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:

$|V|=|\text{Sphere}| + |\text{Cylinder}| - |\text{Portion of Sphere inside Cylinder}|$

  • The first two terms are straightforward: $|\text{Sphere}|=\frac{16}{3} \pi $, $|\text{Cylinder}|=4 \pi$

  • 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.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Commented Jan 4, 2021 at 0:57

3 Answers 3


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 $V_1 + V_2 - V_3$.

  • 1
    $\begingroup$ Actually, I'd compute the volume of $V_3$ with $$\int_0^1 \int_0^{2 \pi} \int_0^{\sqrt{4-\rho^2}} \rho dz d \theta d \rho$$ where I used cylndrical coordinates. Is this fine too? @MathLover $\endgroup$
    – Vefhug
    Commented Jan 4, 2021 at 11:06
  • $\begingroup$ @Vefhug Yes if you are doing in cylindrical coordinates, that will work but in that case your $V_2$ is $4 \pi$. $\endgroup$
    – Math Lover
    Commented Jan 4, 2021 at 11:40

Please, read what you wrote. It is not clear what does the region look like. Draw it and make sure it has finite volume. Than compute the integral in convenient coordinates system, in this case, cylindrical or spherical.


This image may help. (Where's the "cone"??)

enter image description here

  • $\begingroup$ Yes, there's no cone... $\endgroup$
    – Vefhug
    Commented Jan 4, 2021 at 11:07

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