Disk method to find volume of solid revolved around x axis

Question

Use the disk method to find the volume of the solid generated when the region bounded by $\displaystyle y=\frac1{(1-100x^2)^\frac14}$, $y=0$, $x=0$, and $\displaystyle x=\frac1{20}$ is revolved around the x-axis

Answer 1

We get immediately that the volume is equal to $$\int_0^{1/20} \frac{\pi}{\sqrt{1-100x^2}}\,dx.$$ For the integration, I suggest making the change of variable $10x=u$. Then $dx=\frac{1}{10}\,du$. We end up needing the integral $$\int_{u=0}^{1/2} \frac{\pi}{10\sqrt{1-u^2}}\,du.$$

Note that $\int \frac{du}{\sqrt{1-u^2}}=\arcsin u +C$.