$$

Since we are in the first octant, the integral is

$$\int_0^a dr \, r^4 \, \int_0^{\pi/2} d\theta \, \sin^3{\theta} \, \int_0^{\pi/2} d\phi$$