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Homework Help: Rectangle to Cylindrical coordinate question

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data

    can you explain this conversion, I am not sure.

    Rectangle coord :

    [tex]\int^{2}_{-2}[/tex][tex]\int^{sqrt(4-x^2)}_{-sqrt(4-x^2)}[/tex][tex]\int^{2}_{sqrt(x^2 + y^2 )}[/tex] F(x) dzdydx


    cylindrical coord :

    [tex]\int^{2\pi}_{0}[/tex][tex]\int^{2}_{0}[/tex][tex]\int^{2i}_{r}[/tex] r*dzdrd[tex]\theta[/tex]

    I see that x^2 + y^2 = r so the right most integral in cylindrical coordinate is from 2 to r

    The middle integral from symmetry runs from 0 to sqrt(4 - x^2), but they have that as 0 to 2
    So I am assumming this is what they did :

    y = sqrt(4-x^2)

    y^2 = 4 - x^2

    y^2 + x^2 = 4

    r = 2

    thus, 0<= r <= 2 ?

    and the 0 to 2Pi is just the while circle thats begin read.

    Also in the rectangle coordinate, what happened to -2 and 2 in the leftmost integral?
  2. jcsd
  3. Nov 14, 2009 #2


    Staff: Mentor

    I always found it helpful to put in some indication of what the variable was in the limits of integration when dealing with iterated integrals. It helped me keep track of which was the variable in a particular integration, and it helped me better understand the region over which integration was taking place. With that modification, here's your rectangular coordinates iterated integral.

    [tex]\int_{x = -2}^2\int_{y = -\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{z = \sqrt{x^2 + y^2}}^2 F(x)~dz~dy~dx[/tex]

    The conversion is incorrect. For one thing, the F(x) in the original integral doesn't just go away, as it has appeared to do in your iterated integral in cylindrical coordinates. Your differential volume, r dz dr d[itex]\theta[/itex], is correct, though.

    The new limits of integration are close, but not quite right. In the rectangular coordinates iterated integral, think of a stack of blocks, each of volume dz*dy*dx. In the first integral the blocks range from z = sqrt(x^2 + y^2) to z = 2. You show the lower integration limit as z = r, which is incorrect. Also, you show the upper limit of integration as z = 2i, which isn't correct, either. Was the i a typo? What sorts of figures do z = sqrt(x^2 + y^2) and z = 2 represent? The first is not a sphere, as I think you believe.

    In the middle integration, the vertical stack of blocks sweeps from y = -sqrt(4 - x^2) to y = sqrt(4 - x^2). This is a circle of radius 2, centered at the origin, so in the cylindrical form, y does range between 0 and 2.

    Finally, the outer cylindrical limits are correct, because we're dealing with the entire circle.
  4. Nov 14, 2009 #3
    It may help if you start by sketching what the figure looks like. Then from the boundaries you can express it as r, theta, and z. From that you can find your r*d(theta), dr, and dz.
  5. Nov 14, 2009 #4


    Staff: Mentor

    Agreed. One of the most difficult things about these kinds of problems is getting a handle on what the region of integration looks like. Many times the integration itself is pretty easy. It's essential that you draw a sketch of that region.
  6. Nov 14, 2009 #5
    Yea the 2i was a huge typo. I am sorry, F(x) = x^2 + y^2 in the rectangular integral.

    Here are the dimension for the rectangle integral :

    R = { (x,y,z)| -2 <= x <= 2,
    -sqrt(4 - x^2) <= sqrt(4-x^2),
    sqrt(x^2+y^2) <= z <= 2

    and the converted cylindrical coordinate :

    C = { (r,theta,z) | 0 <= theta <= 2pi,
    0 <= r <= 2,
    r <= z <= 2

    Mark, you told me why theta ranges from 0 to 2pi, because we are integrating through
    the whole region.
    I know r <=2 and r>= 0 because of what I should in my first post.
    I know z ranges from r to 2 because r = x^2+y^2.

    But I would like to know what happened to the x limit in the rectangular coordinate.
  7. Nov 14, 2009 #6
    Those are not x, y, and z limits anymore once you convert to cylindrical coordinates. They are limits of theta, r, and z. There is not an x to integrate.
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