# Homework Help: Polar Integral

1. Apr 16, 2015

### whattheheckV

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

Where the region Ω: 1/2≤x≤1 0 ≤ y ≤ sqrt(1-x^2)

2. Relevant equations

3. The attempt at a solution
The question asked to solve the integral using polar coordinates. The problem I have is getting r in terms of θ. I solved the integral in rectangular ordinates using a trig sub and found the answer to be (π/6) - (√3)/8.

Thanks for all input.

Last edited: Apr 16, 2015
2. Apr 16, 2015

### LCKurtz

Have you drawn a picture? Remember $r$ goes from $r$ on the inner curve to $r$ on the outer curve. What is the equation of $x = \frac 1 2$ in polar coordinates? What is the equation of the circle? And from the picture you should be able to see the $\theta$ limits.

3. Apr 16, 2015

### Staff: Mentor

You're not solving an equation -- you're setting up and evaluating an integral
The integrand is pretty straight forward; probably the hardest part is figuring out the limits of integration. A ray extending out from the pole (0, 0) goes through the vertical line, x = 1, to the circle. Convert the equation of the vertical line to polar from. The circle part is simple.

4. Apr 16, 2015

### whattheheckV

I have drawn the picture and it is very straightforward. The problem I am having is that the radius is not constant so i need to get r in terms of theta. http://www5b.wolframalpha.com/Calculate/MSP/MSP17811i1c25cda73i33d700001h169b9cb5h0603d?MSPStoreType=image/gif&s=34&w=200.&h=193.&cdf=Coordinates&cdf=Tooltips [Broken]

Last edited by a moderator: May 7, 2017
5. Apr 16, 2015

### LCKurtz

OK, so you have drawn the picture. Now answer the questions I asked in my reply above.

Last edited by a moderator: May 7, 2017
6. Apr 16, 2015

### SammyS

Staff Emeritus
Do you know how to express x in terms of r and θ ?

Last edited by a moderator: May 7, 2017
7. Apr 16, 2015

### whattheheckV

As in x=rcos (θ) and y=rsin (θ) ?

8. Apr 16, 2015

### SammyS

Staff Emeritus
Good the left hand boundary of the region is x = 2.

That should give you a relation in terms of r and θ for that boundary.

The right hand boundary (curved boundary) should be easy in terms of r .

9. Apr 16, 2015

### LCKurtz

You mean $x = \frac 1 2$.

10. Apr 16, 2015

### SammyS

Staff Emeritus
Right!

11. Apr 17, 2015

### whattheheckV

Ok so using the conversion I found the bounds of r from (1/2)sec(θ) to 1

Thanks for the input

12. Apr 17, 2015

### SammyS

Staff Emeritus
Yes. You're welcome.

13. Apr 17, 2015

### LCKurtz

And what bounds did you get for $\theta$?