I actually have found the solution using double integral in polar coordinate. However, I am curious about whether I could find the same exact solution using triple integral in spherical coordinates but it does not match the one I got from using the double integral. I might have got something wrong with the solution but I couldn't figure out where. My solution goes like this:
- $x^2+y^2+z^2=1$ means that $\rho = 1$.
- Substituting the cone $z=\sqrt{x^2+y^2}$ to the sphere's equation gives $x^2+y^2=\frac{1}{2}$. Since $r=\rho sin(\phi)=\frac{1}{\sqrt{2}}$, then intuitively $\phi=\frac{\pi}{4}$.
- Then, the region bounded by the solid in spherical coordinate is $R=\{(\rho, \theta, \phi)|\space 0\leq \rho \leq 1, 0\leq \theta \leq 2\pi, 0\leq \phi \leq \frac{\pi}{4}\}$.
- $V(R)=\iiint_{R}dV=\int^{2\pi}_{0}\int_{0}^{\frac{\pi}{4}}\int_{0}^{1}\rho^2 sin(\phi)d\rho d\phi d\theta=\frac{2\pi}{3}(-\frac{\sqrt{2}}{2}-1)$But the actual solution is $\frac{2\pi}{3\sqrt{2}}(\sqrt{2}-1)$.
1 Answer
$\begingroup$Your integral is set up correctly. You may have made some mistake in evaluating it.
$ \displaystyle \int_0^{2\pi} \int_0^{\pi/4} \int_{0}^{1} \rho^2 \sin\phi \ d\rho \ d\phi \ d\theta$
$ \displaystyle = \int_0^{2\pi} \int_0^{\pi/4} \left[\cfrac{\rho^3}{3}\right]_0^1 \sin\phi \ d\phi \ d\theta$
$ \displaystyle = \cfrac{1}{3} \int_0^{2\pi} \left[- \cos \phi\right]_0^{\pi/4} \ d\theta$
$ \displaystyle = \cfrac{2\pi}{3} \left(1 - \cfrac{1}{\sqrt2} \right) = \cfrac{2\pi}{3 \sqrt2} \left(\sqrt2 - 1 \right)$
$\endgroup$ 1
