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September 23, 2014

September 23, 2014

Posted by **Grant** on Monday, August 2, 2010 at 10:10pm.

y=0, y=x(3-x) about the axis x = 0

- Calculus -
**Reiny**, Monday, August 2, 2010 at 10:47pmvolume = π[integral]y^2 dx from x = 0 to x = 3

y=x(3-x)

y^2 = x^2(3-x)^2

= x^2(9 - 6x + x^2)

= 9x^2 - 6x^3 + x^4

the integral of that is

3x^3 - (3/2)x^4 + (1/5)x^5

so volume

= π(81 - 243/2 + 243/5 - 0)

= (17/2)π

check my arithmetic

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