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Math Help - HowTo:upper and lowerlimit of integral from Cartesian coordinate to polar coordinate?

  1. #1
    Senior Member x3bnm's Avatar
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    HowTo:upper and lowerlimit of integral from Cartesian coordinate to polar coordinate?

    I'm stuck in finding integration of a integral. I'm given a integral and asked to change the Cartesian
    integral into an equivalent polar integral and then evaluate the resulting integral.


    The integral is:
    <br />
 \int_0^6\int_0^y\,\, x\,\, dx\,\, dy<br />

    So i replaced x\,\,dx\,\,dy with r^2\,\, \cos(\theta)\,\,dr\,\,d\theta.

    The book says:
    <br />
 \int_0^6\int_0^y\,\, x\,\, dx\,\, dy = \int_\frac{\pi}{4}^\frac{\pi}{2}\int_0^{6\csc(\the  ta)}\,r^2\,\cos(\theta)\,dr\,d\theta <br />


    But how do you find the upper and lower boundary of r and \theta in polar coordinate from Cartesian coordinate?

    The book I'm following is vague about the procedure.

    There must be a way to find it. But can't seem to figure it out.

    Can anyone kindly explain the method?
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by x3bnm View Post
    I'm stuck in finding integration of a integral. I'm given a integral and asked to change the Cartesian
    integral into an equivalent polar integral and then evaluate the resulting integral.


    The integral is:
    <br />
 \int_0^6\int_0^y\,\, x\,\, dx\,\, dy<br />

    So i replaced x\,\,dx\,\,dy with r^2\,\, \cos(\theta)\,\,dr\,\,d\theta.

    The book says:
    <br />
 \int_0^6\int_0^y\,\, x\,\, dx\,\, dy = \int_\frac{\pi}{4}^\frac{\pi}{2}\int_0^{6\csc(\the  ta)}\,r^2\,\cos(\theta)\,dr\,d\theta <br />


    But how do you find the upper and lower boundary of r and \theta in polar coordinate from Cartesian coordinate?

    The book I'm following is vague about the procedure.

    There must be a way to find it. But can't seem to figure it out.

    Can anyone kindly explain the method?

    first of all , when change to polar coordinates, you must know that there are "natural" limits of new (how u wrote) r and \theta
    but that only means that boundary can't go beyond the "natural" limits, but can go under...

    if there is no "restriction" then they will be natural but if u have let's say some conditions then they probably will be changed

     x =\rho \cos \varphi
     y =\rho \sin \varphi
     J = \rho

    I prefer to use  \rho and \varphi but it's the same

    natural limits
    \rho \ge 0
    \varphi \mid _0 ^{2\pi}

    most times your boundary will be shown with some region, like few lines, parabola, circle's or anything else, and now u put in that conditions your polar x,y, and solve ... to find how does your boundary change


    in your example there is 2 boundary's , first is line x=y or y=x , and second y=6... so what that means it means that your integral will look like on the graph put the "new" x,y, in that equations and see what conditions will u get


     y=x

     \rho \cos \varphi = \rho \sin \varphi

    \cos \varphi = \sin \varphi

    1=\tan \varphi

    \displaystyle \varphi = \frac {\pi}{4}

    now secon...
    y=6

    can u do it ? and conclude something try and we'll see are you on the right track
    Attached Thumbnails Attached Thumbnails HowTo:upper and lowerlimit of integral from Cartesian coordinate to polar coordinate?-xy_polar.jpg  
    Last edited by yeKciM; August 10th 2010 at 04:08 PM.
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  3. #3
    Senior Member x3bnm's Avatar
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    First of all thank you for taking the time and effort for explaining my question. The method you showed works for certain kind of integral. Say we have
    <br />
 \int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,\,(x^2+y^2)\,\,dx\,\,dy<br />

    Now:
    <br />
x = \sqrt{1-y^2}\,\,\,\,\,\,\,\,\,         (1)<br />

    <br />
x = -\sqrt{1-y^2}\,\,\,\,\,\,\,\,\,\         (2)<br />

    <br />
y = 1\,\,\,\,\,\,\,\,\,\                            (3)<br />

    <br />
y = -1\,\,\,\,\,\,\,\,\,\                           (4)<br />


    if we plug in x =\rho \cos \varphi and y =\rho \sin \varphi into equation (1) and solve for \rho we get:
    <br />
\rho = 1<br />

    Then plug in y = \rho \sin \varphi and \rho = 1 into equation (3) and solve for \varphi we get:
    <br />
 \varphi = \frac{\pi}{2}<br />

    Again if we plug in y = \rho \sin \varphi and \rho = 1 into equation (4) and solve for \varphi we get:
    <br />
 \varphi = \frac{3\pi}{2}<br />

