anonymous 3 years ago What is the volume of a solid obtained by rotating the region between the graph of y = (1/2)sin^2 (x^2) and the x-axis for 0 ≤x≤sqrtpi about the y-axis?

1. anonymous

The main problem I have is that I can't solve for x.

2. anonymous

Are you there?

3. anonymous

@jim_thompson5910

4. Callisto

Put y =0 0 = (1/2)sin^2 (x^2) sin^2 (x^2) = 0 sin (x^2) = 0 x^2 =...???

5. anonymous

0?

6. Callisto

and??

7. anonymous

0 is all I can think of.

8. Callisto

|dw:1358903484725:dw|

9. anonymous

Oh, right! Sine hits zero on every multiple of pi

10. anonymous

So x = sqrt(pi), 0?

11. Callisto

For that domain, yes.

12. anonymous

Right. The domain was a clue.

13. anonymous

I need to use the washer method right?

14. Callisto

I think we need to use shell method?!

15. anonymous

oh, right. just a sec....

16. anonymous

x = sqrtpi for the radis, y = (1/2)sin^2 (x^2) for the height?

17. Callisto

x = radius... y = height

18. anonymous

$2\pi \int\limits_{0}^{\sqrt{\pi}} (\sqrt{\pi}/2) \sin^2 (x^2)$ ?

19. Callisto

$V = 2 \pi \int_a^brh(width)$r = x, h = y, width = dx

20. Callisto

y is a function of x, you know.

21. anonymous

Doesn't x = sqrtpi and y = (1/2)sin^2 (x^2)?