Compute double integral on polar coordinates, find $r(\phi)$

Question

I have the function $f(x,y)=y^2-2x^2y+6x^3-3xy+2y-6x$ and the region $\{y\geq 2x^2-2, y\leq 3x\}$.

The region is:

To compute the integral in cartesian coordinates: $\int_{-\frac{1}{2}}^{2}\int_{2x^2-2}^{3x}f(x,y)dydx$

Now i need to do it on polar coordinates.

What i have so far is:

$\int_{\pi+Arctan(3))}^{Arctan(3)} \int_{0}^{r(\phi ))} f(r,\phi ))rdrd\phi$

Where mi problem is to determine $r(\phi)$ from:

$y=2x^2-2 \Rightarrow r sin(\phi)=2r^2cos(\phi)^2-2$

But im stuck here.

Answer 1

You are almost there.

$r \sin \phi = 2r^2\cos^2\phi - 2$

$(2\cos^2\phi)r^2 - (\sin \phi)r - 2 = 0$

Now, you have a quadratic in terms of $r$. Use the quadratic formula to get:

$r = \dfrac{-(-\sin \phi) \pm \sqrt{(\sin\phi)^2 - 4(2\cos^2\phi)(-2)}}{2(2\cos^2\phi)}$

Pick the correct $\pm$ sign to get $r(\phi)$.