0
$\begingroup$

I have the unit hemisphere centred at $z=1$ and a cone right under it with a point at $(0,0,0)$ and I am trying to find the volume with a triple integral. I already know that the volume is $\frac{2\pi}{3}+\frac{\pi}{3}=\pi$. So I can use that as a reference if my iterated integral is correct. The cartesian triple integral I got is $$\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{1-x^2-y^2}+1} dzdydx$$ I already calculated this and it gives me the right answer. But if I want to transform this to spherical coordinates, the bounds for $\theta$ and $\phi$ are really easy, they're just $0 \leq \theta \leq 2\pi$ and $0 \leq \phi \leq \frac{\pi}{4}$. But I have no idea how to get the bounds for $\rho$. I so far tried turning the bounds into spherical with the change of variables and then I have to solve for $\rho$ using the quadratic formula, but it isn't giving me the correct volume. How would I get the bounds for $\rho$ in this case?

$\endgroup$

1 Answer 1

0
$\begingroup$

For all $\theta$, $\phi$ the value of $\rho$ varies from the origin to the hemisphere: $$x^2 + y^2 + (z-1)^2 = 1,$$ $$ \rho^2\cos^2\theta\sin^2\phi + \rho^2\sin^2\theta\sin^2\phi + (\rho\cos\phi - 1)^2 = 1, $$ $$\rho^2\sin^2\phi + \rho\cos\phi = 2,$$ $$\rho = -\frac{\cos\phi}{2\sin^2\phi}\pm\frac{\sqrt{8-7\cos^2\phi}}{2\sin^2\phi}.$$ As $\rho>0$ is required, $$\rho = \frac{\sqrt{8-7\cos^2\phi}}{2\sin^2\phi}-\frac{\cos\phi}{2\sin^2\phi}.$$ An the integral in spherical coordinates is ugly and nasty. Surely cylindrical coordinates is better.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .