Calculate the triple integral using spherical coordinates: $\int_C z^2dxdydz$

Question

Calculate the triple integral using spherical coordinates: $\int_C z^2dxdydz$ where C is the region in $R^3$ described by $1 \le x^2+y^2+z^2 \le 4$

Here's what I have tried:

My computation for $z$ is: $\sqrt{1-x^2-y^2} \le z \le \sqrt{4-x^2-y^2}$, as for y I get: $-\sqrt{1-x^2}\le y \le \sqrt{4-x^2}$ and for x I get: $1 \le x \le 2$

The triple integral becomes:

$$\int_{1}^2 dx \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dy \int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}}z^2dz$$

The way I have pictured theta is as so:

Where The Red + Green is equal to $\frac{\pi}{2}$ But because we're only interested in the region from $1 \le x \le 2$ this covers only $\frac{\pi}{4}$

The integral becomes:

I presumed $\phi$ only goes from $\frac{\pi}{4}$ also, and we know that $r$ goes from $1\le r \le 2$

So the final integral becomes:

$$\int_0^{\frac{\pi}{4}} d\theta \int_0^{\frac{\pi}{4}} \cos^2(\phi)\sin(\phi) \space d\phi \int_1^2 r^4 dr$$

Because $z^2 = r^2 \cos^2(\phi)$

However my answer that I get is $2\pi(\sqrt{2}-4)$ but the answer should be $\frac{124\pi}{15}$. I would greatly appreciate the communities assistance

Answer 1

The region $C$ is $1 \le x^2+y^2+z^2 \le 4$, which is the entire region between spherical surfaces $x^2 + y^2 + z^2 = 1$ and $x^2 + y^2 + z^2 = 4$.

So clearly, $0 \leq \phi \leq \pi$ and $0 \leq \theta \leq 2 \pi$. Also, $ 1 \leq \rho \leq 2$.

So the integral should be,

$ \displaystyle \int_0^{2\pi} \int_0^{\pi} \int_1^2 \rho^4 \cos^2 \phi \sin \phi ~ d\rho ~ d\phi ~ d\theta$

Alternatively, for ease of calculation, note that due to symmetry of the region $C$, we must have

$ \displaystyle \int_C x^2 ~ dV = \int_C y^2 ~ dV = \int_C z^2 ~ dV$

So, $ ~ \displaystyle \int_C z^2 ~ dV = \frac{1}{3} \int_C (x^2 + y^2 + z^2) ~ dV$

$ \displaystyle = \frac{1}{3} \int_0^{2\pi} \int_0^{\pi} \int_1^2 \rho^4 \sin \phi ~ d \rho ~ d\phi ~ d\theta$

Answer 2

From the conditions you have stated above $C: 1 \leq x^2 + y^2 + z^2 \leq4$ by using a substitution we can see that $1 \leq r \leq 2$. However there are no conditions on the other spherical components so we can see that $$\int_C z^2dxdydz=\int_0^\pi\int_0^{2\pi}\int_1^2 z^2|J|dr d\theta d\phi$$ We still need to find $z$ in terms of spherical coordinates which is $z=r\cos\phi$. The jacobian for spherical coordinates is $r^2\sin\phi$ substituting this in we find that $$\int_0^\pi\int_0^{2\pi}\int_1^2 (r\cos\phi)^2 r^2\sin\phi dr d\theta d\phi$$ $$\int_0^\pi\int_0^{2\pi}\int_1^2 r^2\cos^2\phi \, r^2\sin\phi dr d\theta d\phi$$ $$\int_0^\pi\int_0^{2\pi}\int_1^2 r^4\cos^2\phi \,\sin\phi dr d\theta d\phi$$ $$\int_0^\pi\int_0^{2\pi} \frac{31}{5}\cos^2\phi \,\sin\phi d\theta d\phi$$ $$\int_0^\pi \frac{62\pi}{5}\cos^2\phi \,\sin\phi \,d\phi$$ $$ \frac{62\pi}{5}\int_0^\pi\cos^2\phi \,\sin\phi \,d\phi$$ Use $u$ substitution. Let $u = \cos\phi$ which means $du = -\sin\phi d\phi$. This means that the integral ranges from $u=1$ to $u=-1$ $$ -\frac{62\pi}{5}\int_1^{-1} u^2 du$$ $$ \frac{62\pi}{5}\int_{-1}^{1} u^2 du$$ $$ \frac{62\pi}{5} \left(\frac{1^3}{3} - \frac{-1^3}{3}\right) $$ $$ \frac{62\pi}{5} \left(\frac{2}{3}\right) $$ $$ \frac{124\pi}{15} $$

Answer 3

This is the kind of exercise that is perfect to use the spherical coordinates. Lets start with

$f(x,y,z)=z^2 \rightarrow f(r,\theta,\phi)=(rcos(\phi))^2=r^2cos^2(\phi)$

$C=\{(x,y,z)\in\mathbb{R^3} | 1\leq x^2+y^2+z^2\leq4\} \rightarrow C=\{(r,\theta,\phi)\in\mathbb{R^3} | 1\leq r^2\leq4\}$

In fact $\theta\in[0,2\pi]$ and $\phi\in[0,\pi]$ and $r\in[1,2]$. With this I mean that $\phi$ is the angle between the vector $(x,y,z)$ and positive semi axis $Oz$ and $\theta$ is the angle between the projection of $(x,y,z)$ in $xOy$ and positive semi axis $Ox$.

Therefore the integral you want to calculate is

$\int_0^\pi\int_0^{2\pi}\int_1^2r^2cos^2(\phi)r^2sin(\phi)drd\theta d\phi$

Ps: What does the red and green areas mean? In fact that area is $\frac{\pi2^2}{4}=\pi$. How did you get the red area is half of the red+green area? The domain of the integral is the space between the spheres. I post it so it is a bit more clear what the angles are.