Chapter 6, Problem 4RE

To determine

The area of the region enclosed by the curves.

Answer to Problem 4RE

The area of the region enclosed by the curves is $\underset{\_}{\frac{32}{3}}$.

Explanation of Solution

Given information:

The two curves has a function of $x+y=0$ and $x=y^2+3y$.

Find the intersection points of the curves by equating the curves as shown below:

$f\left(x\right)=g\left(x\right)$ (1)

Here, the top curve function is $f\left(x\right)$ and the bottom curve function is $g\left(x\right)$.

Substitute $y^2+3y$ for $f\left(x\right)$ and $-y$ for $g\left(x\right)$ in Equation (1).

$\begin{array}{c}{y}^2+3y=-y\\ y^2+4y=0\\ y\left(y+4\right)=0\end{array}$

$\begin{array}{c}y=0\\ y=-4\end{array}$

Procedure to sketch the region bounded by the two curves is explained below:

• Draw the graph for the function $x=-y$ by substituting different values for x.
• Similarly in the same graph plot for the function $x=y^2+3y$ by substituting different values for x.
• Shade the region lies between $x=0$ and $x=2$.

The region enclosed by the curves $x=-y$ and $x=y^2+3y$ is shown in Figure (1).

Refer to Figure (1).

The curves are bounded by the top and bottom curve, so the integration can be done with respect to x.

Find the area of the region bounded by the curves using the relation:

$A={\displaystyle {\int }_a^b\left(f\left(x\right)-g\left(x\right)\right)}dx$ (2)

Here, the lower limit is a and the upper limit is b.

Substitute $-y$ for $f\left(x\right)$, $y^2+3y$ for $g\left(x\right)$, $-4$ for a, and 0 for b in Equation (2).

$\begin{array}{c}A={\displaystyle {\int }_{-4}^0\left[\left(-y\right)-\left(y^2+3y\right)\right]}dy\\ ={\displaystyle {\int }_{-4}^0\left[-y-\left(y^2+3y\right)\right]}dy\\ ={\displaystyle {\int }_{-4}^0\left(-y^2-4y\right)}dy\\ ={\left[-\frac{1}{3}{y}^3-2y^2\right]}_{-4}^0\end{array}$

$\begin{array}{c}=\left[0-\left(-\frac{1}{3}{\left(-4\right)}^3-2{\left(-4\right)}^2\right)\right]\\ =0-\left(\frac{64}{3}-32\right)\\ =-\left(\frac{64-96}{3}\right)\\ =\frac{32}{3}\end{array}$

Therefore, the area of the region enclosed by the curves is $\underset{\_}{\frac{32}{3}}$.