The volume of the solid obtained by rotating the region bounded by
y=x^2, \ y = 6 x
about the line
y = 36
can be computed using the method of washers or disks via an integral

Question

The volume of the solid obtained by rotating the region bounded by
y=x^2, \ y = 6 x
about the line
y = 36
can be computed using the method of washers or disks via an integral

anonymous · Answer

Have you done it?  You'll have to find the radius of the washer..

anonymous · Answer

yeah...well i know the limit of integration it is 0 to 6

anonymous · Answer

i'm trying to find the V

anonymous · Answer

not in figure but just how to set it up

anonymous · Answer

this i my answer so far 

pi [(6 - (1/6)y)^2-(6-sqrt(y))^2]

anonymous · Answer

but not sure

anonymous · Answer

the answer ask for that kind of set up :)

anonymous · Answer

my answer is slightly different though. Why do you take dy instead of dx considering that it is rotating about horizontally?

anonymous · Answer

sorry it dx...

anonymous · Answer

you're right

anonymous · Answer

It simplifies the equation alot :)

anonymous · Answer

?

anonymous · Answer

what's ur answer?

anonymous · Answer

I'm getting something like $V=\pi \int\limits_{0}^{6}( 72+6x-x^2)dx$

dumbcow · Answer

hmm... 
$$V = \pi \int\limits_{0}^{6}(36-x^{2})^{2} - (36-6x)^{2} dx$$
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