Chapter 8.2, Problem 30E

To determine

To Find:

To find the area of the given regions enclosed by the lines $y=\cos \left(\frac{\pi x}{2}\right)$ and $y=1-x^2$ .

Answer to Problem 30E

The area of the given regions enclosed by the lines is $\frac{4}{4}-\frac{4}{\pi }$ .

Explanation of Solution

Given information:

The given function is $y=\cos \left(\frac{\pi x}{2}\right)$ and $y=1-x^2$ .

The red line is the graph of $y=\cos \left(\frac{\pi x}{2}\right)$ which is below the graph of $y=1-x^2$ between $x=\pm 1$

  

Both the functions are symmetrical across the y-axis. Therefore, we can find the area between the curves in the first quadrant and multiply it by 2 to get the total area.

Also, the functions intersects at $x=\pm 1$ .

If a region is bounded by a right curve $x_R=g_{L\left(y\right)}$ and a left curve $x_L=g_{R\left(y\right)}$ from $y=c$ to $y=d$ , then the area of this region is

  $A={\displaystyle \underset{c}{\overset{d}{\int }}\left(x_R-x_L\right)}dy$

  $A=2{\displaystyle \underset{0}{\overset{1}{\int }}\left(1-x^2-\cos \left(\frac{\pi x}{2}\right)\right)dx}$

  $\begin{array}{l}A=2{\left[x-\frac{x^3}{3}-\frac{2}{\pi }\sin \left(\frac{\pi x}{2}\right)\right]}_0^1\\ A=2\left(1-\frac{1}{3}-\frac{2}{\pi }\sin \frac{\pi }{2}-\left(0-0-\frac{2}{\pi }\sin 0\right)\right)\\ A=2\left(\frac{2}{3}-\frac{2}{\pi }\right)\\ A=\frac{4}{3}-\frac{4}{\pi }\end{array}$

Therefore, The area of the given regions enclosed by the lines is $\frac{4}{4}-\frac{4}{\pi }$ .