Saturday

April 19, 2014

April 19, 2014

Posted by **Liz** on Sunday, May 27, 2012 at 4:22pm.

y= sqrt(x), y=x^2, in the first Quadrant

- Calculus -
**Steve**, Sunday, May 27, 2012 at 4:32pmThe curves intersect at (0,0) and (1,1).

Using shells,

v = ∫2πrh dx [0,1]

where r = x and h=√x-x^2

v = 2π∫x(√x-x^2) dx [0,1]

= 3π/10

Using discs,

v = ∫π(R^2-r^2) dy [0,1]

where R=√y and r=y^2

v = π∫(y-y^4) dy [0,1]

= 3π/10

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