# Thread: Slightly different area between curve

1. ## Slightly different area between curve

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

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

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

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

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

This can be solved for y in terms of x.

we will use the negative part of the square root.

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

The other can be solved for y.

This gives the integral....

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

3. Hello, polymerase!

It certainly is "different" . . .

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

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

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

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

Bon voyage!