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:

enter image description here

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.


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 $r(\phi)$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.