- #1
fleazo
- 81
- 0
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?
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?