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Double Integral in Polar Coordinates

  1. Apr 1, 2006 #1
    I am having trouble with this seemingly easy problem.

    Evaluate the double integral (sin(x^2+y^2)) , where the region is 16=<x^2+y^2=<81.

    I find the region in polar coordinates to be 4=<r=<9 0=<theta=<2pi
    I find the expression to be sin(rcos^2theta+rsin^2theta) r dr dtheta , which is equal to sin(r) r dr dtheta

    Is what I have done right? I evaluate the integral, first with respect to r over using 4 and 9, then with respect to theta using 0 and 2pi. The answer I get is ~42.43989 but that is wrong :cry:

    any ideas, hints would be greatly appreciated thx :tongue2:
  2. jcsd
  3. Apr 1, 2006 #2
    Ill try using this "tex" to make it more understandable

    [tex] \int \int _R sin(x^2+y^2) dA[/tex] where R is the region [tex]16 \leq x^2 + y^2 \leq 81 [/tex]

    I transfer to polar coord. and get the region [tex] 4 \leq r \leq 9 [/tex] and [tex] 0 \leq \theta \leq 2pi [/tex] and the integral [tex] \int^{2pi} _0 \int^9 _4 sin(r) r dr d\theta [/tex]

    I solve the inner integral using the parts method and get [tex] [-9cos(9)+sin(9)]-[-4cos(4)+sin(4)] [/tex] then i solve the outer integral and get [tex] 2\pi [[-9cos(9)+sin(9)]-[-4cos(4)+sin(4)]]\approx 42.44 [/tex]
    Last edited: Apr 1, 2006
  4. Apr 1, 2006 #3


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    The argument of your sin fucntion is r^2, not r.

    So you will need to integrate sin(r^2) r dr. Just define a new variable, say, t=r^2. Then r dr = dt/2 so you will integrate [itex] \int r dr sin(r^2) = {1 \over 2} \int sin(t) dt [/itex]...Simple, isn't?

  5. Apr 2, 2006 #4
    yes, very simple thank you.

    Something I have been wondering about, that my teacher didnt go into much, is why do we add the extra [tex] r [/tex] in the integral after converting to polar coord.?
  6. Apr 2, 2006 #5


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    There are several ways of looking at it.

    The geometric idea is that you're doing an area integral... so what is the area of the small region bounded by [itex]r \in [r_0, r_0 + \Delta r_0][/itex] and [itex]\theta \in [\theta_0, \theta_0 + \Delta \theta_0][/itex]?

    Algebraically, it's just the chain rule, either via the Jacobian, or via arithmetic with differential forms.
    Last edited: Apr 2, 2006
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