Saturday

August 2, 2014

August 2, 2014

Posted by **Anonymous** on Saturday, April 23, 2011 at 4:41pm.

- calculus -
**drwls**, Saturday, April 23, 2011 at 4:53pmThe bounded region is x =0 to x = 1, so those are your limits of integration.

The integral is of

pi*(x^2 - sqrt x)^2 dx.

Visualize it as a stack of washers with holes in the middle.

- calculus -
**Mgraph**, Saturday, April 23, 2011 at 5:17pmI think that

the volume=pi*Int(from 0 to 1) (sqrt(x))^2dx-pi*Int(0 to 1)(x^2)^2dx=

pi(1/2-1/5)

- calculus -
**drwls**, Saturday, April 23, 2011 at 5:23pmMgraph is correct. I should have written

Integral of pi[(sqrtx)^2 - (x^2)^2] dx

0 to 1

which is pi*(1/2 - 1/5)

- calculus -
**Anonymous**, Saturday, April 23, 2011 at 5:26pmThank you so much that makes alot of since.

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