# Double Integrals using Polar Coordinates

1. Apr 28, 2014

### ThiagoG

1. The problem statement, all variables and given/known data
∫∫Rarctan(y/x) dA, where R={(x,y) | 1$\leq$x2+y2$\leq$4, 0$\leq$y$\leq$x

2. Relevant equations
x=rcos(θ)
y=rsin(θ)
x2+y2=r2

3. The attempt at a solution
I know that the range of r is 1 to 2 but I can't figure out how to change the second part into θ. If I change y and x to rsin(θ) and rcos(θ), respectively, I'm not sure what to do after this point. How would I get the theta by iteslf?

2. Apr 28, 2014

### slider142

The limits or intervals of integration are sometimes easier to see if you look directly at the geometry of the region, as opposed to its algebraic representation in terms of coordinate equations.
Can you sketch the region of points that solves the pair of inequalities represented by 0 ≤ y ≤ x ? How would you describe the region you sketched in terms of r and θ ?
Remember the intuitive descriptions of r and θ that lead to the definition of the equations you cited. θ is meant to denote the counterclockwise angle between the point's radius (or position vector) and the positive x-axis. r is just meant to be the distance between the point and the origin, or the length of the point's position vector.
In the region described by those two inequalities, what is the minimum possible value of θ ? The maximum ? Is r restricted in any way ?

Last edited: Apr 28, 2014
3. Apr 28, 2014

### LCKurtz

Draw a picture of the region. Think of a windshield wiper cleaning the region. What values of $\theta$ would it sweep through?

4. Apr 28, 2014

### pasmith

Look at $\frac yx = \tan \theta$.

5. Apr 28, 2014

### ThiagoG

So would it be from 0 to ∏/4?

6. Apr 28, 2014

### ThiagoG

After I drew it, I think it's ∏/4 because y=x bisects the first quadrant which would mean it is at ∏/4. So it ranges from 0 to ∏/4

7. Apr 28, 2014

### slider142

That's it exactly. :) The limits you have for r are perfect as well.