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Angles in polar coordinates

  1. Jun 11, 2010 #1
    I'm doing work on polar coordinates in double integrals. Could someone explain why when circles aren't centered on the origin the angle only varies from (if it is translated above the origin) 0 to pi. I thought the angle was supposed to be the angle in the circle, so if its a full circle then 0 to 2pi. This is the same for when its translated right, the angle only varies from -pi/2 to pi/2.... I would have thought as its a full circle it is again 0 to 2pi.

    Where does this angle come from?
  2. jcsd
  3. Jun 11, 2010 #2


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    Science Advisor

    I am not sure what you are asking. If, for example, you have a circle with center at (1, 0) (on a Cartesian system) and radius 2 then in polar coordinates, [itex]\theta[/itex] certainly does go from 0 to [itex]2\pi[/itex].

    On the other hand, if the center is at (2, 2) and the radius 2, so that the entire circle is in the first quadrant but touches both x and y axes, [itex]\theta[/itex] ranges from 0 to [itex]\pi/2[/itex].

    And if the center is at (2, 2) and the radius 1, [itex]\theta[/itex] ranges between two numbers strictly between 0 and [itex]\pi/2[/itex].
  4. Jun 11, 2010 #3
    The angles are measured from the origin (0,0) and therefore circles that are entirely in one quadrant or half plane will have a smaller angle range.

    You could define polar coordinates which centered about the center of a circle, and then for that circle the angle will vary from 0 to 2pi
  5. Jun 11, 2010 #4
    Oh I see... so is this because when you change to polar coordinates, you go into the r theta plane and things are measured in angles from the origin (not angles in the circle if its not centered on the origin)? I sort of see how this works. When you change to polar coordinates, does the way it looks when graphed change? Or is it just a circle still? We are doing change of variables for double integrals and I noticed that when you change say x to u^2 - v^2 the graph of the domain is transformed to something different on the u v plane.
  6. Jun 12, 2010 #5


    Staff: Mentor

    A curve in the Cartesian (or rectangular) plane will have a different equation in the polar plane. For example, the equation x2 + y2 = 1 in the Cartesian plane has an equation of r = 1 in the polar plane. Each equation represents a circle of radius 1, centered at the origin/pole. In the polar equation, as theta ranges from 0 to 2pi, all points on the circle are swept out.

    The equation x2 + (y - 1)2 = 1 represents a circle of radius in the Cartesian plane, and centered at (0, 1). The same circle has the equation r = 2sin(theta) in the polar plane. In the polar plane, as theta ranges from 0 to pi, all points on the circle are swept out.

    To convert from Cartesian to polar coordinates use these equations:
    r2 = x2 + y2
    [itex]\theta = tan^{-1}(y/x)[/itex]

    To convert the other way, use these equations:
    [itex]x = r cos(\theta)[/itex]
    [itex]y = r sin(\theta)[/itex]
  7. Jun 12, 2010 #6
    How did you get r= 2sin(theta)? What does this tell us? The distance to the edge of the circle has this dependence on theta? Wouldnt there be two 'r' for every theta because if you draw a radial line out from the origin, it would cut the circle twice?
  8. Jun 12, 2010 #7


    Staff: Mentor

    x2 + (y - 1)2 = 1
    <==> x2 + y2 -2y + 1 = 1
    <==> r2 - 2r sin(theta) = 0
    <==> r(r - 2 sin(theta)) = 0
    <==> r = 0 or r = 2 sin(theta)
    We can discard r = 0 because when theta = 0, r = 0.
    So the polar equation of this circle is r = 2 sin(theta).
    It's the equation in polar form of the circle of radius 1 whose center is at (0, 1). My point was that a curve can have different equations in polar form and Cartesian form.
    Yes. Each point on the circle has coordinates (r, theta), so the radial distance from the pole depends on what the value of theta is.
    No, for each value of theta, you get one value of r. For example, if theta = pi/4, r = sqrt(2). To mark this point, draw a ray that extends out from the pole at an angle of pi/4, and measure sqrt(2) units on the ray. Draw a point there.
  9. Jun 12, 2010 #8
    Ok but what if the circle is translated right so you draw a line from the pole and then what represents the inner intersection and outer intersection? Isn't the value of r the distance to a point, so if it intersects the translated circle twice, one value of theta has two distances 'r'.
  10. Jun 12, 2010 #9


    Staff: Mentor

    You might not be thinking about this the right way. If you are thinking about an "r theta" plane that's an analogue to an "x, y" plane, there's an important difference. In the Cartesian or rectangular plane there is an x-axis and there is a y-axis. In the polar plane, there's no r axis and there's no theta axis. The positive x-axis in the Cartesian plane is the ray theta = 0. The positive y-axis is the ray theta = pi/2.
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