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December 22, 2014

December 22, 2014

Posted by **sara** on Sunday, August 3, 2014 at 6:47pm.

- math -
**Steve**, Sunday, August 3, 2014 at 7:00pmusing discs, of thickness dx

v = ∫[-1,1] πr^2 dx

where r = y = 1-x^2

So,

v = π∫[-1,1] (1-x^2)^2 dx = 16/15 π

Using shells, of thickness dy, we have to account for the two branches of the parabola, so it's easier just to use symmetry and get

v = 2∫[0,1] 2πrh dy

where r = y and h = x = √(1-y)

v = 4π∫[0,1] y√(1-y) dy = 16/15 π

- math -
**Damon**, Sunday, August 3, 2014 at 7:04pmpi y^2 dx from -1 to +1

same as 2 pi y^2 dx from 0 to 1

2 pi (1-2 x^2 + x^4) dx from 0 to 1

2 pi (1-0) - 4 pi (1^3/3-0) + 2 pi (1^5/5-0)

2 pi - 4 pi/3 + 2 pi/5

(pi/15) ( 30 - 20 + 6)

(16/15) pi

Check my arithmetic !

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