Double Integrals in polar coordinates: Calculus 3

[timmastny]

Homework Statement

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

Homework Equations

We know that x=rcos\theta and y=rsin\theta and r =x^2+y^2

The Attempt at a Solution

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

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

 

[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.

 

[Mark44]

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.

You're right. I'm trying to multitask, and not doing to well at it. θ
is exactly as you say.

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

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.

 

[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.