# Calculus

The area bounded by the curve y^2 = 12x and the line x=3 is revolved about the line x=3. What is the volume generated?

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1. x = 3 and y^2 = 12x intersect at (3,6) and (3,-6)
so we can take horizontal slices, and because of the symmetry just go from y = 0 to y = 6 and double the result.
we also have a radius of 3 - y^2/12
V = 2π ∫r^2 dy from 0 to 6
= 2π ∫(3 - y^2/12)^2 dy from 0 to 6
= 2π ∫(9 - (1/2)y^2 + y^4/144) dy
= 2π [ 9y - (1/6)y^3 + (1/720)y^5] from 0 to 6
= 2π ( 54 - 216/6 + 7776/720 - 0 )
= 2π(144/5)
= 288π/5 or appr 180.96

confirmation:
http://www.wolframalpha.com/input/?i=integral+2%CF%80%283+-+x%5E2%2F12%29%5E2++from+0+to+6

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2. Hi, can anyone do it it disk method in terms of dx plss

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