# Double Integrals in polar coordinates: Calculus 3

1. Feb 27, 2013

### timmastny

1. The problem statement, all variables and given/known data

Given $\int^{\sqrt{6}}_{0}$$\int^{x}_{-x}$dydx, convert to ploar coordinates and evaluate.

2. Relevant equations
We know that x=rcos$\theta$ and y=rsin$\theta$ and r =x^2+y^2

3. The attempt at a solution

First, I defined the region of the original integral: R = 0 $\leq$x$\leq$ $\sqrt{6}$ and -x$\leq$y$\leq$x.

In other words, we know that y is bounded on the lower limit by the line -x and the upper limit x. Because the line x produces a 45 degree angle, -$\pi$/4 $\leq$ $\theta$ $\leq$ $\pi$/4.

However, I do not understand how to find domain of r. The region the original integral covers is not circular. We can prove this because at ($\sqrt{6}$,0), r = $\sqrt{6}$ and at ($\sqrt{6}$, $\sqrt{6}$), which is still in the region defined by the original integral, r = $\sqrt{12}$ ($\sqrt{\sqrt{6}^2+\sqrt{6}^2}$)

So that is where I am stuck. The answer to the problem according to the book is 6 but I do not know how to get there. All the examples we worked were circular regions.

Additionally, assuming the region is 0 $\leq$ r $\leq$ $\sqrt{6}$ does not work either. $\int^{pi/4}_{-pi/4}$ $\int^{sqrt{6}}_{0}$ r dr d$\theta$ = (3$\pi$)/2

Last edited: Feb 27, 2013
2. Feb 27, 2013

### timmastny

3pi/4 is 135 degrees. According the limits of integration, x only varies from 0 to $\sqrt{6}$. $\theta$ cannot go past the limits set by x.

If $\theta$ varied from pi/4 to to 3pi/4 wouldn't the limits of x be different?

However, I agree it is a triangle but I think the triangle is in positive x region, and the lower side of the triangle is the line -x.

I am fairly confident on the bounds of $\theta$, but very confused by the bounds of r.

3. Feb 27, 2013

### Staff: Mentor

You're right. I'm trying to multitask, and not doing to well at it. θ
is exactly as you say.
The limit for r is the vertical line x = √6. Going to polar coordinates, this is r cos(θ) = √6. You can solve this equation for r. The lower limit for r is, of course, 0.

4. Feb 27, 2013

### timmastny

No problem, multitasking makes breathing hard, haha.

That makes perfect sense though. That means the upper limit of integration depends on θ. So many problems in polar coordinates have static bounds, I completely forgot about that.

Thanks!

P.S. I just calculated the new integral, it is indeed correct.