Triple integrals: Range of integrals

Question

Use spherical coordinates to evaluate: The volume of the part of the sphere $x^2$ + $y^2$+$z^2$=16 that lies between the planes z=2 and z= $2{\sqrt 3}$

What are the range of integration? My solution:

From $x^2$ + $y^2$+$z^2$ =$\rho^2$

$\rho^2$ = 16, hence $\rho$=4, implying that 0 $\leq$$\rho$ $\leq$4

Then $\theta$ lies between 0 and 2$\pi$

But is the range of $\phi$? Does $\phi$ lies between 0 and $\pi$?

The fact that in the question it is stated that 'lies between the planes z=2 and z= $2{\sqrt 3}$' is really confusing me.

Can someone please help me in clarifying my doubts?

Answer 1

As the region is bound between planes, $\rho$ is not constant.

Sphere is $\rho \leq 4$.

As $z = \rho \cos\phi, \ $ at $z = 2 \sqrt3, \ \rho = 2 \sqrt3 \sec \phi$

At intersection of plane $z = 2 \sqrt3$ and sphere $\rho = 4$,

$2\sqrt3 = 4 \cos\phi \implies \phi = \frac{\pi}{6}$

Similarly, at $z = 2, \ \rho = 2 \sec \phi$

At intersection of plane $z = 2 \sqrt3$ and sphere, $\phi = \frac{\pi}{3}$

Now to integrate in the order $d\rho$ first, note that -

• For $0 \leq \phi \leq \frac{\pi}{6}$, $\rho$ is bound between planes.
• For $\frac{\pi}{6} \leq \phi \leq \frac{\pi}{3}$, $\rho$ is bound between plane $z = 2$ and sphere.

So integral to find volume is,

$\displaystyle \int_0^{2\pi} \int_0^{\pi/6} \int_{2 \sec \phi}^{2\sqrt3 \sec\phi} \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta \ \ + $

$\displaystyle \int_0^{2\pi} \int_{\pi/6}^{\pi/3} \int_{2 \sec \phi}^{4} \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta$

Answer 2

The $z$-coordinate belongs to the interval $[2,2\sqrt 3]$.
If you want to determine the constraints for $x$ and $y$ you have to look at the the intersections of the sphere with the two planes you wrote: $$\begin{cases}z=\sqrt{16-x^2-y^2}\\z=2 \end{cases}\implies 16-x^2-y^2=4\iff x^2+y^2=(2\sqrt 3)^2$$ $$\begin{cases}z=\sqrt{16-x^2-y^2}\\z=2\sqrt 3 \end{cases}\implies 16-x^2-y^2=12\iff x^2+y^2=(2)^2$$ Now you can find the volume of the solid integrating constant function $f(x)=1$ as $$\underset{{\mathbb B_{2\sqrt 3}(0)\setminus\mathbb B_2(0)}}{\iint}\int_2^{\sqrt{16-x^2-y^2}}dz dxdy+\underset{\mathbb B_2(0)}\iint\int_2^{2\sqrt 3}dzdxdy.$$ Observe that integrating with "$\le$" or $<$ is the same thing since the two sets will differ for a zero measure set, which gives zero contribute also to the integral.

Answer 3

No, $\varphi$ cannot take any value from $0$ to $\pi$. See the picture below, which is the intersection of you region with the plane $y=0$. So, yes, $\theta$ can take any value from $0$ to $2\pi$. And, as you can see from the picture, the range of $\phi$ is $\left[0,\frac\pi3\right]$. Now, concerning $\rho$, there are two possibilities:

• If $\phi\in\left[0,\frac\pi6\right]$, then, since $z=\rho\cos(\phi)$ and since $2\leqslant z\leqslant2\sqrt3$, the range of $\rho$ is $\left[\frac2{\cos\phi},\frac{2\sqrt3}{\cos\phi}\right]$.
• If $\phi\in\left[\frac\pi6,\frac\pi3\right]$, then then range of $\rho$ is $\left[\frac2{\cos\phi},4\right]$.

So, compute$$\int_0^{2\pi}\int_0^{\pi/6}\int_{2/\cos(\phi)}^{2\sqrt3/\cos(\phi)}\rho^2\sin(\phi)\,\mathrm d\rho\,\mathrm d\phi\,\mathrm d\theta+\int_0^{2\pi}\int_{\pi/6}^{\pi/3}\int_{2/\cos(\phi)}^4\rho^2\sin(\phi)\,\mathrm d\rho\,\mathrm d\phi\,\mathrm d\theta.$$