Integrating in Polar Coordinates: Ω Region

In summary: And what bounds did you get for ##\theta##?In summary, the conversation involved solving an integral using polar coordinates for a given region. The main difficulty was getting the radius in terms of theta, but by drawing a picture and using the equations x=rcos(theta) and y=rsin(theta), the bounds for r were determined to be (1/2)sec(theta) to 1. The bounds for theta were not explicitly stated in the conversation.
  • #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##?
 

Related to Integrating in Polar Coordinates: Ω Region

1. What is the purpose of integrating in polar coordinates?

Integrating in polar coordinates allows us to solve problems involving circular or symmetrical regions. This method is particularly useful for calculating volumes, areas, and other quantities that are difficult to calculate using Cartesian coordinates.

2. How do you convert Cartesian coordinates to polar coordinates?

To convert from Cartesian coordinates (x,y) to polar coordinates (r,θ), you can use the following formulas: r^2 = x^2 + y^2 and tan(θ) = y/x. Once you have r and θ, you can plug them into the appropriate polar coordinate expressions for the quantity you are trying to calculate.

3. What is the difference between integrating in rectangular coordinates vs polar coordinates?

Integrating in rectangular coordinates involves dividing a region into small rectangles and summing their areas to find the total area. Integrating in polar coordinates involves dividing a region into small sectors and summing their areas to find the total area. While both methods can be used to solve similar problems, polar coordinates are often more efficient for dealing with circular or symmetrical regions.

4. Can you integrate in polar coordinates for regions that are not circular or symmetrical?

Yes, you can integrate in polar coordinates for non-circular or non-symmetrical regions. However, the equations and methods for calculating the necessary integrals may be more complex than in rectangular coordinates.

5. Are there any limitations to using polar coordinates for integration?

Polar coordinates are not suitable for all types of integrals. They are most effective for solving problems involving circular or symmetrical regions. In some cases, the equations and methods for calculating integrals in polar coordinates may be more difficult to work with than in rectangular coordinates. It is important to consider the nature of the problem before deciding whether to use polar or rectangular coordinates for integration.

Similar threads

Integration of acceleration in polar coordinates
    • Calculus and Beyond Homework Help
Replies
24
Views
1K
Double Integral Problem: Incorrect Jacobian Calculation for Polar Coordinates
    • Calculus and Beyond Homework Help
Replies
4
Views
913
Calculate this Integral around the Circular Path using Green's Theorem
    • Calculus and Beyond Homework Help
Replies
3
Views
307
Change of Variables: Integrals w/Polar Coordinates
    • Calculus and Beyond Homework Help
Replies
2
Views
869
Where to use polar (cylindrical coor.) in double and triple integrals
    • Calculus and Beyond Homework Help
Replies
4
Views
863
Computing the polar moment of inertia (calculus)
    • Calculus and Beyond Homework Help
Replies
1
Views
961
Evaluating Cartesian integral in polar coordinates
    • Calculus and Beyond Homework Help
Replies
11
Views
2K
Region bounded by a line and a parabola (polar coordinates)
    • Calculus and Beyond Homework Help
Replies
22
Views
1K
Trying to find this double integral using polar coordinates
    • Calculus and Beyond Homework Help
Replies
14
Views
2K
Polar coordinates of the centroid of a uniform sector
    • Calculus and Beyond Homework Help
Replies
5
Views
1K

Hot Threads

Recent Insights

Back
Top