Chapter 6, Problem 1CRE

To determine

Compute the area of the region in Figure 1(A) enclosed by $y=2-x^2$ and $y=-2$.

Answer to Problem 1CRE

Solution:

The area of the enclosed region is $\frac{32}{3}$.

Explanation of Solution

Given:

The figure is:

The functions are $y=2-x^2\text{and}y=-2$.

Formula used:

Area of the region $A={\displaystyle {\int }_a^b\left[\left(\text{Upper}\text{function}\right)-\left(\text{Lower}\text{function}\right)\right]dx}$

Where $a\le x\le b$

Calculation:

The functions are $y=2-x^2\text{and}y=-2$.

The graphs of $y=2-x^2\text{and}y=-2$ intersect where

$\begin{array}{c}2-x^2=-2\\ -x^2=-2-2\\ -x^2=-4\\ x^2=4\end{array}$

Square root on both sides.

$\begin{array}{c}\sqrt{x^2}=\sqrt{4}\\ x=\pm 2\end{array}$

Thus, the enclosed area lies over the interval $\left[-2,2\right]$.

The region enclosed by the graphs lies below $y=2-x^2$ and above $y=-2$.

Now, find the area.

$\begin{array}{c}\text{Area}={\displaystyle {\int }_{-2}^2\left[\left(2-x^2\right)-\left(-2\right)\right]dx}\\ ={\displaystyle {\int }_{-2}^2\left[2-x^2+2\right]dx}\\ ={\displaystyle {\int }_{-2}^2\left(4-x^2\right)dx}\end{array}$

Solve the integral.

${\displaystyle {\int }_{-2}^2\left(4-x^2\right)dx}={\left[4x-\frac{x^3}{3}\right]}_{-2}^2$

Substitute the limits.

$\begin{array}{c}{\left[4x-\frac{x^3}{3}\right]}_{-2}^2=4\left(2\right)-\frac{{\left(2\right)}^3}{3}-4\left(-2\right)+\frac{{\left(-2\right)}^3}{3}\\ =8-\frac{8}{3}+8-\frac{8}{3}\\ =\frac{24-8+24-8}{3}\\ =\frac{32}{3}\end{array}$

Hence, the area of the enclosed region is $\frac{32}{3}$.

Conclusion:

With the help of the formula, we could figure out the area of the enclosed region.