anonymous
  • anonymous
[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?
MIT 18.02 Multivariable Calculus, Fall 2007
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1332519642789:dw|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. \[dA=r dr d \theta\]
anonymous
  • anonymous
Ok, but my question was slightly different. We should be able to compute dA everywhere in the quadrant, shouldn't we? So, let's say we want to compute dA here: |dw:1332522845996:dw| Let's say that its distance from the origin is r/2, now for a given theta, its area is: \[dA = r/2 dr d \theta\] What I'm trying to say is that if we consider dA not on the border of the circle, but inside it, we should reduce the lenght of r and so dA. Where am I wrong?
anonymous
  • anonymous
But dA is inside the circle. Not sure what you mean by "on the border" of the circle? For any dA in the circle the r value varies..but its area it still rdrd(theta)...but r (and theta) differs depending on where the region is.

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anonymous
  • anonymous
And the region does not need to be a circle...any region conveniently described in polar coordinates in fine as long as it is a continuous region.
anonymous
  • anonymous
So, yes, the values of r and theta are different for every dA within a region. Thus the value of dA is different. But the formula is still rdrd(theta) for all subregions within the overall region. Not sure if I understand what you are asking?
anonymous
  • anonymous
I think I see what you mean. Yes, as r increase each dA contribues more and more to the total area of the region. An extreme case, the origin, r=0 so dA=0, which contributes nothing to the total area. Yes, dA depends on how far away you are from the origin, in direct proportion to r.
anonymous
  • anonymous
Oh, thank you, I finally understood! r and theta are variables...
anonymous
  • anonymous
yep :)

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