Chapter 7, Problem 10RE

To determine

To determine the area of the region enclosed by the lines and curves.

Answer to Problem 10RE

  $\text{Area}=30.375$

Explanation of Solution

Given:

The given equations are:

  $4x=y^2-4\text{ and }4x=y+16$

Calculation:

The limits of integration are found by solving the equations.

  $\begin{array}{c}{y}^2-4=y+16\\ y^2-y-4-16=0\\ y^2-y-20=0\\ y^2-5y+4y-20=0\\ y\left(y-5\right)+4\left(y-5\right)=0\\ \left(y-5\right)\left(y+4\right)=0\\ y-5=0\text{ and }y+4=0\\ y=5\text{ and }y=-4\end{array}$

The solutions are $y=-4\text{ and }y=5$ and the limits are $\left[-4,5\right].$

Rewrite the given equations as:

  $\begin{array}{c}4x=y^2-4\\ x=\frac{y^2-4}{4}\\ x=\frac{y^2}{4}-1\end{array}$

And

  $\begin{array}{c}4x=y+16\\ x=\frac{y+16}{4}\\ x=\frac{y}{4}+4\end{array}$

The region enclosed by the line and curve is shown as:

  

Now, calculating the area between curves is given by:

  $\begin{array}{l}A={\displaystyle \underset{a}{\overset{b}{\int }}f\left(y\right)-g\left(y\right)dy}\\ A={\displaystyle \underset{-4}{\overset{5}{\int }}\left[\left(\frac{y}{4}+4\right)-\left(\frac{y^2}{4}-1\right)\right]}dy\\ A={\displaystyle \underset{-4}{\overset{5}{\int }}\left(\frac{y}{4}+4-\frac{y^2}{4}+1\right)dy}\\ A={\displaystyle \underset{-4}{\overset{5}{\int }}\left(-\frac{y^2}{4}+\frac{y}{4}+5\right)\cdot dy}\\ A={\left[\frac{-y^3}{12}+\frac{y^2}{8}+5y\right]}_{-4}^5\\ A=\left[\left(\frac{-{\left(5\right)}^3}{12}+\frac{{\left(5\right)}^2}{8}+5\left(5\right)\right)-\left(\frac{-{\left(-4\right)}^3}{12}+\frac{{\left(-4\right)}^2}{8}+5\left(-4\right)\right)\right]\\ A=\left[\left(\frac{-125}{12}+\frac{25}{8}+25\right)-\left(\frac{64}{12}+\frac{16}{8}-20\right)\right]\\ A=\left[\frac{-125}{12}+\frac{25}{8}+25-\frac{16}{3}-2+20\right]\\ A=\left[\frac{-125}{12}+\frac{25}{8}-\frac{16}{3}+43\right]\\ A=\left[\frac{-250+75-128+1032}{24}\right]\\ A=\left[\frac{-378+1107}{24}\right]\\ A=\frac{729}{24}\\ A=30.375\end{array}$

Hence, the area of the region is $\mathrm{30.375.}$