Find the volume of the solid obtained by revolving the region under the graph of y=sin(x/2) from x=0 to x=2pi about the y-axis

Question

anonymous · Answer

I tried doing $$\int\limits_{0}^{2\Pi}2\Pi rh$$ which simplifies to 
$$2\Pi\int\limits_{0}^{2\Pi} h dx$$
but that didn't give the right answer

rational · Answer

x=0 to x=2

rational · Answer

may i knw why you're doing it 0 to 2pi ?

anonymous · Answer

sorry it should've said from 0 to 2pi

rational · Answer

okay, wat u had setup is for revolving around x-axis, but the question is asking u to revolve it around y-axis right ?

anonymous · Answer

2pi*r is the formula for circumference.  So I am taking the integral of a whole bunch of tiny circumferences with height sin(x/2) from 0 to 2pi.  Please tell me if this is wrong.

rational · Answer

try this : http://www.wolframalpha.com/input/?i=%5Cint+%5Climits_%7B0%7D%5E%7B2%5Cpi%7D+2%5Cpi+x+%5Csin%28x%2F2%29+dx

rational · Answer

i have used shells...

rational · Answer

I think u just missed multiplying wid "x"

rational · Answer

otherwise, your integrand is also looking good :)

rational · Answer

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