Question

Find the volume of the solid obtained by rotating the region enclosed by the curves y=x^(3/2) and y = x^2 about the x-axis

Answer 1

The formulas intercept at 0 and 1 so that is the range. Per the washer method, the outer radius is y=x^(3/2). The inner radius is y=x^2.
Solving using an intergral we get:
\[\pi \int\limits_{0}^{1}[(x^{\frac{3}{2}})^2-(x^2)^2]dx=\pi \int\limits_{0}^{1}[x^{3}-x^4]dx\]
\[\pi [\frac{1}{4}x^4-\frac{1}{5}x^5]_{0}^{1}=\pi [\frac{1}{4}-\frac{1}{5}]=\frac{\pi}{20}\]