Chapter 7.2, Problem 40E

To determine

To find: The area of the region which is enclosed by the given curve and line $(a)$ in the term of $x$ and $(b)$ in the term of $y$ .

Answer to Problem 40E

The area:

  $\begin{array}{c}(a)A=10.667\\ (b)A=10.667\end{array}$

Explanation of Solution

Given information:

The curve is $y=3-x^2$

And line is $y=-1$ .

Calculation:

  $(a)$

The limits of integration can be discovered by first locating the spots where these two functions overlap. Consequently, the equation below:

  $\begin{array}{c}3-x^2=-1\\ \Rightarrow x^2=4\\ \Rightarrow x=-2\text{ or }2\end{array}$

Notice that, $3-x^2\ge -1$ for each $x\in [-2,2]$ .

So, the area of the region is:

  

  

Let express the given curve in terms of $y$ .

Thus, $x=\sqrt{3-y}$

And

  $x=-\sqrt{3-y}$

For each

  $y\in [-1,3]$

Hence, the area of the region is:

  

Therefore, the required area of the region which is enclosed by the given curve and line $(a)$ in the term of $x$ is $A=10.667$ and

  $(b)$ in the term of $y$ is $A=10.667$ .