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Trouble converting Cartesian to Polar

  1. Mar 30, 2003 #1
    Well, it is not really hard to convert them. My main problem is thinking in Polar coordinates. Cartesian coordinates are really easy to think about for me (after how many years of experience) but then I get to Calc 3 and I hit a brick wall. Does anyone have some insight on how to get past this problem?

    Thanks in advance!


    x^2 + y^2 = 9 gives a cylinder of radius 3 in 3-D

    the same equation is r = 3, right? That's pretty simple, but once it gets more complicated Boom, I'm lost.
  2. jcsd
  3. Mar 30, 2003 #2


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    Ellipse (center at origin, axes of ellipse along coordinate axes)

    (x^2/a^2) + (y^2/b^2) = 1

    Where a and b are the two semi-axes.

    Hyperbola (similar conditions)

    (x^2/a^2) - (y^2/b^2) = 1

    Parabola (vertex at origin, axis along y axis opening up)

    y = ax^2
  4. Mar 30, 2003 #3
    Just lots of practice to get familiar with them, just keep tredging through it and do your homework and after a while it will all start making sense. In fact for many problems the polar, or spherical coordinate systems will be come your coordinates of choice becuase they make many things much much simpler. Have you gotten to the change of variables topic yet in calc 3? If not, you'll see what I mean once you get there, many integrals are much easier to perform in sperical or cylindrical coordinates rather then in cartesian.
  5. Mar 30, 2003 #4
    Actually, the substitution part is what we're working on right now. That is why I asked the question. It's killing me. :) I'll just keep trudging. Thanks!
  6. Mar 30, 2003 #5


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    Well, substitution can mean different things... which operation gives you trouble when substituting?

    Can you apply the chain rule for differentiation when substitutions are involved? e.g.:

    Letting @ be the symbol for partial differentiation:

    @f/@r = @f/@x * @x/@r + @f/@y * @y/@r
    = cos &theta * @f/@x + sin &theta * @f/@y

    Did you understand substitution in integrals in calc 2? i.e.:

    &int sqrt(1 - x^2) dx = &int sqrt(1 - (sin &theta)^2) cos &theta d&theta = ...

    and by understand I mean that you could replace all occurances of one variable with the other (x -> sin &theta) in the integral, compute the equation for replacing the differential (dx = cos &theta d&theta), and you could compute the new bounds for your integral if it happened to be a definite integral.

    Do you understand how to perform substitution in a multiple integral, meaning you can both replace the variables and the differentials? E.G.

    via Jacobian:
    x = r cos &theta, y = r sin &theta
    dx dy = J(x, y / r, &theta) dr d&theta
    Code (Text):

     = det /@x/@r   @x/@&theta\ dr d&theta
           \@y/@r   @y/@&theta/
     = det /cos &theta   -r sin &theta\ dr d&theta
           \sin &theta    r cos &theta/
    = (r (cos &theta)^2 + r (sin &theta)^2) dr d&theta
    = r dr d&theta

    or via algebra: (I'm lazy so I won't compute things, just label 'em with capital variables)
    dx dy = (@x/@r dr + @x/@&theta d&theta) * (@y/@r dr + @y/@&theta d&theta)
    = A dr dr + B dr d&theta + C d&theta dr + D d&theta d&theta
    = 0 + B dr d&theta - C dr d&theta + 0
    = (B - C) dr d&theta = r dr d&theta

    Do you understand how to recompute the limits of integration in a multiple integral, or equivalently being able to translate a description of a region in one set of coordinates into another set of coordinates? e.g. going from:

    x^2 + y^2 <= 9


    r <= 3 and 0 <= &theta <= 2&pi

  7. Mar 30, 2003 #6
    Hurkyl, thanks, I am working through some topics you listed now. I did well in Calc 2 with the substitution. It is just stuff like the following:

    "Use a double integral in polar coordinates to find the volume of the solid that is described."

    x^2 + y^2 + z^2 = 9 (inside)
    x^2 + y^2 = 1 (outside)

    The surface is a circle with a cylinder through it (duh). I know I want the area from the cylinder wall out to the edge of the circle. I know that the lower bound for r will be 1. But the upper bound of r and the bounds for theta are kind of beyond me at the moment.

