Solve ∫ (from r to sqrt { alpha ^ 2) of - r}r wrt z

$$\int _ { r } ^ { \sqrt { \alpha ^ { 2 } - r } } r d z$$

Evaluate

$$r\left(\sqrt{\alpha ^{2}-r}-r\right)$$

$$r\leq \alpha ^{2}$$

Solution Steps

Evaluate the indefinite integral first.

$$\int r\mathrm{d}z$$

Find the integral of r using the table of common integrals rule \int a\mathrm{d}z=az.

$$rz$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

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

Simplify.

$$r\left(\sqrt{\alpha ^{2}-r}-r\right)$$

Differentiate w.r.t. r

$$\frac{\alpha ^{2}}{\sqrt{\alpha ^{2}-r}}-2r-\frac{3r}{2\sqrt{\alpha ^{2}-r}}$$