Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Double Integral in Polar Coordinates problem

  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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook