Volume of rotation - my answer seems off a bit

lucidlobster
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Find the area of the given function, rotated about the y axis

The Area below the line y=2 and above y=sin(x) from 0-\pi

I did this rotated around the X AXIS no problem,

I found the area of the disk to be \pi(4-sin^{2}(x))

The volume around the X AXIS to be\frac{7\pi^{2}}{2}



The Attempt at a Solution



For the Y axis I have used this:\int^{\pi}_{0} 2\pi xf(x)dx
Plugging in:

\int^{\pi}_{0} 2\pi x(2-sin(x))dx

I solved this to be 2\pi^{3}-2\pi^{2}


Does this look right?
 
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Really? That is awesome!

Any second opinions?
 
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