I have to calculate integral:
$$ \iiint (x^2+y^2+z^2)\,\mathrm dx\,\mathrm dy\,\mathrm dz$$
on the volume bound between the following surfaces:
$$y^2+z^2=x^2$$ $$x^2+y^2+z^2 = a^2$$ $$x \ge 0$$
Now, I substituted spherical coordinates. From the second equation I got that $r \in [0,a]$. From the third equation I got that $\phi \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
But, as for the third equation, I am only getting this:
$$\sin^2\phi \sin^2\theta + \cos^2\theta = \cos^2\phi \sin^2\theta$$
Now, the best thing I can do here is divide both sides by $\sin^2\theta$ and I get that
$$\sin^2\phi + \cot^2\theta = \cos^2\phi$$
Which brings me to
$$\cot\theta = \sqrt{\cos(2\phi)}.$$
Is this correct? But even if it is, I'd have to take the inverse cotangent function to get the upper bound for $\theta$.
Since the axis of symmetry for the cone is along x-axis, please define spherical coordinates with second polar axis being x-axis.
$y = \rho \cos\theta \sin \phi, z = \rho \sin\theta \sin\phi, x = \rho \cos\phi$
Then cone is $\tan \phi = 1 \implies 0 \leq \phi \leq \dfrac{\pi}{4}$.
Bounds of $\rho$ from the equation of the sphere: $0 \leq \rho \leq a$
$0 \leq \theta \leq 2\pi$
So the integral becomes,
$\displaystyle \int_0^{2\pi} \int_{0}^{\pi/4} \int_0^a \rho^4 \sin \phi \ d\rho \ d\phi \ d\theta$