Question

Find the area of the region bounded by the given curves.

\(x + y = 0,\;\;\;x = {y^2} + 3y\)

Short answer

The area of the shaded region is \(\frac{{32}}{3}\)

Step-by-step solution

:Given information

The given value is:

\(x + y = 0,\;\;\;x = {y^2} + 3y\)

Sketch the curves

Find the Intersections

\(\begin{aligned}{\rm{x + y = 0}} \to x = - y\\x = {y^2} + 3y\end{aligned}\)

Find the intersections

\(\begin{aligned} {y^2} + 3y = - y\\{y^2} + 4y = 0\\{\rm{y(y + 4) = 0}}\\{\rm{y = - 4,0}}\end{aligned}\)

These will be the integral limits.

Find the area

Because we want a positive response, subtract the left from the right curve. (deduct a lower value from a higher value)

\(\begin{aligned}A &= \int_{ - 4}^0 {\left( { - y - \left( {{y^2} + 3y} \right)} \right)} dy\\ &= \int_{ - 4}^0 {\left( { - y - {y^2} - 3y} \right)} dy\end{aligned}\)

\(\begin{aligned}&= \int_{ - 4}^0 {\left( { - {y^2} - 4y} \right)} dy\\ &= \left( { - \frac{1}{3}{y^3} - 2{y^2}} \right)_{ - 4}^0\\ &= 0 - 0 - \left( { - \frac{1}{3}{{( - 4)}^3} - 2{{( - 4)}^2}} \right)\end{aligned}\)

\(\begin{aligned}&= - \left( {\frac{{64}}{3} - 32} \right)\\ &= \frac{{ - 64 + 96}}{3}\\ &= \frac{{32}}{3}\end{aligned}\)

The area of the shaded region is \(\frac{{32}}{3}.\)