Chapter 6.3, Problem 8E

To determine

Find the volume enclosed by the given curves.

Answer to Problem 8E

The volume of the curve is $\frac{3}{10}\pi$

Explanation of Solution

Given information: Given curves $y=x^2,y=\sqrt{x}$

Calculation:

The region is shown in the picture:

  

Now, rotate the region about $y-$ axis get:

  

Circumference is $2\pi x$ . Height is $\sqrt{x}-x^2.$

To find the volume to find:

  $\begin{array}{l}V={\displaystyle \underset{0}{\overset{1}{\int }}\left(2\pi x\right)\left(\sqrt{x-x^2}\right)}dx\\ =2\pi {\displaystyle \underset{0}{\overset{1}{\int }}x\left(x^{\frac{1}{2}}-x^2\right)dx}\\ =2\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left(x^{\frac{3}{2}}-x^3\right)dx}\\ =2\pi \left[{\left[\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right]}_0^1-{\left[\frac{x^4}{4}\right]}_0^1\right]\\ =2\pi \left[\frac{2}{5}-\frac{1}{4}\right]\\ =2\pi \left[\frac{8}{20}-\frac{5}{20}\right]\\ =2\pi \frac{3}{20}\\ =\frac{3}{10}\pi \end{array}$

Thus, the volume of the curve is $\frac{3}{10}\pi$