Chapter 6.2, Problem 17E

To determine

Find the volume enclosed by the given curves.

Answer to Problem 17E

The volume of the solid is $\frac{13\pi }{30}$

Explanation of Solution

Given information: Given curves $y=x^3,y=\sqrt{x}$

Calculation:

First convert both function in terms of x since were reflect across the $y-axis$

  $\begin{array}{l}y=x^3\\ x=y^{\frac{1}{3}}\\ y=\sqrt{x}\\ x=y^2\end{array}$

Now graph the function and insert your horizontal disk for reference.

  

Notice that the function for the large radius of the

Disk is $1-y^2$ and the function for the radius is $1-y^{1/3}$

Put these in the area formula and subtract the

The smaller area from the large area to get our function

For the area of the disk given any $y$ between $0\text{ and }1$ .

  $\begin{array}{l}A=\pi r^2\\ =\pi {\left(1-y^2\right)}^2-\pi {\left(1-y^{\frac{1}{3}}\right)}^2\\ =\pi \left(y^4-2y^2+2y^{\frac{1}{3}}-y^{\frac{2}{3}}\right)\\ =\pi \left(y^4-2y^2+2y^{\frac{1}{3}}-y^{\frac{2}{3}}\right)\Delta y\end{array}$

Now take the integral keep in mind that the $\Delta y$ Becomes infinity small as we integrate thus allow

Us to express it as out $dy$ in the integrate.

  $\begin{array}{l}V=\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left(y^4-2y^2+2y^{\frac{1}{3}}-y^{\frac{2}{3}}\right)}dy\\ =\pi {\left(\frac{1}{5}{y}^5-\frac{2}{3}{y}^3+\frac{3}{2}{y}^{\frac{4}{3}}-\frac{3}{5}{y}^{\frac{5}{3}}\right)}_0^1\\ =\pi \left[\frac{1}{5}-\frac{2}{3}+\frac{3}{2}-\frac{3}{5}\right]\\ =\frac{13\pi }{30}\end{array}$

Thus the volume of the solid is $\frac{13\pi }{30}$