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Mathematics 19 Online
OpenStudy (anonymous):

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.

OpenStudy (anonymous):

y=?????

OpenStudy (anonymous):

Just updated, sorry.

OpenStudy (agent0smith):

Using the disk method http://library.thinkquest.org/3616/Calc/S3/TDM.html \[\huge V = \pi \int\limits_{1}^{6} (\sqrt x )^2 dx\]

OpenStudy (anonymous):

not very helpful mate,

OpenStudy (agent0smith):

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.

OpenStudy (anonymous):

|dw:1368248461290:dw|

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