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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:

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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.

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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 $r(\phi)$.

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  • $\begingroup$ How to know wich is the correct $\pm$?, I thought it was $+$, but it tours that out it should be $-$. $\endgroup$ – Wyvern666 Jul 7 '14 at 21:08
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    $\begingroup$ Try plugging in $\phi = 0$ or $\phi = -\pi/2$ and see which one gives you the correct result. Also, your lower bound should be $-\pi+\arctan 3$ not $\pi+\arctan 3$. $\endgroup$ – JimmyK4542 Jul 7 '14 at 21:18

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