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Expert: - 2/15/2010


Question
What is the volume of the solid obtained?

A region R is bounded by y=0, y=f(x)>0, x=o, x=t>0. If R is rotated around the x-axis, the volume of the resulting solid is (π/3)*(t^3+3t).

If the same region is rotated around the y-axis, what is the volume of the solid obtained?

I am completely stumped on this problem and was looking for some help in order to solve it. I am not sure if I should start with the given volume that is obtained when R is rotated around x-axis and work backwards to find out the volume when R is rotated around y-axis, OR if there is another way to go about finding the solution.

Answer
To find the volumne when rotated about the x-axis, that would be ∫πf²(x)dx from 0 to t.

This means that we need to differentiate the result to find f(x).
Doing this, we get π(x² + 3) = πf²(x), so x² + 3 = f²(x), so f(x) = √(x²+3).
Solving y = √(x²+3) for x in terms of y gives us y² = x² + 3, so x² = y²  3, so x = √(y²-3).

When x is 0, y is √3.  When x is t, y is √(t²-3).

To find the volume when rotated about the y-axis, do the following:
∫π(y²-3)dy from √(t²-3) downto √3.

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