    So the upper and lower bound of \rho is 1 and 0 respectively and upper and
    lower bound for \varphi is \frac{3\pi}{2} and \frac{\pi}{2} respectively.
    <br />
 \int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,\,(x^2+y^2)\,\,dx\,\,dy = \int_{\frac{\pi}{2}}^\frac{3\pi}{2}\int_0^1\,\,r^3  \,\,d\rho\,\,d\varphi<br />

    But the answer is:
    <br />
\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,\,(x^2+y^2)\,\,dx\,\,dy = \int_0^{2\pi} \int_0^1\,\,r^3\,\,d\rho\,\,d\varphi<br />

    Why is the upper limit for \varphi is 2\pi instead of \frac{3\pi}{2}(what i found) and lower bound 0 instead of
    \frac{\pi}{2}(what i found)? Why didn't the method you showed didn't work here? Did i make a mistake here?

    Note: I edited this post couple of times to be correct. Sorry about that.
    Last edited by x3bnm; August 10th 2010 at 07:37 PM.
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  4. #4
    Senior Member x3bnm's Avatar
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    I found the solution to my problem. For those who are interested don't use x boundary equations with y's
    boundary equations(don't mix them) to find r and \theta.

    So solve two sets of equations separately(so there will be two equations for each x and y) for r
    and \theta. Again I thank yeKciM for expletive answer.
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  5. #5
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by x3bnm View Post
    First of all thank you for taking the time and effort for explaining my question. The method you showed works for certain kind of integral. Say we have
    <br />
 \int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,\,(x^2+y^2)\,\,dx\,\,dy<br />



    Now:
    <br />
x = \sqrt{1-y^2}\,\,\,\,\,\,\,\,\,         (1)<br />

    <br />
x = -\sqrt{1-y^2}\,\,\,\,\,\,\,\,\,\         (2)<br />

    <br />
y = 1\,\,\,\,\,\,\,\,\,\                            (3)<br />

    <br />
y = -1\,\,\,\,\,\,\,\,\,\                           (4)<br />


    if we plug in x =\rho \cos \varphi and y =\rho \sin \varphi into equation (1) and solve for \rho we get:
    <br />
\rho = 1<br />

    Then plug in y = \rho \sin \varphi and \rho = 1 into equation (3) and solve for \varphi we get:
    <br />
 \varphi = \frac{\pi}{2}<br />

    Again if we plug in y = \rho \sin \varphi and \rho = 1 into equation (4) and solve for \varphi we get:
    <br />
 \varphi = \frac{3\pi}{2}<br />

    So the upper and lower bound of \rho is 1 and 0 respectively and upper and
    lower bound for \varphi is \frac{3\pi}{2} and \frac{\pi}{2} respectively.
    <br />
 \int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,\,(x^2+y^2)\,\,dx\,\,dy = \int_{\frac{\pi}{2}}^\frac{3\pi}{2}\int_0^1\,\,r^3  \,\,d\rho\,\,d\varphi<br />

    But the answer is:
    <br />
\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,\,(x^2+y^2)\,\,dx\,\,dy = \int_0^{2\pi} \int_0^1\,\,r^3\,\,d\rho\,\,d\varphi<br />

    Why is the upper limit for \varphi is 2\pi instead of \frac{3\pi}{2}(what i found) and lower bound 0 instead of
    \frac{\pi}{2}(what i found)? Why didn't the method you showed didn't work here? Did i make a mistake here?

    Note: I edited this post couple of times to be correct. Sorry about that.

    wait, u have there circle  x^2+y^2=1 and how u done is that you look at it like 2 functions, first  \sqrt{1-y^2} and second -\sqrt{1-y^2}. Let's look how that looks (drawing)

    u looked at function like that, and you did get that function goes from \pi /2 to 3\pi /2... that is true but u (if u look at it like that) have still and another function (right one) which will go from 3\pi/2 to \pi/2.... that's why you get that


    now for that integral you wrote :
    \displaystyle \int _{-1} ^1 dy \int _{-\sqrt{1-y^2}} ^{\sqrt{1-y^2}}(x^2+y^2) \; dx


    hehehehehe that limit from -1 to 1 you didn't get from equation of line y=1 and y=-1 so that u put and do that what you have done ... that -1 to 1 is from the equation of circle , so that \varphi is on his natural limits from zero to 2\pi

    you don't have any another condition by which your \varphi will change limits


    Edit: sorry i didn't see your post, been writing this
    Attached Thumbnails Attached Thumbnails HowTo:upper and lowerlimit of integral from Cartesian coordinate to polar coordinate?-krugpopola.jpg  
    Last edited by yeKciM; August 10th 2010 at 10:32 PM.
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  6. #6
    Senior Member x3bnm's Avatar
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    That makes a lot of sense. You are absolutely right. Isn't math fun or what? Thanks for extending a helping hand.
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