Chapter 6.2, Problem 52E

To determine

Find the volumeof the remaining portion of the sphere.

Answer to Problem 52E

The volume of the sphere is $-\frac{4}{3}\pi {\left(R^2-r^2\right)}^{\frac{3}{2}}$

Explanation of Solution

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

Calculation:

The washer method, to integrate along the $y-axis$ ,subtracting the volume between the

Red lines from the volume of the sphere. Use symmetry and double the integral from $y=0\text{ to }y=\sqrt{R^2-r^2}$ .the upper limits is found by Pythagorean Theorem on the right

Triangle in the picture.

  

To solve the circle equation for $x$ .where is the positive square root represents the right half of the circle

  $\begin{array}{l}{R}^2=x^2+y^2\\ x^2=R^2-y^2\\ x=\sqrt{R^2-y^2}\\ r_1=\sqrt{R^2-y^2}\left(\text{outer radius}\right)\\ r_2=r\left(\text{inner radius}\right)\end{array}$

Now, to integrate with washer formula:

  $\begin{array}{l}V=\pi {\displaystyle \underset{a}{\overset{b}{\int }}{r}_1{}^2-r_2{}^2}dx\left(\text{put outer and inner radius}\right)\\ =2\pi {\displaystyle \underset{0}{\overset{\sqrt{R^2-y^2}}{\int }}\left(\left(\sqrt{R^2-y^2}\right)-\left(r^2\right)\right)}dy\\ =2\pi {\displaystyle \underset{0}{\overset{\sqrt{R^2-y^2}}{\int }}\left(R^2-y^2-r^2\right)}dy\\ =2\pi {\left[\left(R^2-r^2\right)y-\frac{1}{3}{y}^3\right]}_0^{\sqrt{R^2-y^2}}\\ =2\pi \left[\left(R^2-r^2\right)\left(\sqrt{R^2-r^2}\right)-\frac{1}{3}{\left(\sqrt{R^2-r^2}\right)}^3-\left(0-0\right)\right]\\ =2\pi \left[{\left(R^2-r^2\right)}^{\frac{3}{2}}-\frac{1}{3}{\left(R^2-r^2\right)}^{\frac{3}{2}}\right]\\ =2\pi \left(\frac{2}{3}{\left(R^2-r^2\right)}^{\frac{3}{2}}\right)\\ =\frac{4}{3}{\left(R^2-r^2\right)}^{\frac{3}{2}}\\ =\frac{5}{12}\pi r^3\end{array}$

Use the cylinder method:

  $\begin{array}{l}{R}^2=x^2+y^2\\ Y^2=R^2-x^2\\ y=\sqrt{R^2-x^2}\\ y=2\sqrt{R^2-x^2}\end{array}$

Now, volume is :

  $\begin{array}{l}V=2\pi {\displaystyle \underset{r}{\overset{R}{\int }}\left(x\right)\left(2\sqrt{R^2-r^2}\right)}dx\\ \text{ =4}\pi {\left[-\frac{1}{3}{\left(R^2-r^2\right)}^{\frac{3}{2}}\right]}_r^R\\ =-\frac{4}{3}\pi {\left(R^2-r^2\right)}^{\frac{3}{2}}\end{array}$

Thus, the volume of the sphere is $-\frac{4}{3}\pi {\left(R^2-r^2\right)}^{\frac{3}{2}}$