Chapter 7, Problem 13RE

To determine

To calculate: The area of the region

Answer to Problem 13RE

The area of the region is $8.90226$

Explanation of Solution

Given information:

  $y=\cos x,$

  $y=4-x^2$

Calculation:

To find the area of the region which is enclosed by the cosine curve $y=\cos x$ and by the parabola $y=4-x^2$ .

The sketch of these graphs will help us to more easily spot a given problem.

Let’s sketch the graphs into Desmos. Further by entering the code, see in the red rectangle below, mark the required region with the color.

  

By observing the given picture, notice that our area is bounded above with the $y=4-x^2$ and below with the curve $y=\cos x$ from $x=-2.12813$ to $x=2.12813$ . Let’s show how to get the limits of integration (approximately).

The points of intersection i.e. the limits of integration get by equalizing:

  $\cos x=4-x^2$

Since there is no simple method for finding solutions to this equation, use the Wolfram Alpha. By entering the

  $x^2+\cos x-4=0$

Into program the required values is

  $x\approx -2.12813$

Let’s find the area of the green region. Therefore

  $\text{Area}={\displaystyle {\int }_{-2.12813}^{2.12813}4-x^2-\cos x\text{d}x}$

To solve this obtained integral more simple, note that area is symmetric with the respect to the $y$ -axis and multiply area by 2, where lower limit is now $0$ .

Hence,

  $\text{Area}=2{\displaystyle {\int }_0^{2.12813}4-x^2-\cos x\text{d}x}$

So let’s proceed with the calculation by using the Power formula and Trigonometric formula:

  $\begin{array}{c}\text{ Area}=2{\displaystyle {\int }_0^{2.12813}4-x^2-\cos x\text{d}x}\\ =2\cdot 4x-\frac{x^3}{3}-{\left.\sin x\right|}_0^{2.12813}\\ \approx 2\cdot (8.51252-3.21272-0.848668)-0\end{array}$

The area of the region equals

  $\text{ Area }\approx 8.90226$

Where used previously that

  $\sin (2.12813)=0.848668$