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January 30, 2015

January 30, 2015

Posted by **Anonymous** on Thursday, May 22, 2008 at 1:53am.

He carved a parabola out of the cover of his teacher’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 5:19amx=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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