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

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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

Mathematics
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I used the washer method, and for some reason, I am continuing to do it wrong
\(\int_1^3 \pi(6^2-(6-\frac{6}{x^2})^2)\,dx\)
yes, that's what I started out with, using (top- bottom)

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ok, then how do you know that you continuing it wrong?
\[\pi \int\limits_{1}^{3} 36 - ((36/x^4) - (72/x^2) +36) dx\]
when I simplify, I am getting negative numbers, or no right answer at all
\[\pi \int\limits_{1}^{3}((36/x4)−(72/x2))dx\]
\[\pi \int\limits_{1}^{3}(-36/x^4) + (72/x^2) dx\]
\(\pi\int_1^3 72x^{-2}-36x^{-4}\,dx\) \(\pi(-72x^{-1}+12x^{-3}\mid_1^3)\) Did you get that?
no, I think maybe I did not integrate properly
Everything you put up so far was right ilovemath
I think the answer is 328/9 pi, so I will check it really quick
Thanks for helping! The answer is correct!

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