Posted by Anonymous on Thursday, May 22, 2008 at 1:53am.
Can you check if I set up the integral correctly? Thanks!
He carved a parabola out of the cover of his teacherâ€™s edition. It spins around the antenna in the wind. Find the volume of this solid of revolution if the equation of the parabola is x=1y^2 and the antenna is the yaxis.
2Pi integral (0 to 1) (1x^2)^2 dy =16pi/15

Calculus  drwls, Thursday, May 22, 2008 at 5:19am
x=0 at y = 1 and +1
The base of the parabola is at x =0 and it extends to a vertex at (x=1, y=0)
As you rotate about x=0, you generate cylindrical shell elements of the solid of revolution with volume
dV = (2 pi x) * 2 sqrt (1x)dx
V = (Integral of) 4 pi x*sqrt(1x) dx
0 to 1
substitute 1x = u
= (Integral of) 4 pi (u1) *sqrt u du
1 to 0
= (4/15)(4 pi) = 16 pi/15
I agree with your answer but see that you integrated slabs running in the y direction. That would work, but it seems to me that your (1x^2)^2 in the integral should be (1y^2)^2
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