If we place the sphere so that its center is at the origin, then the plane $P_x$ intersects the sphere in a circle whose radius is $y = \sqrt{r^2-x^2}$.So the cross-sectional area is $$A(x) = \pi y^2 = \pi (r^2 - x^2)$$
Where I get lost is in the integration. How does $2\pi\int_{0}^{r} (r^2-x^2)dx$ integrate to $2\pi \bigg[ r^2x - \frac{x^3}{3}\bigg]_{0}^{r}$ ? In my mind it integrates to $2\pi \bigg[ \frac{1}{3} r^3 - \frac{1}{3}x^3\bigg]_{0}^{r}$
What am I overlooking here? I am really confused why $r^2 \to r^2x$ if it's just a simple power series integration.
$\endgroup$ 42 Answers
$\begingroup$As was mentioned in the comments, $r$ is a constant, so $$\int_0^r(r^2-x^2)dx=r^2\int_0^rdx-\int_0^rx^2dx=r^3-\frac{r^3}3=\frac23r^3$$
$\endgroup$ $\begingroup$Assume area of spherical shell is $A = 4\pi r^2$. Expand the shell by thickness $dr$. Then $dA = 4\pi r^2 dr$. Integrate from 0 to R. Get $\frac{4}{3}R^3$.
$\endgroup$ 1
