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Thread: Use Polar Coordinates to Evaluate the double Integral

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    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
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    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 =].
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    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
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    Re: Use Polar Coordinates to Evaluate the double Integral

    Quote Originally Posted by math951 View Post
    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.
    Thanks from math951
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    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
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    Re: Use Polar Coordinates to Evaluate the double Integral

    Quote Originally Posted by math951 View Post
    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.
    Attached Thumbnails Attached Thumbnails Use Polar Coordinates to Evaluate the double Integral-polar.jpg  
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