0
$\begingroup$

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 $\theta, r, \phi$.

$\endgroup$
1
$\begingroup$

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, $0 \leq \theta \leq 2\pi$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.