Chapter 6, Problem 10RE

(a)

To determine

The volume of the solid.

Answer to Problem 10RE

The total volume is $\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\pi {\sin }^2{m}_i\left(\Delta x\right)}$

Explanation of Solution

Given information:

The curve y = sin x is revolved around the x-axis

Formula used:

Calculation:

Surface of sphere were generated by revolving the graph of the function “f (x) = sin x" about the x-axis. Partition the interval into n subintervals of equal length “ $\Delta x$ ". Then 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 our case h is $\Delta x$ while r varies according to where are on x-axis. A logical radius to choose for each cylinder is $f\left(m_i\right)=\sin \left(m_i\right)$ , Where m_iis the midpoint of the interval where the i^th, slice intersects the x-axis.

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

  $\begin{array}{l}\text{volume of one cylinder}=\pi r^{2}h\\ =\pi {\left(\sin \left(m_i\right)\right)}^2\Delta x\\ \text{volume of one cylinder}=\pi {\sin }^2{m}_i\left(\Delta x\right)\end{array}$

Now, using Riemann sum approximation

  $\text{total volume}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\pi {\sin }^2{m}_i\left(\Delta x\right)}$

Conclusion:

The total volume is $\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\pi {\sin }^2{m}_i\left(\Delta x\right)}$

(b)

To determine

The volume using NINT.

Answer to Problem 10RE

The value is ${\displaystyle \underset{0}{\overset{\pi }{\int }}\left(\pi {\sin }^{2}x\right)dx=4.9348}$

Explanation of Solution

Given information:

The curve y = sin x is revolved around the x-axis

Formula used:

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=\text{NINT}\left(f\left(x\right),x,a,b\right)$

Calculation:

Calculator has a numerical integration capability, which will denote as NINT. In particular, use $\text{NINT}\left(f\left(x\right),x,a,b\right)$ to denote a calculator approximation of ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ . When write

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=\text{NINT}\left(f\left(x\right),x,a,b\right)$

The understanding that the right-hand side of the equation is an approximation of the left-hand side

Now, use above expression with this function

  $\begin{array}{l}f\left(x\right)=\pi {\sin }^{2}x\\ {\displaystyle \underset{0}{\overset{\pi }{\int }}\left(\pi {\sin }^{2}x\right)dx=\text{NINT}\left(\pi {\sin }^{2}x,x,0,\pi \right)}\\ {\displaystyle \underset{0}{\overset{\pi }{\int }}\left(\pi {\sin }^{2}x\right)dx=4.9348}\end{array}$

Conclusion:

The value is ${\displaystyle \underset{0}{\overset{\pi }{\int }}\left(\pi {\sin }^{2}x\right)dx=4.9348}$