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Double Integral with Polar Coordinates

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int^{0}_{-3}\int^{\sqrt{9 - x^2}}_{- \sqrt{9 - x^2}} \sqrt{1 + x^2 + y^2} dy dx[/tex]

    2. Relevant equations

    x = rcos(theta)
    y = rsin(theta)


    3. The attempt at a solution

    By making [tex]\sqrt{9 - x^2} = y[/tex] then changing it to polar coordinates, I got r to be +/-3

    but I'm not sure how to find what theta is bounded by. From what I read, since its [tex]\sqrt{1 + x^2 + y^2}[/tex] it's in the first quadrant. Not sure what to do now.
     
  2. jcsd
  3. Feb 12, 2009 #2

    Mark44

    Staff: Mentor

    Your region of integration is the left half of a circle of radius 3, centered at the origin. I would let r run from 0 to 3, theta from pi/2 to 3pi/2.
     
  4. Feb 12, 2009 #3
    How did you know that theta runs from pi/2 to 3pi/2?
     
    Last edited: Feb 12, 2009
  5. Feb 12, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    He just told you:
    [itex]\theta= 0[/itex] is in the direction of the positive x-axis, [itex]\theta= \pi/2[/itex] points in the direction of the positive y-axis, [itex]\theta= \pi[/itex] points in the direction of the negative x-axis, and [itex]\theta= 3\pi/2[/itex] points in the direction of the negative x-axis. The cover the left half of the circle, you go counterclockwise from the positive y-axis to the negative x-axis: from [itex]\pi/2[/itex] to [itex]3\pi/2[/itex].
     
  6. Feb 12, 2009 #5
    Can't believe I didn't realise that; I'm such an idiot. Thank you both.

    I'm probably wrong, but wouldnt [itex]\theta= 3\pi/2[/itex] point to the negative y - axis, since it's the left half of a circle?
     
    Last edited: Feb 12, 2009
  7. Feb 12, 2009 #6

    Mark44

    Staff: Mentor

    Right. I'm sure that's what Halls meant.
     
  8. Feb 12, 2009 #7
    So I tried another one, but the book says I'm wrong:


    [tex]\int^{1}_{0}\int^{x}_{0} \frac{x + y}{\sqrt{x^2 + y^2}} dx dy = \int^{\pi / 2}_{0}\int^{1}_{0} \frac{rcos\vartheta + rsin \vartheta}{\sqrt{r^2 cos^2\vartheta + r^2 sin^2 \vartheta}} r dr d \vartheta = \int^{\pi / 2}_{0}\int^{1}_{0} \frac{rcos\vartheta + rsin \vartheta}{r} r dr d \vartheta = 1/2 \int^{\pi / 2}_{0} (sin\vartheta - cos \vartheta ) r^2 |^1_0 d \vartheta = [/tex]

    [tex] 1/2 \int^{\pi / 2}_{0} sin \vartheta - cos \vartheta d\vartheta[/tex]

    when I change the boundry for x to polar coordinates, would it go from 0 - 1 or 0 - rcos?
     
    Last edited: Feb 12, 2009
  9. Feb 12, 2009 #8

    Mark44

    Staff: Mentor

    Your limits on the the first two iterated integrals are wrong. One of the key points in switching from Cartesian to polar is understanding exactly what the region is over which integration takes place.

    For the first integral, it looks like you copied the order of integration incorrectly, and it should by dy, then dx, not the other way around as you have them.

    Can you describe the region on which integration takes place in the first integral? It's a very simple geometric object. When you have that figured out, you should be able to get the right limits on your polar integral.
     
  10. Feb 12, 2009 #9
    It's the way in my book, unless I have to change them.

    For the first integral, y is bounded between 0 - 1, while x is bounded on the positive x-axis (since it's 0 - x).

    dx dy = r dr d(theta) or is it dy dx = r dr d (theta)?
     
  11. Feb 12, 2009 #10

    Mark44

    Staff: Mentor

    That's a typo in your book. If it were as you are interpreting it, the region would be infinitely large. x is bounded by what, x?
    That's the same as saying that the region for the inner integral is from x = 0 to x = x, and for the outer integral, y = 0 to y = 1.

    I'm sure this is what they meant:
    [tex]\int_{x = 0}^1 \int_{y = 0}^x <your f(x, y)> dy dx[/tex]
    and got the differentials in the wrong order.
    Yes.
    As far as areas are concerned dx*dy = dy*dx, and they both equal r*dr*d(theta). The difference is that the order of dx and dy controls the order in which integration is done.
     
  12. Feb 12, 2009 #11
    so y is bounded between 0 and x, while x is bounded by 0 and 1. That does make more sense. It looks like it's in the first quadrant. So changing it to polar coodrinates:

    [tex]\int_{0}^{\pi / 4} \int_{0}^{cos\vartheta} r dr d \vartheta[/tex]
     
  13. Feb 12, 2009 #12

    Mark44

    Staff: Mentor

    The limits look good, so I trust that you understand the integration region. Don't forget your integrand, though. The integral you have just computes the area of the region, and that's not what you want.
     
  14. Feb 12, 2009 #13
    Thanks for your help. Just got one more question. I'm wondering if how I changed the boundries from rectangular coordiantes to polar coordinates is the correct way, and I didnt get the right answer by some fluke.

    in [tex]\int_{0}^{\pi / 4} \int_{0}^{cos\vartheta} r dr d \vartheta[/tex]

    I did this: y = x => rsin = rcos => sin = cos (and this can only be true when the angle is (pi/4)

    for [tex]\int^{0}_{-3}\int^{\sqrt{9 - x^2}}_{- \sqrt{9 - x^2}} \sqrt{1 + x^2 + y^2} dy dx[/tex]

    I let [tex]y = \sqrt{9 - x^2}\Rightarrow rsin = \sqrt{9 - r^2 cos^2}[/tex] which eventually gave me r = +/- 3.

    Sorry for being so annoying.
     
  15. Feb 12, 2009 #14

    Mark44

    Staff: Mentor

    On thinking about it some more, your upper limit of integration needs to be r = 1/cos(theta). The limits of integration x = 0 to x = 1 and y = 0 to y = x represent a right triangle with vertices at (0,0), (1, 0) and (1, 1). It's isosceles with angles of 45 degrees, for which sin(theta) = cos(theta) or tan(theta) = 1. In the polar integration r ranges from 0 to the line x = 1, or rcos(theta) = 1, or r = 1/cos(theta).

    For the other (actually the first integral in this thread), the region is the left half of a circle whose radius is 3 and where theta ranges between pi/2 and 3pi/2 (already discussed).

    In converting from one type of iterated integral to another, the limits of integration depend completely on the region, so if you don't understand what the region looks like, you won't get them right in the new integral.
     
  16. Feb 12, 2009 #15
    but I thought r ranges from 0 to the line y=x (or is the same as y = x = 1?)
     
  17. Feb 12, 2009 #16
    If r varied from 0 to 1 and theta varied from 0 to pi/4, you would have a region that is the sector of a circle. It would have the curvature because r would include all values between 0 and 1 on this eighth of a circle.

    Your region, on the other hand, is a right triangle. Do you see the difference?
     
  18. Feb 12, 2009 #17

    Mark44

    Staff: Mentor

    Not the same. y = x = 1 is the point (1, 1).
     
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