Chapter 6.1, Problem 10E

To determine

To sketch: the region enclosed by the given curves and decides whether to integrate with respect to x or y . Draw a typical approximating rectangle and label its height and width. Then find the area of the region.

Answer to Problem 10E

The area of the shaded region is 21.3333.

Explanation of Solution

Given information: The function of the given curves is:

  $4x+y^2=12,x=y.$

Calculation:

The sketch of the region enclosed by the given curves is given below.

  

The top and bottom boundary functions of the shaded region are:

  $x_T=\frac{12-y^2}{4}\text{and}{x}_B=y.$

So a typical rectangle will have a height of,

  $x_T-x_B=\frac{12-y^2}{4}-y.$

To find the area of the shaded region, integrate between y −values, y = -6 and y = 2.

  $\begin{array}{c}{\displaystyle \underset{-6}{\overset{2}{\int }}(\frac{12-y^2}{4}-y)dy}={\displaystyle \underset{-6}{\overset{2}{\int }}(\frac{12-y^2}{4})dy}-{\displaystyle \underset{-6}{\overset{2}{\int }}ydy}\\ =\frac{1}{4}{\displaystyle \underset{-6}{\overset{2}{\int }}12dy}-\frac{1}{4}{\displaystyle \underset{-6}{\overset{2}{\int }}{y}^{2}dy}-{\displaystyle \underset{-6}{\overset{2}{\int }}ydy}\\ =3y|\begin{array}{c}2\\ -6\end{array}-\frac{1}{4}\frac{y^3}{3}|\begin{array}{c}2\\ -6\end{array}-\frac{y^2}{2}|\begin{array}{c}2\\ -6\end{array}\\ =3(2)-3(-6)-\frac{1}{4}\frac{2^3}{3}+\frac{1}{4}\frac{{(-6)}^3}{3}-\frac{2^2}{2}+\frac{{(-6)}^2}{2}\\ =6+18-\frac{2}{3}-\frac{216}{12}-2+\frac{36}{2}\\ =24-\frac{2}{3}-18-2+18\\ =21.3333\end{array}$

The area of the shaded region is 21.3333.