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Find the volume of solid that is obtained when region Under the curve y=sqrt(x) over the interval {1,6}is revolved around the x-axis.
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y=?????
Just updated, sorry.
Using the disk method http://library.thinkquest.org/3616/Calc/S3/TDM.html \[\huge V = \pi \int\limits_{1}^{6} (\sqrt x )^2 dx\]
not very helpful mate,
Umm..okay... if you read the link it might have been. The radius of the disk would be sqrtx, thickness is dx, the area of the integral is under the curve sqrtx from x=1 to x=6.
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