anonymous 4 years ago Calculus: Rotating volume

1. anonymous

2. anonymous

# 22

3. anonymous

so?

4. dumbcow

x = y^3, x = 4y y^3 = 4y y^3 -4y = 0 y(y^2 -4) = 0 y(y-2)(y+2) = 0 they intersect at y= 0,2, -2 to graph: y = cubert(x), y = x/4 intersect at x = 0, 8, -8 |dw:1328001094873:dw|

5. anonymous

ok

6. anonymous

ummm lol How wld u set up? Sorry I need giuded help :D

7. anonymous

* set up the integral?

8. dumbcow

$V =\pi \int\limits_{-8}^{8}R^{2} -r^{2} dx$ so you are integrating over the circular areas of each cross-section R is outer radius from axis r is inner radius from axis

9. anonymous

ohhhh i see Thanks :D

10. anonymous

Thanks Dumbcow

11. dumbcow

welcome make sure you get R and r in terms of x, also this set up only works if you are revolving around a horizontal axis

12. anonymous

well it says the y axis

13. dumbcow

i just noticed something, because of the shape of the region, the outer and inner radius flip when x is negative. so just integrate from 0 to 8 but multiply by 2. due to symmetry the two regions will have identical volume

14. anonymous

ok I see Thanks :D