I'm trying to solve the following exercise about triple integrals, but I need a check.
Let $V \subset \mathbb{R}^3$ which is made by one sphere of radius $2$ centered at $(0,0,0)$ and a cylinder with base at $(0,0,0)$ and radius $1$. Formally, this is
$$V=\{x^2+y^2+z^4 \leq 4 \} \cup \{(x,y,z): x^2 + y^2 \leq 1, z \in [0,4] \}$$
Goal: Compute the Volume of $V$
My attempt:
$|V|=|\text{Sphere}| + |\text{Cylinder}| – |\text{Portion of Sphere inside Cylinder}|$
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The first two terms are straightforward: $|\text{Sphere}|=\frac{16}{3} \pi $, $|\text{Cylinder}|=4 \pi$
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The tricky part for me is $|\text{Portion of Sphere inside Cylinder}|$
To do that, I consider the $Oxz$ plane: here I can see that cylinder intersects the sphere at $(1,\sqrt{3})$ and the angle $\varphi = \frac{\pi}{6}$ ($\varphi$ is intended with this convenction)
Now I want to use spherical coordinates to compute the volume of that region: to this aim, I let the $\rho$ of the spherical coordinates range in $(0,1)$, in order to be inside the cylinder. All in all, the triple integral do to is the following:
$$\int_{0}^{2 \pi} \int_0^{\frac{\pi}{6}} \int_{0}^{1} \rho^2 \sin(\varphi) d \rho d \varphi d \theta$$
which can easily be solved.
Best Answer
Your approach is fine but the integral does not give the volume of the sphere inside the cylinder - i) the way polar angle works, it will leave out parts of the sphere inside the cylinder on the sides unless you specifically address it in your integral. ii) you take $0 \leq \rho \leq 1$ but that will not take out the correct volume for this sphere.
As you are trying to find the volume of the combined region between $0 \leq z \leq 4$, here is what I would suggest.
You have already found that the cylinder and sphere intersect at $z = \sqrt3$. So we can add the volume of the hemisphere ($V_1$) AND the volume of the cylinder between $\sqrt3 \leq z \leq 4 \, (V_2)$. We have to then just subtract from the hemisphere the volume of the spherical cap that is inside the cylinder above $z = \sqrt3 \,, \, (V_3)$.
$V_3$ can be calculated with the below integral,
$V_3 = \displaystyle \int_{0}^{2 \pi} \int_0^{\pi /6} \int_{(\sqrt3 / \cos \phi)}^{2} \, \rho^2 \sin(\phi) \, d \rho \, d \phi \, d \theta$
$V_1 = \frac{16 \pi}{3} \,$ as you mentioned
$V_2 = \pi \times 1^2 \times (4 - \sqrt3) = (4 - \sqrt3) \pi$
Total volume of the combined solid is $V_1 + V_2 - V_3$.