Chapter 7.3, Problem 79E

To determine

To Derive: The volume of a hemisphere is $\frac{2}{3}\pi R^3$ .

Answer to Problem 79E

The volume of a hemisphere $\frac{2}{3}\pi R^3$ is derived.

Explanation of Solution

Given information: The radius of hemisphere is R. The radius and height of circular cylinder and cone is R.

Calculation:

The hemisphere is generated when the area is bounded by the following equation;

  $x^2+y^2=R^2$

  $y^2=R^2-x^2$

Now, use disk method with radius r to find out the integral volume:

  $r=y=\sqrt{R^2-x^2}$

  $\begin{array}{l}dv=\pi r^{2}dx\\ dv=\pi (R^2-x^2)dx\end{array}$

Now, to find the volume within the hemisphere, integrate the above differential volume with the limits $0\le x\le R$ .

  $v={\displaystyle \underset{0}{\overset{R}{\int }}\pi (R^2-x^2)dx}$

Now, find out the volume of solid which is cylinder from which cone is removed. Use Washer method to find the volume.

Suppose, outer radius (R₂) is R and inner radius (R₁) is x. Then, differential volume of a solid is;

  $\begin{array}{l}dv=\pi (R_2^2-R_1^2)dx\\ dv=\pi (R^2-x^2)dx\end{array}$

Now, integrate this differential volume with the limits $0\le x\le R$ :

  $v=\pi {\displaystyle \underset{0}{\overset{R}{\int }}(R^2-x^2)dx}$

From this, it is clear that volume within the hemisphere is same as that of volume of solid.

Now, volume of cone.

  $\begin{array}{l}{v}_{cone}=\frac{1}{3}\pi r^{2}h\\ v_{cone}=\frac{1}{3}\pi R^2(R)\\ v_{cone}=\frac{1}{3}\pi R^3\end{array}$

Volume of cylinder:

  $\begin{array}{l}{v}_{cyl}=\pi r^{2}h\\ v_{cyl}=\pi R^2(R)\\ v_{cyl}=\pi R^3\end{array}$

Hence, volume of solid can be find out as;

  $\begin{array}{l}{v}_s=v_{cyl}-v_{cone}\\ v_s=\pi R^3-\frac{1}{3}\pi R^3\\ v_s=\frac{2}{3}\pi R^3\end{array}$

Therefore, volume within the hemisphere:

  $v=\frac{2}{3}\pi R^3$