Computing volume inside a ball and outside a cylinder

Question

I try to calculate the volume inside the ball $\ x^2 + y^2 + z ^2 = 4 $ the outside the cylinder $\ x^2+y^2=2x $ using double integral. Since the shape is symmetric I chose $\ z = 0 $ as bottom limit and the ball is the upper limit. I try to convert to polar form and so I get

$$\ x^2 + y^2 + z ^2 = 4 \Rightarrow z = \sqrt{4-r^2} \\x^2 + y^2 = 2x \Rightarrow r = 2 \cos \theta $$

therefore the integral should be

$$\ 4 \cdot \int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ r \ dr \ d \theta$$

but this integral is way to messy for me to calculate. I mean first step of calculating integral for $\ r $ is okay but then the value I get and calculating integral of $\ \theta $ is beyond me and I believe it is beyond the scope of the course. I guess I'm missing something in the process here?

Answer 1

You've correctly set up the integral representing the volume of the inside cylinder. The scalar $4$ on the left represents

1. the reflectional symmetry about the $xy$-plane; and
2. the reflectional symmetry about the $yz$-plane.

\begin{align} & 4 \int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ r \ dr \ d \theta \\ =& 2 \int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \, d(r^2) \ d \theta \\ =& 2 \int_0^{\pi/2} \left[- \frac{(4-r^2)^{3/2}}{3/2} \right]_0^{2\cos\theta} \, d(r^2) \ d \theta \\ =& \frac{32}{3} \int_0^{\pi/2} (1-\sin^3\theta) \, d \theta \\ =& \frac{32}{3} \cdot \frac{\pi}{2} - \frac{32}{3} \cdot \frac{2}{3} \\ =& \frac{16(3\pi - 4)}{9} \end{align}

The second last equality is due to Wallis's integrals.

Answer 2

Let $I = \int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ r \ dr \ d \theta$

$I = \frac{-1}{2}\int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ d(4-r^2) \ d \theta$

$I = \frac{-1}{2}.\frac{2}{3}\int_0^{\pi/2}\big[(4-r^2)^{3/2}\big]^{2cos\theta}_0d\theta$

$I = \frac{-1}{2}.\frac{2}{3}\int_0^{\pi/2}\big[(4-4cos^2\theta)^{3/2} - 4^{3/2}\big]d\theta$

$I = \frac{-1}{3}.2^3\int_0^{\pi/2}\big[(sin^2\theta)^{3/2} - 1\big]d\theta$

$I = \frac{-1}{3}.2^3\int_0^{\pi/2}\big[(sin^3\theta)- 1\big]d\theta$

$\bigg[ \int^{\pi/2}_0sin^mxdx = \frac{m-1}{m}.\frac{m-3}{m-2}...1 \bigg]$, if m is odd

$I = \frac{-2^3}{3}\bigg[\big[\frac{2}{3}.1\big] -\frac{\pi}{2}\bigg]$

Thus, $I = \frac{8}{18}[3\pi - 4] = \frac{4}{9}[3\pi - 4]$

Can you proceed now?

Answer 3

Let's call $V_o$=volume outside cylinder and $V_i$=volume inside cylinder. Then volume you request is volume ball ($\frac{4}{3}\pi\,2^3$) minus$V_i$, that is, $V_o=\frac{32}{3}\pi-V_i$.

By symmetry

$$V_i=2\int\int_{\{(x,y)\,|\,x^2+y^2=2x\}}\int_{z=0}^{z=\sqrt{4-x^2-y^2}}1dz\,dx\,dy$$

Doing the first integral and changing to polar variables and even function in $\theta$ variable we obtain

$$V_i=4\int_{0}^{\pi/2}\int_0^{2\,\cos\theta}r\,\sqrt{4-r^2}\,dr\,d\theta=\frac{16(3\pi - 4)}{9}$$

as colleagues posted before. Finally, volume desired is

$$V_o=\frac{32}{3}\pi-\frac{16(3\pi - 4)}{9}=\frac{4}{9}\left(12\,\pi+1\right)$$

Answer 4

Of course I cannot comment directly about the requirements for your class, but in my Vector Calculus class I would expect the students to be able to calculate that integral without much issue.

You have \begin{align} \int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ r \ dr \ d \theta &=-\frac13\int_0^{\pi/2} [(4-4\cos^2\theta)^{3/2}- 4^{3/2}]\,d\theta\\ &=\frac83\,\int_0^{\pi/2}[1-(1-\cos^2\theta)^{3/2}]\,d\theta\\ &=\frac83\,\int_0^{\pi/2}[1-\sin^3\theta]\,d\theta\\ &=\frac83\,\left[\frac\pi2-\int_0^{\pi/2}(1-\cos^2\theta)\,\sin\theta\,d\theta \right]\\ &=\frac83\,\left[\frac\pi2-1-\left.\frac{\cos^3\theta}3\right|_0^{\pi/2} \right]\\ &=\frac83\,\left[\frac\pi2-1-\frac13(0-1) \right]\\ &=\frac83\,\left[\frac\pi2-\frac23\right]\\ &=\frac49(3\pi-4) \end{align}