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Multi-variable integration with polar coordination

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex][tex]\int[/tex] of R ( sin( x^2 + y^2) ) dA where the region 4[tex]\leq[/tex] x^2+y^2 [tex]\leq[/tex] 49

    2. Relevant equations

    not too sure but i know that dy dx = r d(r) d(theta)

    3. The attempt at a solution

    i don't understand how to change into polar coordinates in order to integrate. i'm unsure of how the 4[tex]\leq[/tex] x^2+y^2 [tex]\leq[/tex] 49 becomes the r i need to integrate with. and if i do get that then how do i change the other bounds of integrate for the d (theta)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 21, 2009 #2

    Mark44

    Staff: Mentor

    What does "of R" mean in your integral? Does it represent the region you are describing, or is it some factor in the integrand?
    The region described by the inequalities is an annulus (a shape like a washer) that lies outside the circle x^2 + y^2 = 4 and inside the circle x^2 + y^2 = 49. In this region, r is going to be between two values.

    Your integrand is pretty simple to change to polar coordinates, since r^2 = x^2 + y^2.
     
  4. Oct 21, 2009 #3
    the R refers to the region that is being described. sorry for not clarifying that earlier.

    so with r^2 = x^2 + y^2, does that mean the bounds of d(r) would be from 4 to 49? and if so then would the bound of the integral for d(theta) be from 0 to 2(pi) since it is a complete circle?
     
  5. Oct 21, 2009 #4

    Mark44

    Staff: Mentor

    r does NOT range from 4 to 49. Draw the circles and it should be clear to you why this is so.
    [itex]\theta[/itex] does range from 0 to 2[itex]\pi[/itex].
     
  6. Oct 21, 2009 #5
    ah ok. i believe i see what you mean.

    after looking at the circles i think i see that r ranges from 2 to 7 which would be the radius of the circles. and so from there on do i just need to a double integral of sin(r^2) r dr d(theta)? or is there some other little step i'm missing?
     
  7. Oct 21, 2009 #6

    Mark44

    Staff: Mentor

    That's pretty much it. Your integral should look like this:
    [tex]\int_{\theta = 0}^{2\pi}\int_{r = 2}^7 sin(r^2) r dr d\theta[/tex]
     
  8. Oct 21, 2009 #7
    ok awesome. i got that integral and then attempted to solve for it.

    i used a u substitution for the sin(r^2) where:
    u=r^2
    du= 2rdr

    and i ended up with

    -(1/2) [tex]\int cos(u)d(theta) [/tex]
    where the integral is bounded from 0 to 2(pi) and the cos (u) goes from 4 to 49 because of the u substitution i used. i put the answer in and it said that it was incorrect
     
  9. Oct 21, 2009 #8

    Mark44

    Staff: Mentor

    Your second integral should look like this:
    [tex](-1/2)\int_{\theta = 0}^{2\pi}\left(cos(49) - cos(4)\right) d\theta[/tex]

    Edit: Added multiplier to front of integral.
     
    Last edited: Oct 21, 2009
  10. Oct 21, 2009 #9
    ok i got that but shouldn't there be a (-1/2) out in front of the integral to account for the u substitution?

    when i solve for the integral should i get

    [(cos(49)-cos(4))* 2pi]

    since the there isn't originally a [tex]\theta[/tex] so by taking the integral with respect to [tex]\theta[/tex] a [tex]\theta[/tex] should appear which can be bounded by the 2pi and 0
     
  11. Oct 21, 2009 #10

    Mark44

    Staff: Mentor

    Yes, it was in my notes, but I neglected to put that in my post.
    Yes, but you need to put in the missing (-1/2). If you evaluate your answer, remember that the numbers are real numbers (i.e., radians), not degrees.
     
  12. Oct 21, 2009 #11
    ah ok i finally got the right answer! it turns out that the answer is negative but i didn't think that was possible since we're computing an area so wouldn't the answer be a positive number?
     
  13. Oct 21, 2009 #12

    Mark44

    Staff: Mentor

    No, you're not computing the area of the region R. If you were, the integrand would be 1, not sin(x^2 + y^2). A lot of the surface z = sin(x^2 + y^2) must lie below the x-y plane for the integral to come out negative.
     
  14. Oct 21, 2009 #13
    o ok that makes sense! thank you very much for helping me with this problem and helping me to understand the concept better :smile:
     
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