Integrating in Polar Coordinates: Ω Region

  • #1
whattheheckV
7
0

Homework Statement


∫∫dydx

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

Homework Equations

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:
Physics news on Phys.org
  • #2
whattheheckV said:

Homework Statement


∫∫dydx

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

Homework Equations

The Attempt at a Solution


The question asked to solve the equation 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.

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
whattheheckV said:

Homework Statement


∫∫dydx

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

Homework Equations

The Attempt at a Solution


The question asked to solve the equation using polar coordinates.
You're not solving an equation -- you're setting up and evaluating an integral
whattheheckV said:
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.
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
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
 
Last edited by a moderator:
  • #5
LCKurtz said:
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.

whattheheckV said:
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

OK, so you have drawn the picture. Now answer the questions I asked in my reply above.
 
Last edited by a moderator:
  • #6
whattheheckV said:
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
Do you know how to express x in terms of r and θ ?
 
Last edited by a moderator:
  • #7
SammyS said:
Do you know how to express x in terms of r and θ ?
As in x=rcos (θ) and y=rsin (θ) ?
 
  • #8
whattheheckV said:
As in x=rcos (θ) and y=sin (θ) ?
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
SammyS said:
Good the left hand boundary of the region is x = 2.

You mean ##x = \frac 1 2##.
 
  • #10
LCKurtz said:
You mean ##x = \frac 1 2##.
Right!
 
  • #11
Ok so using the conversion I found the bounds of r from (1/2)sec(θ) to 1

Thanks for the input
 
  • #12
whattheheckV said:
Ok so using the conversion I found the bounds of r from (1/2)sec(θ) to 1

Thanks for the input
Yes. You're welcome.
 
  • #13
whattheheckV said:
Ok so using the conversion I found the bounds of r from (1/2)sec(θ) to 1

Thanks for the input

And what bounds did you get for ##\theta##?
 
Back
Top