Integration with Spherical Coordinates

Question

Use spherical coordinates to find the volume of the solid inside both $x^2+y^2+z^2=16$ and $z=(x^2+y^2)^{1/2}$.

Answer 1

The whole is the sum of its parts. Do you understand why dv = $\rho{^2}\sin\theta d\rho d\theta d\phi$? Here I am assuming $\theta$ is the angle between $\rho$ and the z axis and $\phi$ is the angle in the x-y plane.

Your integral is then $\iiint$dv where the limits of integration must be set up in accordance with the given solids. These must first be translated to spherical coordinates.

Answer 2

For problems like this, it helps to draw a picture. Fortunately, programs like Mathematica make tasks like this easy. You may wish to seek out this program or something similar.

From the picture, I can see that the solid is something like an ice cream cone: a cone on the bottom and a spherical cap on top. We can find where it goes from being a cone to being a sphere by setting the $z$ coordinates of the two solids equal. Let $x^2+y^2=r^2$; then

$$r^2 = 16-r^2 \implies r^2=z^2=8 \implies z=2 \sqrt{2}$$

So, when $z \in [0,2 \sqrt{2}]$, the solid is that cone, and when $z \in [2 \sqrt{2},4]$, the solid is the sphere.

Because there is circular symmetry about the $z$ axis, I would rather use cylindrical coordinates. The volume of the cone is

$$V_{\text{cone}} = \int_0^{2 \sqrt{2}} dz \: \int_0^z dr \, r \: \int_0 ^{2 \pi} d\theta$$

The volume of the sphere is

$$V_{\text{sphere}} = \int_{2 \sqrt{2}}^4 dz \: \int_0^{\sqrt{16-z^2}} dr \, r \: \int_0 ^{2 \pi} d\theta$$

The volume you seek is $V_{\text{cone}} + V_{\text{sphere}}$.

ADDENDUM

If you must use sphericals, you need to define the range of coordinates. Here, $r \in [0,4]$, and the azimuth angle $\phi \in [0,2 \pi]$. This leaves the polar angle $\theta$, whose range may be found by considering a cross-section of the geometry. In this case, we know from above that the $z$ coordinate of the intersection of $z=2 \sqrt{2}$, so that the extent of the polar angle is given by

$$\sin{\theta_0}=\frac{2 \sqrt{2}}{4} \implies \theta_0 = \frac{\pi}{4}$$

The volume of this ice cream cone is then

$$V = \int_0^4 dr \, r^2 \: \int_0^{\pi/4} d\theta \, \sin{\theta} \: \int_0^{2 \pi} d\phi$$

You may check to see that the volumes computed in cylindricals and sphericals are the same.

Answer 3

Let $(r,\phi,\theta)$ denote spherical coordinates in $(x,y,z)$-space, i.e., $$x=r\cos\phi\cos\theta,\quad y=r\sin\phi\cos\theta,\quad z=r\sin\theta\ ;\tag{1}$$ where $-{\pi\over2}\leq\theta\leq{\pi\over2}$. Then $${z\over\sqrt{x^2+y^2}}=\tan\theta\ ;$$ therefore the condition $z\geq\sqrt{x^2+y^2}$ amounts to $\theta\geq{\pi\over4}$. It follows that your solid $B$ appears in spherical coordinates as a box $$\hat B:=\left\{(r,\phi,\theta)\biggm|\ 0\leq r\leq4,\ 0\leq\phi\leq 2\pi, \ {\pi\over4}\leq\theta\leq{\pi\over2}\right\}\ .$$ In order to compute ${\rm vol}(B)$ we need the Jacobian of the transformation $(1)$. It computes to $$\det\left[{\partial(x,y,z)\over\partial(r,\phi,\theta)}\right]=r^2\cos\theta\ .$$ We then obtain $${\rm vol}(B)=\int_{\hat B} r^2\cos\theta\ {\rm d}(r,\phi,\theta)=2\pi \int_0^4\int_{\pi/4}^{\pi/2} r^2 \ \cos\theta \ d\phi\ dr={128\pi\over3}\left(1-{1\over\sqrt{2}}\right)\ .$$ If you want to avoid Jacobians you can argue in a more leisurely fashion as follows: The solid $B$ can be regarded as a pyramid of height $4$ erected over a spherical cap $C$. Therefore one has $${\rm vol}(B)={1\over3}\cdot\omega(C)\cdot 4\ ,\tag{2}$$ where $\omega(C)$ denotes the surface area of $C$. The cap $C$ can be regarded as a union of infinitesimal lampshades of radius $4\cos\theta$ and width $4d\theta$. It follows that $$\omega(C)=\int_{\pi/4}^{\pi/2}2\pi\cdot 4\cos\theta\cdot 4d\theta=32\pi\left(1-{1\over\sqrt{2}}\right)\ ,$$ which leads to the same end result as before.