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December 19, 2014

December 19, 2014

Posted by **carlton** on Wednesday, May 2, 2012 at 7:31pm.

Y=sqrt(x+2), y=x,y=4

- calculus -
**Reiny**, Wednesday, May 2, 2012 at 9:50pmA rough sketch and some quick easy algebra shows that the curve intersects the lines at (2,2) and (14,4)

and the two straight lines intersect at (4,4)

so in 2 parts ....

V = π∫( x^2 - (x+2) ) dx from 2 to 4 + π∫(16 - (x+2))dx from 4 to 14

= π [(1/3)x^3 - (1/2)x^2 - 2x] from 2 to 4 + π[ 16x - (1/2)x^2 - 2x] from 4 to 14

I will let you do the arithmetic

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