Evaluation of $\int (r^2-x^2) dx$

Question

I am learning how to approximate the volume of a sphere with integrals. I am confused on how $(1/3)x^3$ became $(1/3)r^3$, and why this being an even function allows you to pull a $2$ out front. Please help, many thanks

Answer 1

The volume is $$\int_{-r}^r \pi(r^2-x^2)dx=[\pi(r^2x-\tfrac{1}{3}x^3)]_{-r}^r=\tfrac{4\pi}{3}r^3.$$Since the integrand is even, we could alternatively write $$\int_{-r}^r \pi(r^2-x^2)dx=\int_0^r 2\pi(r^2-x^2)dx=[2\pi(r^2x-\tfrac{1}{3}x^3)]_0^r=\tfrac{4\pi}{3}r^3.$$

Answer 2

$$F(x)|_{x=0}^{x=r}=F(r)-F(0)$$ (by definition of the left term)

$$f(-x)=f(x) \implies \int_{-r}^{r}f(x)\ dx=2\int_{0}^{r}f(x)\ dx$$ (consider the change of the lower bound too!)

Answer 3

The volume of sphere is found by $$ V=2\int _0^r \pi (r^2-x^2) dx= 2\pi \int _0^r (r^2-x^2) dx$$

When you integrate you get $$r^2x -(1/3) x^3$$ and when you evaluate from $0$ to $r$ you get $$(2/3) r^3$$

Thus the total volume will be $$V= (4/3) \pi r^3 $$