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=1-y^2 and the antenna is the y-axis.

2Pi integral (0 to 1) (1-x^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 (1-x)dx

V = (Integral of) 4 pi x*sqrt(1-x) dx

0 to 1

substitute 1-x = u

= (Integral of) 4 pi (u-1) *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 (1-x^2)^2 in the integral should be (1-y^2)^2

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