# Area Problem

• Jan 5th 2010, 05:37 PM
Em Yeu Anh
Area Problem
Q: Find the area of the plane region bounded by the curves $\displaystyle x=y^2$, $\displaystyle x=2y^2-y-2$

First I set the graphs equal to each other to find the intersection points and obtained (4,2) and (1,-1). My difficulty for these ones that are stated in terms of x is drawing the graph out to figure out which graph to subtract from the other, $\displaystyle x=y^2$ is fairly easy but for the other one I'm stuck. Greatly appreciate your help! Thanks!
• Jan 5th 2010, 06:14 PM
Abu-Khalil
It's the same area between $\displaystyle y=x^2$ and $\displaystyle y=2x^2-x-2$.
• Jan 5th 2010, 06:18 PM
tonio
Quote:

Originally Posted by Em Yeu Anh
Q: Find the area of the plane region bounded by the curves $\displaystyle x=y^2$, $\displaystyle x=2y^2-y-2$

First I set the graphs equal to each other to find the intersection points and obtained (4,2) and (1,-1). My difficulty for these ones that are stated in terms of x is drawing the graph out to figure out which graph to subtract from the other, $\displaystyle x=y^2$ is fairly easy but for the other one I'm stuck. Greatly appreciate your help! Thanks!

Interchange variables: $\displaystyle x\mapsto y\,,\,y\mapsto x$ , and then you want the area enclosed by the curves $\displaystyle y = x^2\,,\,y = 2x^2-x-2$, which have intersection points $\displaystyle (-1,1)\,,\,(2,4)$.
We do this so that we'll get functions of the variable x, which is what most of us is way more used to, instead of the given functions of the variable y. This is the only reason.

Now, $\displaystyle 2x^2-x-2<x^2\Longleftrightarrow (x-2)(x+1)<0\Longleftrightarrow -1<x<2$ $\displaystyle \Longrightarrow$ in our integration interval $\displaystyle [-1,2]$ , the graph of the parabola $\displaystyle y=x^2$ is over the graph of the parabola $\displaystyle y=2x^2-x-2$ , and then the wanted area is

$\displaystyle S=\int\limits_{-1}^2\left(2x^2-x-2-x^2\right)\,dx$

Now end the exercise

Tonio