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coverband
Aug14-08, 10:31 AM
If you can imagine two curves
y = x^2
y=x
between y=[0,1] and x=[0,1] and you are asked to perform a double integration (compute the area encloed by the two curves) you can perfom this with x as the inner integral or y as the inner integral.
When x is the inner integral the limits are from y to sqrt(y)
When y is the inner integral the limits are from x^2 to x

My question is why do they go from y to sqrt(y) and x^2 to x and not sqrt(y) to y and x to x^2?

For a picture of the above see http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/255doub/255doub.html final example

NoMoreExams
Aug14-08, 11:04 AM
If you can imagine two curves
y = x^2
y=x
between y=[0,1] and x=[0,1] and you are asked to perform a double integration (compute the area encloed by the two curves) you can perfom this with x as the inner integral or y as the inner integral.
When x is the inner integral the limits are from y to sqrt(y)
When y is the inner integral the limits are from x^2 to x

My question is why do they go from y to sqrt(y) and x^2 to x and not sqrt(y) to y and x to x^2?

For a picture of the above see http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/255doub/255doub.html final example

Because you are going "left to right" when your inner integral is wrt dy and from "bottom" to "up" when you do it wrt to dx. So when you do it wrt dy, note that you are going between y = x and y = x^2 (or x = sqrt(y) since we need bounds in terms of some function of y). When you are doing it wrt dx, you are going from y = x^2 to y = x. Visually you want to see which function is "greater". So when dealing with dy, rotate your graph and you will see that y = x^2 is "above" y = x and thus your lower bound is x = y and your upper bound is x = sqrt(y). Now when you do it wrt dx, you should see that y = x is "above" y = x^2 (at least on your interval of [0,1] x [0,1]) and that is why dx goes from y = x^2 to y = x.

coverband
Aug14-08, 11:20 AM
you're a genius

coverband
Aug14-08, 01:46 PM
Actually i don't know if this is correct. See http://www.libraryofmath.com/double-integral-over-a-more-general-region.html example (d)

NoMoreExams
Aug14-08, 02:20 PM
Actually i don't know if this is correct. See http://www.libraryofmath.com/double-integral-over-a-more-general-region.html example (d)

Which part of their work do you disagree with? See the response to the thread you started. (you don't need to start a thread if you have a question for me, a PM would do or just asking in the thread that already exists).