Hello, polymerase!
It certainly is "different" . . .
Find the area between the curves $\displaystyle xy+y^2\:=\:0$ and $\displaystyle x+x^2+2y\:=\:0$.
They intersect at: $\displaystyle (0,0)$ and $\displaystyle (2,1)$ Code:
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The upper parabola is: .$\displaystyle y^2y+x\:=\:0\quad\Rightarrow\quad y \;=\;\frac{1 \pm \sqrt{14x}}{2}$
. . and we want the "lower half": .$\displaystyle y \;=\;\frac{1  \sqrt{14x}}{2}$
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}(14x)^{\frac{1}{2}}\right)\bigg]\,dx $
Bon voyage!