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

Question

anonymous · Answer

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

watchmath · Answer

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

anonymous · Answer

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

watchmath · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

watchmath · Answer

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

anonymous · Answer

no, I think maybe I did not integrate properly

anonymous · Answer

Everything you put up so far was right ilovemath

anonymous · Answer

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

anonymous · Answer

Thanks for helping! The answer is correct!