Let $D$ be the region bounded by the surface of the hemisphere $z = \sqrt {1-x^2-y^2}$ and the plane $z=0$. Use spherical polar coordinates to find $\iiint \limits _D z^3 \Bbb dV$.

I'm fine with the actual integration, but I'm struggling to establish the limits for $\theta, r, \phi$.


Since $z = \sqrt{1-x^{2} - y^{2}}$, we can square both sides, and rearrange to get, $x^{2} + y^{2} + z^{2} = 1$.

Now use the fact that $\rho^{2} = x^{2} + y^{2} + z^{2}$. Therefore, $\rho$ goes from $0\leq \rho \leq 1$

Since we're bounded by the plane $z = 0$, $\phi$ will have the bounds, $0 \leq \phi \leq \frac{\pi}{2}$

And we "sweep" the entire circle, so $\theta$ will have the bounds, $0 \leq \theta \leq 2\pi$


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