    Though, I am working these a little easier now.
  8. Mar 31, 2003 #7


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    So the description of your volume is:

    x^2 + y^2 > 1
    x^2 + y^2 + z^2 < 9

    Are you having trouble simple converting these equations into polar form, or can you get that far and just have trouble figuring out what the bounds of integration are?

  9. Mar 31, 2003 #8
    I am mainly having trouble figuring out the bounds. I can draw the pictures and see them. I cannot figure out the bounds. That is what is holding me up.
  10. Mar 31, 2003 #9
    Hmmm, OK, lemme try.

    x = r cos[psi] cos[the]
    y = r sin[psi] cos[the]
    z = r sin[the]

    I. r^2 < 9
    II. r^2 cos^2[the] > 1, or r^2 > 1/cos^2[the]

    dV = r^2 cos [the] dr d[the] d[psi]
    V = Int [0 ; 2[pi]] Int [-[pi]/2 ; [pi]/2] cos [the] Int [1/cos^2[the] ; 9] r^2 dr d[the] d[psi]

    OK so far?
    Last edited: Mar 31, 2003
  11. Mar 31, 2003 #10

    My Theta limits are wrong. We've got to keep 1/cos^2[the] smaller than 9, which means
    1/cos^2[the] < 9
    cos^2[the] > 1/9
    |[the]| < arccos 1/3

    So, should read Int [-arccos 1/3 ; arccos 1/3] ... d[the]
  12. Mar 31, 2003 #11
    Well, it's in polar coordinates, not cylindrical... But I assume those are correct. What I calculated:

    r^2 = 9 - 9cos^2(theta)
    r = 3 - 3cos(theta)

    Now, the book claims that the bounds are:

    8Int[0, pi/2]Int[1,3]...(dr)(d(theta))

    I understand the first bounds (eight quadrants for the full circle), but why is it Int[1,3] for r?

    Thanks for all your help!
  13. Mar 31, 2003 #12


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    There are only 4 quadrants in the full circle (I imagine the extra factor of 2 was pulled from inside the integral to outside).

    As for the bounds on r, go back to the inequalities I listed:

    x^2 + y^2 > 1 (outside the cylinder)
    x^2 + y^2 + z^2 < 9 (inside the sphere)

    Convert those to cylindrical coordinates (cylindrical coordinates are the same thing as polar coordinates, just with z as a third coordinate) and you might see where the bounds on r come from! (If not, show me what you got as the result from converting both equations and I'll go from there)

  14. Mar 31, 2003 #13
    I meant octants, not quadrants. Sorry! That explains the 8.

    In cyclindrical coordinates, I got the following for x^2 + y^2 + z^2 = 9:

    So, r goes from 1 to 3 - 3cos(theta) right?

    Thanks a lot for your help, I know I am being a bit dense.
  15. Apr 1, 2003 #14


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    Where are you getting cos theta? z should just remain z when you do the conversion

  16. Apr 1, 2003 #15
    Mixing up Spherical with Cylindrical for some reason. So it should remian r^2 + z^2 = 9?
  17. Apr 1, 2003 #16


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    Right, it should remain r^2 + z^2 < 9 (don't forget the inequality; it specifies you care about the interior of the sphere rather than just its surface. More complicated problems can be made a little easier if you don't forget this)

    From this inequality, you can find the largest possible value of r (which is 3 when z = 0), and then once you have your bounds on r you can solve for z to get the bounds on it.

  18. Apr 1, 2003 #17
    Thanks! I am working on a lot more of these problems now. Better safe than sorry. I just cannot understand why only Odd answers are given in supplemental answer books. If professors do collect homework, work has to be shown... Just doesn't make sense. :)

    Thanks again!
  19. Apr 2, 2003 #18
    one of these should be a symbol for partial differencation

    delta &delta;
    Delta &Delta;
  20. Apr 2, 2003 #19
    actually it's this one [pard]
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