## anonymous one year ago 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?

1. anonymous

my main trouble is finding the area bounded by the functions

2. freckles

|dw:1438886124740:dw| looks something like that

3. freckles

well the area bounded by x=0 and x=pi and y=sin(sin(x) and y=0

4. anonymous

x = pi !!! i see for some reason i just assumed it was y = pi

5. freckles

y=sin(sin(x))* forgot to close the parenthesis

6. anonymous

okay so this area right |dw:1438886325934:dw|

7. freckles

they want you to find the volume of that thing rotated about y=0? or just find the integral representation fro the volume?

8. anonymous

the former

9. freckles

have you ever heard of the bessel function?

10. anonymous

i have now

11. anonymous

let me try to do the problem and tell me if i did something wrong

12. freckles

alright just so we are clear that one equation is y=sin(sin(x))?

13. freckles

if so i have some research to do :p unless we can use some form of approximating maybe like taylor series

14. anonymous

yes it is y= sin(sin(x)) but if your wondering i'm rather certain the x intercept happens to be at pi

15. anonymous

its just calc 1 so should not involve anything too complex

16. freckles

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

17. anonymous

well ill just use the trusty ti-84 :D

18. freckles

ok so that is awesome if we can use calculator

19. anonymous

1.219

20. freckles

I'm not getting that

21. anonymous

what are you getting ?

22. anonymous

oh woops forgot to multiply by pi

23. freckles
24. anonymous

close one 1.219 is an answer choice too

25. freckles

lol they were probably like oooo let's throw in a trick choice the answer sorta you know just without being multiplied by pi

26. anonymous

yep almost fell right for it , thanks for catching me :D