# Thread: Slightly different area between curve

1. ## Slightly different area between curve

Find the area between the curves $\displaystyle x-y+y^2=0$ and $\displaystyle x+x^2+2y=0$.

They intersect at $\displaystyle (0,0)$ and $\displaystyle (-2,-1)$

2. Originally Posted by polymerase
Find the area between the curves $\displaystyle x-y+y^2=0$ and $\displaystyle x+x^2+2y=0$.

They intersect at $\displaystyle (0,0)$ and $\displaystyle (-2,-1)$
Completing the square on the first equation gives

$\displaystyle \left(y-\frac{1}{2}\right)^2=-x+\frac{1}{4}$

This can be solved for y in terms of x.

we will use the negative part of the square root.

$\displaystyle y=\frac{1}{2}-\sqrt{\frac{1}{4}-x}$

The other can be solved for y.

This gives the integral....

$\displaystyle -\int_{-2}^0\left(\frac{1}{2}-\sqrt{\frac{1}{4}-x}\right)-\frac{-1}{2}(x^2+x)dx$

3. Hello, polymerase!

It certainly is "different" . . .

Find the area between the curves $\displaystyle x-y+y^2\:=\:0$ and $\displaystyle x+x^2+2y\:=\:0$.

They intersect at: $\displaystyle (0,0)$ and $\displaystyle (-2,-1)$
Code:
*                     |
*            |
*    |
*
|  *
|    *
*     |     *
*:::::*  |    *
*:::::::::*|  *
- - - - *:-:-:-:-:-:* - - - -
:::::::*    |
*            |*
*                     |
|
*             | *
|

The upper parabola is: .$\displaystyle y^2-y+x\:=\:0\quad\Rightarrow\quad y \;=\;\frac{1 \pm \sqrt{1-4x}}{2}$
. . and we want the "lower half": .$\displaystyle y \;=\;\frac{1 - \sqrt{1-4x}}{2}$

The lower parabola is: .$\displaystyle y \;=\;-\frac{1}{2}x^2-\frac{1}{2}x$

The area is given by: .$\displaystyle A \;=\;\int^0_{-2}\bigg[\left(-\frac{1}{2}x^2-\frac{1}{2}x\right) -$ $\displaystyle \left(\frac{1}{2} - \frac{1}{2}(1-4x)^{\frac{1}{2}}\right)\bigg]\,dx$

Bon voyage!