Integral of sqrt(1-r^2)*r dr

1. Let $u = 1 - r^{2}$.

   Then let $du = - 2 r dr$ and substitute $- \frac{du}{2}$:

   $$\int \left(- \frac{\sqrt{u}}{2}\right)\, du$$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \sqrt{u}\, du = - \frac{\int \sqrt{u}\, du}{2}$$

      1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int \sqrt{u}\, du = \frac{2 u^{\frac{3}{2}}}{3}$$

      So, the result is: $- \frac{u^{\frac{3}{2}}}{3}$

   Now substitute $u$ back in:

   $$- \frac{\left(1 - r^{2}\right)^{\frac{3}{2}}}{3}$$

2. Add the constant of integration:

   $$- \frac{\left(1 - r^{2}\right)^{\frac{3}{2}}}{3}+ \mathrm{constant}$$

The answer is:

$$- \frac{\left(1 - r^{2}\right)^{\frac{3}{2}}}{3}+ \mathrm{constant}$$