$Exercise$.
$a)$ Calculate the volume of the solid limited by the surfaces: $x^2+y^2+z^2=9$, $x^2+y^2+z^2=16$, $x^2+y^2=z^2$, $z\geq0$.
$b)$ Calculate $\int\int\int_A \sqrt{x^2+y^2}dV$, where $A={\{(x,y,z)\in\mathbb{R}^3: x^2+y^2+z^2\leq1, x^2+y^2\leq z^2,z\geq0 }\}$
I could already solve the part $ b) $, and I did it as follows:
Using the transformation $T(r,\theta,\phi)=(r\text{cos($\theta$)},r\text{sin($\theta$)sin($\phi$)},r\text{cos($\phi$)})$ with Jacobian $J(T)=r^2\text{sin($\phi$)}$ It is true that $$0\leq r\leq 1$$ $$0\leq \phi \leq \frac{\pi}{4}$$ $$0\leq \theta \leq 2\pi$$
Hence,
$$\int\int\int_A \sqrt{x^2+y^2}dV=\int_0^1 \int_0^{2\pi}\int_0^{\frac{\pi}{4}}r^3\text{sin$^2(\phi$)}d\phi d\theta dr=\cdot \cdot \cdot = \frac{\pi^2}{16}-\frac{\pi}{8}$$ Is my solution correct, how can I solve point $a)$? I do not know how to start. Thanks for your attention.
Best Answer
Note that the volume is the portion of the hollow sphere with inner and outer radii 3 and 4 respectively and inside the cone of 45 degrees and with the vertex at the origin.
In spherical coordinates, the volume integral is set up as,
$$\int_V dV= \int_0^{2\pi}\int_0^{\frac{\pi}{4}}\int_3^4 r^2\sin\theta\> dr d\theta d\phi =\frac{37\pi}3(2-\sqrt2)$$