I know that the integral for the volume of a sphere has been done many times on this site, and I have even done it myself using a different method in class. I know there are better ways, but I want to know why the way I'm about to show is giving me an incorrect answer.
Consider the sphere as the infinite sum of circular discs, of area $\pi r^2$, for some $r$ varying from the tip of the sphere ~ conceptually at $(0, 0, R)$ ~ to the base, at $(0, 0, -R)$, with $R$ being the full radius of the sphere. Letting the angle swept from the $z$ axis to the point on the sphere's surface that also lies on the circumference of one of these discs be $\theta$, the radius of our element discs $r$ would then be $R \cdot \sin(\theta)$, and therefore:
\begin{align*}V &= \int_0^\pi \pi (R\cdot \sin(\theta))^2 \space d\theta \\
V &= \pi R^2\int_0^\pi \sin^2(\theta) \space d\theta \\
V &= \pi R^2\int_0^\pi \frac{1}{2}(1 – \cos(2\theta))\space d\theta \\
V &= \pi R^2 \cdot \left[\frac{1}{2}\theta – \frac{\sin(2\theta)}{4}\space \right]_0^\pi\end{align*}
which is clearly false, and not at all $\frac{4}{3} \cdot \pi r^3$!
Where's my mistake? Thanks.
Best Answer
Let's take a horizontal disk. Its radius is $r$, and its thickness is $dt$.
Thus, its volume is $dV=\pi r^2 dt$
The total volume of the sphere is $V= \int{\pi r^2 dt}$
Now, taking angle $\theta$ counter-clockwise
$r= R cos\theta \;\Rightarrow dr= -Rsin\theta d\theta$
Notice $dt^2+dr^2=(Rd\theta)^2 \;\Rightarrow \;dt = \sqrt{R^2d\theta^2 - R^2sin^2\theta d\theta^2} \;=\;Rd \theta \sqrt{1-sin^2 \theta} \;=\;Rcos\theta d \theta$
So, $$V= 2 \int _0 ^{\pi/2}{\pi R^2·cos^2(\theta)·R·cos(\theta)\;d\theta }$$ $$V= 2 \pi R^3 \int _0 ^{\pi/2}{cos^3(\theta)\;d\theta} \;=\; 2 \pi R^3 \int _0 ^{\pi/2} \left( \frac{3cos \theta + cos 3\theta}{4} \right) d \theta$$
$$V= \frac{1}{2} \pi R^3 \left[3sin \theta + \frac {1}{3} sin 3 \theta \right]_0 ^{\pi/2}$$
$$V= \frac{1}{2} \pi R^3 \left[3- \frac{1}{3}- 0 -0\right] \;\;=\;\frac{1}{2} \pi R^3 \left[\frac{8}{3} \right] \;\;=\; \frac{4}{3} \pi R^3$$