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Area under the curve using polar coordinates - help!

  1. Aug 26, 2012 #1
    Hi, I have a pretty simple question but I'm not certain I know how to phrase it properly. I will try.

    When we are integrating using cartesian coordinates to find the area under a curve, area under the x axis is negative and area above the x axis is positive. This makes sense when I think of the integral in terms of reimann sums because because we are just summing areas of rectangles using the formula f(t)*(t-a) for some t in the interval we're integrating over. If we have an f(t) under the x axis, that means f(t) is negative, and since dx is "positive", we would get a negative number for the area of the rectangle.

    But when thinking of polar coordinates I'm confused. In polar coordinates, θ is like our "x" and r is like our "y". So it seems the analogue to this situation would be when r is negative. But when thinking of this in terms of reimann sums, that doesn't seem like the case. Since the area of a sector of a circle is (1/2)r2θ, if r is negative it just becomes positive when we square it. So the only thing that would make this amount negative is if θ is negative. in the integral, dθ takes the place of θ in this equation. Does dθ ever become "negative"? Do we have to worry about negative area when dealing with integrating using polar coordinates?
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  3. Aug 26, 2012 #2


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    Imagine integrating around a circle that does not touch or contain the origin to find its area. Suppose you go anticlockwise around it. On the section that's further from the origin, dθ is positive, and it gives you not just the area of the circle but also the area of the triangle it subtends at the origin; on the return section, it is negative, and subtracts the extra area to leave you with just the area of the circle.
  4. Aug 26, 2012 #3
    wow thank you, that makes perfect sense!

    I have one doubt though. How is it that dθ is negative? When I think of dθ, I think of it like a metric, like it's measuring the distance between two points and that it's always positive. So I'm not quite grasping what it is that makes dθ negative. Is it because we are going anti clockwise? If so, why is it that the direction we go has an impact on the sign of dθ?

    Thanks again for your explanation
  5. Aug 26, 2012 #4
    Can you give an example where you would use polars to evaluate the area under a graph?

    I generally only use it to evaluate an area within some sort of boundary.
  6. Aug 26, 2012 #5
    Sorry, it was bad wording on my part. That's what i meant, was the interior of say a circle, or a cardioid or something like that.
  7. Aug 27, 2012 #6


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    Remember that θ is an angle measured at the origin. On the 'return' part, although you are still going anticlockwise around the circle, θ is decreasing.
  8. Aug 27, 2012 #7
    You can also think of it in terms of a surface integral. dA=dxdy=rdrdθ. This is calculated through the determinant of the Jacobian. Now, if dxdy is a small rectangle what is rdrdθ?

    Well, rdθ is an arc length, therefore you have small convex rectangles. However, because both dr and dθ are infinitesimal in size, you can consider that dθ(r) ~ dθ(r+dr), therefore they degenerate to normal rectangles like in the cartesian case.
  9. Aug 27, 2012 #8
    I think you are mixing together a number of different things.

    When you think of an integral in terms of "area under the curve", you think of a "simple" integral (of f(x) on an interval). When you think of polar coordinates, you think either about a contour integral or a planar double integral.

    They have somewhat different forms, but all of them can be used to compute "area under the curve", in which case one form can always be transformed into another.
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