Question

[Polar Coordinates] Why dA = r dr d(theta)? I can't understand why is r dr. D. Auroux says it's because r d(theta) is the base of a rectangle, but doesn't this works only if we consider dA near the border? We do want to compute dA everywhere. What happens if dA is not on border, but somewhere else in the quadrant? Shouldn't we reduce the length of r?

Answer 1

Ok, in the diagram above we want to compute the area labeled dA. This is a polar "rectangle" of height delta r and width r*delta (theta). So, we are going to approximate this area by using the usual area=base*height formula for normal rectangles. Now, r of course is a little larger on the top of the "rectangle" than at the bottom. This means that the width of the top of the polar rectangle is a bit larger than the width at the bottom. But, picking either the top width or the bottom width we can say that \[\Delta A \approx \Delta r*r \Delta \theta\] where r is the distance from the origin to the bottom of the rectangle OR the top of the rectangle. We don't care which one you choose since they differ by such a small amount and this area is only an approximation. Now we allow delta r to shrink and shrink and shrink and we see that this area approximation becomes better and better. If the approximation approaches some limiting value we define the area to be the limit of this process and the formula above is no longer an approximation but becomes exact in the limit. \[dA=r dr d \theta\]