Results 1 to 8 of 8

Math Help - 2-dimensional integral

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    77

    2-dimensional integral

    Region A is the area between the unit circle and the circle of radius 2, in the top right quadrant only. Find

    \int_A{xy}~dA

    Ok so that's the question. We're doing the theory behind changing variables from x,y to u,v or r,θ.

    In standard x,y coordinates:

    \int_0^2\int_{\sqrt(1-y^2)}^{\sqrt(4-y^2)}{xy}~dx~dy

    which comes out as 3. I'd write it all out but I'm not fluent in Latex and it would take hours.

    In r,θ:

    \int_1^2\int_0^{\pi/2}{sin\theta cos\theta r^3}~dx~dy which comes out as 15/8. I'm pretty sure the second one is the one intended to be correct, but not sure what I'm missing. Perhaps because the x,y integral doesn't limit to the top right quadrant? I just can't work it out.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by Thomas154321 View Post
    Region A is the area between the unit circle and the circle of radius 2, in the top right quadrant only. Find

    \int_A{xy}~dA

    Ok so that's the question. We're doing the theory behind changing variables from x,y to u,v or r,θ.

    In standard x,y coordinates:

    \int_0^2\int_{\sqrt(1-y^2)}^{\sqrt(4-y^2)}{xy}~dx~dy

    which comes out as 3. I'd write it all out but I'm not fluent in Latex and it would take hours.

    In r,θ:

    \int_1^2\int_0^{\pi/2}{sin\theta cos\theta r^3}~dx~dy which comes out as 15/8. I'm pretty sure the second one is the one intended to be correct, but not sure what I'm missing. Perhaps because the x,y integral doesn't limit to the top right quadrant? I just can't work it out.
    Perhaps it would be best to work out the area enclosed by the circle of radius 2 and the coordinate axes, then subtract the area enclosed by the 1 radius circle and the axes. I believe that will yield the correct answer.

    So:

    Circle radius 2:

     \int_0^2 \int_0^{\sqrt{4-y^2}} xy dx dy

    Circle radius 1:

     \int_0^1 \int_0^{\sqrt{1-y^2}} xy dx dy

     \text{Total Area} = \int_0^2 \int_0^{\sqrt{4-y^2}} xy dx dy-\int_0^1 \int_0^{\sqrt{1-y^2}} xy dx dy
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,384
    Thanks
    51
    Quote Originally Posted by Thomas154321 View Post
    Region A is the area between the unit circle and the circle of radius 2, in the top right quadrant only. Find

    \int_A{xy}~dA

    Ok so that's the question. We're doing the theory behind changing variables from x,y to u,v or r,θ.

    In standard x,y coordinates:

    \int_0^2\int_{\sqrt(1-y^2)}^{\sqrt(4-y^2)}{xy}~dx~dy

    which comes out as 3. I'd write it all out but I'm not fluent in Latex and it would take hours.

    In r,θ:

    \int_1^2\int_0^{\pi/2}{sin\theta cos\theta r^3}~dx~dy which comes out as 15/8. I'm pretty sure the second one is the one intended to be correct, but not sure what I'm missing. Perhaps because the x,y integral doesn't limit to the top right quadrant? I just can't work it out.
    You need to switch dx\;dy\; = \; r \; dr\; d\theta (you have the r)

    so

    \int_0^{\pi/2} \int_1^2 \; r sin\theta \;r cos\theta \;r \; dr\; d\theta = \frac{15}{8}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2006
    Posts
    77
    I did that, hence my r^3. I'm more confident that my polar integral is correct than the x,y one.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by danny arrigo View Post
    You need to switch dx\;dy\; = \; r \; dr\; d\theta (you have the r)

    so

    \int_0^{\pi/2} \int_1^2 \; r sin\theta \;r cos\theta \;r \; dr\; d\theta = \frac{15}{8}
    The OP already did this. The 2nd integral was not the problem. It yielded the correct answer. The 1st integral was the anomaly.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,384
    Thanks
    51
    Quote Originally Posted by Thomas154321 View Post
    I did that, hence my r^3. I'm more confident that my polar integral is correct than the x,y one.
    Sorry. To do your problem in x\; \text{and}\;y you'll need two integrals. The problem is that your left boundary changes from y = [0,1]\;\text{and}\; [1,2]

    so

    <br />
\int_0^1\int_{\sqrt{1-y^2}}^{\sqrt{4-y^2}}{xy}~dx~dy\;\;+\;\;\int_1^2\int_{0}^{\sqrt{4-y^2}}{xy}~dx~dy
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by Thomas154321 View Post
    I did that, hence my r^3. I'm more confident that my polar integral is correct than the x,y one.
    It is. See my post.

    The reason your first integral didn't work out is because the x limits are not ALWAYS true for the area we are considering.

    Consider the small slice of the region ABOVE the point where the smaller circle meets the y axis, and BELOW the point there the big circle meets the y axis. Is it true for that region, to say that the you enter the region through the boundary of the smaller circle and exit through the boundary of the larger circle? No, it is not. In fact, for that small portion, you enter through the y axis, and exit through the boundary of the larger circle. Your limits do not account for that small slice of the region.

    Employ my method.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Sep 2006
    Posts
    77
    Both your methods work, thanks very much.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. N-dimensional spheres
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 25th 2011, 11:07 PM
  2. Dimensional
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: November 30th 2010, 06:07 PM
  3. Replies: 1
    Last Post: February 25th 2010, 02:15 AM
  4. 3 Dimensional Volume Integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 14th 2007, 08:34 PM
  5. A one dimensional...
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 22nd 2006, 12:18 AM

Search Tags


/mathhelpforum @mathhelpforum