The discussion revolves around evaluating a double integral in polar coordinates over a specified region Ω defined by the inequalities 1/2 ≤ x ≤ 1 and 0 ≤ y ≤ sqrt(1-x^2). Participants are exploring how to express the variables in polar coordinates and determine the limits of integration.
There is ongoing exploration of the problem, with participants sharing insights about the conversion process and the relationships between x, y, r, and θ. Some have drawn diagrams to aid in understanding the setup, while others are clarifying the boundaries and limits of integration. Guidance has been offered regarding the expressions needed for the boundaries in polar coordinates.
Participants are navigating the complexities of converting the boundaries of the region into polar coordinates, which includes addressing the non-constant nature of r and the need to determine the limits for θ. There is an emphasis on ensuring that the setup aligns with the original problem constraints.
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.
You're not solving an equation -- you're setting up and evaluating an integralwhattheheckV 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 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.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.
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
Do you know how to express x in terms of r and θ ?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
As in x=rcos (θ) and y=rsin (θ) ?SammyS said:Do you know how to express x in terms of r and θ ?
Good the left hand boundary of the region is x = 2.whattheheckV said:As in x=rcos (θ) and y=sin (θ) ?
SammyS said:Good the left hand boundary of the region is x = 2.
Right!LCKurtz said:You mean ##x = \frac 1 2##.
Yes. You're welcome.whattheheckV said:Ok so using the conversion I found the bounds of r from (1/2)sec(θ) to 1
Thanks for the input
whattheheckV said:Ok so using the conversion I found the bounds of r from (1/2)sec(θ) to 1
Thanks for the input