Double Integration Using Polar Coordinates

[MisterMan]

Homework Statement

$$\int\int \frac{x^3}{x^2 + y^2}\,dxdy$$

Use polar coordinates to evaluate the triangle R, with vertices (0,0), (1,0) and (1,1)

Homework Equations

$$\int\int f(r,\theta) r\,drd\theta$$

$$r^2 = x^2 + y^2$$

$$x = rcos\theta$$

$$y = rsin\theta$$

The Attempt at a Solution

I drew the triangle and got the upper limit of r to be 1 and the lower limit 0. I think the limits for theta are pi over 4 and 0, but I'm not sure, I got stuck on the integration part:

$$\int\,d\theta\int_0^1 \frac{(rcos\theta)^3}{r^2}r\,dr$$

$$\int cos^3\theta\,d\theta\int_0^1 r^2\,dr$$

$$\frac{1}{3}\int cos^3\theta\,d\theta$$

At which point, I wasn't sure how to proceed. I tried to integrate it by splitting it into $$cos^2\theta$$ and $$cos\theta$$ and using $$\frac{1}{2}(1 + cos2\theta)$$, but I never got the correct answering ( I'm looking for pi over 12 ) since I believe I need theta on its own.

Any help will be appreciated, thanks.

 

[vela]

Your limits for theta are correct, but your limits for r are wrong. The limits change depending on what theta equals. For theta=0, for example, r would go from 0 to 1. When theta=pi/4, however, r would go from 0 to sqrt(2). You want to determine how the limits vary as a function of theta.

To integrate cos³ x, you do the following:

$$\int \cos^3 \theta\,d\theta = \int (\cos^2 \theta)\cos \theta\,d\theta = \int (1-\sin^2 \theta)\cos\theta\,d\theta$$

Then use the substitution $u=\sin \theta$. Whenever you have an odd power of cosine, you can use this technique. I'm not sure if you'll need it for this problem though.

 

[hunt_mat]

Vela, beaten me to it!

 

[MisterMan]

MAJOR EDIT: I am so so sorry to both of you for wasting your time. Me and my infinite stupidity didn't take a good enough look at the answer at the back of the book. The answer at the book did not give r = 1, but stated the upper limit for r was the line x = 1 ( Ah, how stupid of me ). This gives:

$$r = \frac{1}{cos\theta}$$

Cancelling gives a third and evaluating the integral gives the required answer of pi over 12.

In hindsight, the problem was extremely simple if it was not for my lack of brain functionality. No, excuses I attempted this problem during the day, when I should be awake!

Anyway, apologises for wasting your time and thanks for for the help nonetheless.