How to solve this triple integral (Solid bounded by cone and sphere)

Question

Does anyone know how to solve this triple integral question? I been working on it for such a long time, and have no idea how to solve it.

Answer 1

Based on my comment we use spherical coordinates. Also to help you visualize, here is a "slice" of the domain. You should imagine that this "roundy cone" revolves 360° around the $z$-axis.

$$0 \leq r \leq 1 \\ 0 \leq \theta \leq 2 \pi \\ 0 \leq \phi \leq \pi/4$$

we know that $z =r \cos \phi$ and the jacobian is $r^2 \sin \phi$ therefore the integral of $\iiint f(x,y,z)$ gives : $$ \int_0^1 dr \int_0^{2\pi}d \theta \int_0^{\pi/4} d \phi\ r \cos \phi \cdot r^2 \sin \phi = \int_0^1 r^3 dr \int_0^{2\pi} d\theta \int_0^{\pi/4} \sin\phi \cos\phi\ d\phi = \frac14 \cdot 2\pi \cdot \frac14 = \frac{\pi}{8} $$

Now for the volume we use exactly the same parametrization and simply integrate the function $1$ : $$ \int_0^1 dr \int_0^{2\pi}d \theta \int_0^{\pi/4} d \phi\ 1 \cdot r^2 \sin \phi = \int_0^1 r^2 dr \int_0^{2\pi} d\theta \int_0^{\pi/4} \sin\phi\ d\phi = \frac13 \cdot 2 \pi \cdot (1- 1/\sqrt{2}) $$ Putting all together and doing some algebra gives you the answer.