Use the spherical coordinates to compute the integral $\int\limits_{B} z^2 dx dy dz$ where B is defined by $1\leq x^2 + y^2 + z^2 \leq 4$

Question

, however the answer I got to is different than the answer sheet. The answer sheet says that it should be $\frac{62}{15}$ Am I making some mistake or is the answer sheet incorrect?

Answer 1

Neither your answer nor the suggested one seem correct to me... Using the standard conversion to spherical coordinates, namely $$\begin{cases}x=r\cos\theta\sin\varphi\\y=r\sin\theta\sin\varphi\\z=r\cos\phi\end{cases}$$ we get the integral $$\begin{align*} \iiint_Bz^2\,\mathrm dx\,\mathrm dy\,\mathrm dz&=\int_{r=1}^{r=2}\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi}(r\cos\varphi)^2r^2\sin\varphi\,\mathrm d\varphi\,\mathrm d\theta\,\mathrm dr\\[1ex] &=2\pi\int_{r=1}^{r=2}\int_{\varphi=0}^{\varphi=\pi}r^4\cos^2\varphi\sin\varphi\,\mathrm d\varphi\,\mathrm dr\\[1ex] &=\frac\pi2\left(\int_{r=1}^{r=2}r^4\,\mathrm dr\right)\left(\int_{\varphi=0}^{\varphi=\pi}\sin\varphi+\sin3\varphi\,\mathrm d\varphi\right) \end{align*}$$ which ultimately evaluates to

$$\frac{124\pi}{15}$$

Answer 2

Your set-up for polar coordinates is not the standard one. That doesn't mean you can't adjust from there. However, you have then gone on to use the limits and Jacobian associated with the standard set-up.

You have

$x = r\cos \theta\sin \phi\\ y = r\cos\theta\cos\phi\\ z = r\sin\theta$

The more conventional configuration:

$x = r\cos \theta\sin \phi\\ y = r\sin\theta\sin\phi\\ z = r\cos\phi$

In the conventional configuration

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

In the non-standard configuration.

The Jacobian will be $r^2\cos\theta$

The limits will change as well.

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

Answer 3

According to spherical coordinates system with should have

• $x=r\sin\theta \cos \phi$

• $y=r\sin \theta \sin \phi$

• $z=r\cos \theta$

and the set up should be sligtly different

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

and the result is equal to $\frac{\pi124}{15}$.

Answer 4

Your work can be remedied. The correct volume form of your version of spherical coordinates is $$r^2\,{\color{red}{\cos(\theta)}}\,\text{d}\phi\,\text{d}\theta\,\text{d}r\,.$$ The angle $\theta$ runs from $-\dfrac{\pi}{2}$ to $+\dfrac{\pi}{2}$, not $0$ to $\pi$. Therefore, $$I:=\iiint_B\,z^2\,\text{d}x\,\text{d}y\,\text{d}z=\int_1^2\,r^4\,\text{d}r\,\int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}}\,\sin^2(\theta)\,\cos(\theta)\,\text{d}\theta\,\int_0^{2\pi}\,\text{d}\phi\,.$$ Ergo, $$I=\left(\frac{2^5-1^5}{5}\right)\,\left(\frac{(+1)^3-(-1)^3}{3}\right)\,(2\pi)=\frac{124\pi}{15}\,.$$

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Alternatively, by symmetry, $$I=\iiint_B\,x^2\,\text{d}x\,\text{d}y\,\text{d}z=\iiint_B\,y^2\,\text{d}x\,\text{d}y\,\text{d}z\,.$$ Thus, $$I=\frac13\,\iiint_B\,\left(x^2+y^2+z^2\right)\,\text{d}x\,\text{d}y\,\text{d}z\,.$$ Let $\text{d}^2\Omega$ denote the solid angle element, then $$I=\frac13\,\iiint_B\,r^2\cdot r^2\,\text{d}r\,\text{d}^2\Omega=\frac13\,\int_1^2\,r^4\,\text{d}r\,\iint_{\Sigma}\,\text{d}^2\Omega\,,$$ where $\Sigma$ is the surface of the unit $2$-dimensional sphere centered at the origin. Thus, $$I=\frac13\,\left(\frac{2^5-1^5}{5}\right)\,(4\pi)=\frac{124\pi}{15}\,.$$