Chapter 7.2, Problem 30E

To determine

To find: The area of the region enclosed by the lines and curve.

Answer to Problem 30E

The area is: $A=\frac{4}{3}-\frac{4}{\pi }$

Explanation of Solution

Given information:

The curves are:

  $\begin{array}{l}y=\cos \left(\frac{\pi x}{2}\right)\\ y=1-x^2\end{array}$

Calculation:

The graph is shown below:

  

The red line is the graph of $y=\cos \left(\frac{\pi x}{2}\right)$ that is graph of

  $y=1-x^2$

between

  $x=\pm 1$ .

The $\text{y}$ -axis is symmetrical for both functions. As a result, calculate the first quadrant's area between curves and multiply it by two to obtain the overall area. The evaluation of the integral is slightly facilitated by doing this. Also take note of the intersection of the functions at $x=\pm 1$ .

Hence, the area is:

  

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

Therefore, the required area of the region enclosed by the lines and curve is $\frac{4}{3}-\frac{4}{\pi }$ .