Complementary Coffee Cups?

Question

Hi, I can't seem to get this problem, can anybody help me out?

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Suppose you have a choice of two coffee cups of the typeshown, one that bends outward and one inward, and you notice thatthey have the same height and their shapes fit together snugly. Youwonder which cup holds more coffee. Of course you could fill onecup with water and pour it into the other one but, being a calculusstudent, you decide on a more mathematical approach. Ignoring thehandles, you observe that both cups are surfaces of revolution, soyou can think of the coffee as a volume of revolution.



1.Suppose the cups have height h,cup A is formed by rotatingthecurve X = f(y) aboutthe Y-axis, and cup B is formed by rotatingthe same curve about the line X = K. Find the value of K suchthat the two cups hold the same amount of coffee.



2.What does your result from Problem 1 say about the areas A1 andA2 shown in thefigure?

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So, in order to solve this, I made two equations for volumes of revolution,

 π int([f(y)]^2dy)
and
 π int([k-f(y)]^2dy)

And I set them equal to each other, simplifying to get:

K= (2/y) * int (f(y)dy)

Which is all good and well, but what does that mean for the areas?

Mathsorcerer · Accepted Answer

Once you set the volumes equal to each other you have this:

π ∫[f(y)]^2 dy = π ∫[k - f(y)]^2 dy
∫[f(y)]^2 dy = ∫[k - f(y)]^2 dy

[f(y)]^2 = [k - f(y)]^2 (differentiate both sides to get rid of the ∫}

f(y)^2 = k^2 - 2kf*y) + f(y)^2
0 = k^2 - 2kf(y)
0 = k[k - 2f(y)]
either k = 0 (which means the second cup is also rotated around the y-axis, making it the first cup)
or 0 = k - 2f(y) --> k = 2f(y)

When k = 2f(y) both cups will have the same volume.  I don't see any images, so I can't say anything about the surface areas of the two cups.

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Complementary Coffee Cups?

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