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The region bounded above by the line y=6, below by the curve y=sqrt(x) and to the left by the y-axis, rotated around x=36.
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we should get the same results if im thinking of it correctly
but just in case; think of this as a solid cylinder that we are removing the under part of the sqrt(x)
|dw:1327534530288:dw|
so, cylindar itself; r=36, h=6; pi 36^2*6 is total area of this cylindar, now we gut out the area under the sqrt(x)
so it's like an upside down bowl
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yes, and lets disc out the underside, i think it might integrate easier that way
ok after I integrated I got, 18x^2 - 2/3 * x^(3/2)
|dw:1327534763514:dw|
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