Chapter 6.1, Problem 13E

To determine

The MRAM sum that approximate the volume of the sphere.

Answer to Problem 13E

The estimated volume is to be 523.6.

Explanation of Solution

Given information:

The Volume of a sphere of radius is 5. Use n = 10, 20, 40, 80, and 160.

Formula used:

The volume of the sphere is $\pi r^{2}h$

Calculation:

Surface of sphere were generated by revolving the graph of the function $f\left(x\right)=\sqrt{25-x^2}$ about the x-axis. The interval $-5\le x\le 5$ into n subintervals of equal length $\Delta x=\frac{10}{n}$ . The slice the sphere with planes perpendicular to the x-axis at the partition points, cutting it like a round loaf of bread into n parallel slices of width $\Delta x$ .

When n is large, each slice can be approximated by a cylinder of volume $\pi r^{2}h$ . In this case h is $\Delta x$ while r varies according to where we are on the x-axis. A logical radius to choose for each cylinder is $f\left(m_i\right)=\sqrt{25-m_i^2}$ where mi is the midpoint of the interval where the $i^{th}$ , slice intersects the x-axis.

Now approximate the volume of the sphere by using MRAM to sum the cylinder volumes,

  $\pi r^{2}h=\pi \left(25-m_i^2\right)\Delta x$

The function we use in the RAM program is

  $\pi {\left(\sqrt{25-m^2}\right)}^2=\pi \left(25-x^2\right)$

The interval is [-5, 5].

  

So, estimated volume is to be 523.6

Conclusion:

The estimated volume is to be 523.6