## anonymous 5 years ago Use the disk or the shell method to find the volume of the solid generated by revolving the region bounded by the graph of the equations about y = 6. y = 6/x^2 y = 0 x = 1 x = 3

1. anonymous

I used the washer method, and for some reason, I am continuing to do it wrong

2. watchmath

$$\int_1^3 \pi(6^2-(6-\frac{6}{x^2})^2)\,dx$$

3. anonymous

yes, that's what I started out with, using (top- bottom)

4. watchmath

ok, then how do you know that you continuing it wrong?

5. anonymous

$\pi \int\limits_{1}^{3} 36 - ((36/x^4) - (72/x^2) +36) dx$

6. anonymous

when I simplify, I am getting negative numbers, or no right answer at all

7. anonymous

$\pi \int\limits_{1}^{3}((36/x4)−(72/x2))dx$

8. anonymous

$\pi \int\limits_{1}^{3}(-36/x^4) + (72/x^2) dx$

9. watchmath

$$\pi\int_1^3 72x^{-2}-36x^{-4}\,dx$$ $$\pi(-72x^{-1}+12x^{-3}\mid_1^3)$$ Did you get that?

10. anonymous

no, I think maybe I did not integrate properly

11. anonymous

Everything you put up so far was right ilovemath

12. anonymous

I think the answer is 328/9 pi, so I will check it really quick

13. anonymous

Thanks for helping! The answer is correct!