# Thread: Use Polar Coordinates to Evaluate the double Integral

1. ## Use Polar Coordinates to Evaluate the double Integral

double integral 2yDA, where R is the region in the first quadrant bounded above by the circle (x-1)^2+y^2=1

and below by the line y=x

So we know the top limit is going to be the circle, and the bottom limit is going to be the line y=x.

We know that the circle has a radius of 1, and the center is (1,)

I have to evaluate in this in polar coordinates, so 2y turns into 2r^2sin(thetta) drd(thetta)

I just do not know how to find the limits for this, I am able to graph it and see it visually, but I cannot find the limits, nor do I understand why they are.

I guess I can see how thetta starts at pi/4 because of y=x, but how the hell does it end at pi/2???

I think for the dr limits I understand it. Please someone correct me if I am wrong. They found the bottom limit as 0 from y=x, and we are given y=0. then they found the top limit, from (x-1)^2+y^2=1 by converting this into polar coordinates??

The solution has these limits:

Calculus (9780470647721), Pg. 1025, Ex. 26 :: Homework Help and Answers :: Slader

2. ## Re: Use Polar Coordinates to Evaluate the double Integral

I actually enjoy doing these problems. I like to see how integrals are set up and how the boundary points can be viewed =].

3. ## Re: Use Polar Coordinates to Evaluate the double Integral

I have noticed a trend... in all of these problems they are contrived to have x^2+y^2, but I am starting to understand why they convert it to polar coordinates to set up and evaluate the integral, it works out nicely

4. ## Re: Use Polar Coordinates to Evaluate the double Integral

Originally Posted by math951
double integral 2yDA, where R is the region in the first quadrant bounded above by the circle (x-1)^2+y^2=1

and below by the line y=x

So we know the top limit is going to be the circle, and the bottom limit is going to be the line y=x.

We know that the circle has a radius of 1, and the center is (1,)

I have to evaluate in this in polar coordinates, so 2y turns into 2r^2sin(thetta) drd(thetta)

I just do not know how to find the limits for this, I am able to graph it and see it visually, but I cannot find the limits, nor do I understand why they are.

I guess I can see how thetta starts at pi/4 because of y=x, but how the hell does it end at pi/2???

I think for the dr limits I understand it. Please someone correct me if I am wrong. They found the bottom limit as 0 from y=x, and we are given y=0. then they found the top limit, from (x-1)^2+y^2=1 by converting this into polar coordinates??
To answer your last question first, yes. You must convert the circle equation to polar coordinates. Expand out the $(x-1)^2$ and change to polar coordinates and simplify. You should be able to get $r = 2\cos\theta$ for the curve. Now, to see how to get the limits, draw a good sized picture and put your pencil right in the middle of the designated area. Then draw a straight line from that point in the $r$ direction. It should go from the $r$ at origin (=0) to the the $r$ on the curve. That will give the $r$ limits. Now think of sweeping that line in the $\theta$ direction, kind of like a variable length windshield wiper. What values will $\theta$ take to sweep the whole area? That's how you get the $\theta$ limits.

5. ## Re: Use Polar Coordinates to Evaluate the double Integral

Walagster, why is it that when we convert to polar coordinates, and we are solving for what r=, it seems we let an 'r' just disappear or omit.

As seen in this example:
https://imgur.com/a/dp9vrDf

6. ## Re: Use Polar Coordinates to Evaluate the double Integral

Originally Posted by math951
Walagster, why is it that when we convert to polar coordinates, and we are solving for what r=, it seems we let an 'r' just disappear or omit.

As seen in this example:
https://imgur.com/a/dp9vrDf
In polar coordinates curves are most often written in the form $r = f(\theta)$ just as in rectangular coordinates they are often written as $y = f(x)$. So in rectangular coordinates if you are doing a double integral with area element $dA = dydx$, for the $dy$ limits you ask yourself if you move the area element in the $y$ direction, what are its lower and upper limits? Similarly, in polar coordinates where your area element is $rdrd\theta$ and you want the $dr$ limits you ask yourself if you move the area element in the $r$ direction, what are its lower and upper limits. Frequently, as in this problem, but not always, the lower limit for $r$ is $0$ and the upper limit (outer value of $r$) is the $r$ on the curve, which depends on $\theta$.

Here is a picture of your problem illustrating where you can see $r$ goes from $0$ to $2\cos\theta$. Click on it to enlarge it.