The region in the first quadrant bounded by the x-axis, the line x = π, and the curve y = sin(sin(x)) is rotated about the x-axis. What is the volume of the generated solid?

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The region in the first quadrant bounded by the x-axis, the line x = π, and the curve y = sin(sin(x)) is rotated about the x-axis. What is the volume of the generated solid?

Mathematics
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my main trouble is finding the area bounded by the functions
|dw:1438886124740:dw| looks something like that
well the area bounded by x=0 and x=pi and y=sin(sin(x) and y=0

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Other answers:

x = pi !!! i see for some reason i just assumed it was y = pi
y=sin(sin(x))* forgot to close the parenthesis
okay so this area right |dw:1438886325934:dw|
they want you to find the volume of that thing rotated about y=0? or just find the integral representation fro the volume?
the former
have you ever heard of the bessel function?
i have now
let me try to do the problem and tell me if i did something wrong
alright just so we are clear that one equation is y=sin(sin(x))?
if so i have some research to do :p unless we can use some form of approximating maybe like taylor series
yes it is y= sin(sin(x)) but if your wondering i'm rather certain the x intercept happens to be at pi
its just calc 1 so should not involve anything too complex
i'm just extremely curious how we are going to evaluate: \[\int\limits_{0}^{\pi} \pi \sin^2(\sin(x)) dx\] it involves non-elementary functions
well ill just use the trusty ti-84 :D
ok so that is awesome if we can use calculator
1.219
I'm not getting that
what are you getting ?
oh woops forgot to multiply by pi
http://www.wolframalpha.com/input/?i=integrate%28pi*sin%5E2%28sin%28x%29%29%2Cx%3D0..pi%29 approximately 3.83
close one 1.219 is an answer choice too
lol they were probably like oooo let's throw in a trick choice the answer sorta you know just without being multiplied by pi
yep almost fell right for it , thanks for catching me :D

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