Chapter 6.3, Problem 17E

To determine

Find the volume generated by the given curves.

Answer to Problem 17E

The volume of the curve is $\frac{5}{14}\pi$

Explanation of Solution

Given information: Given curves $y=x^3,y=0,x=1;\text{about }y=1$

Calculation:

The region is shown in the picture:

  

Now, rotate the region about $y=1$ ,

  

Circumference is $2\pi \left(1-y\right)$ . Height is $1-y^{\frac{1}{3}}.$

To find the volume to find: $V={\displaystyle \underset{0}{\overset{1}{\int }}2\pi \left(1-y\right)\left(1-y^{\frac{1}{3}}\right)dy}$

  $\begin{array}{l}V={\displaystyle \underset{0}{\overset{1}{\int }}2\pi \left(1-y\right)\left(1-y^{\frac{1}{3}}\right)dy}\\ =2\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left(1-y\right)\left(1-y^{\frac{1}{3}}\right)dy}\\ =2\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left(1-y^{\frac{1}{3}}-y+y^{\frac{3}{4}}\right)dy}\\ =2\pi \left[{\left[y\right]}_0^1-{\left[\frac{y^{\frac{4}{3}}}{\frac{4}{3}}\right]}_0^1-{\left[\frac{y^2}{2}\right]}_0^1+{\left[\frac{y^{\frac{7}{3}}}{\frac{7}{3}}\right]}_0^1\right]\text{ Do partial intigration}\\ =2\pi \left[1-\frac{3}{4}-\frac{1}{2}+\frac{3}{7}\right]\\ =2\pi -\frac{3}{4}\pi -\pi +\frac{6}{7}\pi \\ =\frac{5}{14}\pi \end{array}$

Thus, the volume of the curve is $\frac{5}{14}\pi$ .