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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
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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I used the washer method, and for some reason, I am continuing to do it wrong

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1\(\int_1^3 \pi(6^2(6\frac{6}{x^2})^2)\,dx\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, that's what I started out with, using (top bottom)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1ok, then how do you know that you continuing it wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\pi \int\limits_{1}^{3} 36  ((36/x^4)  (72/x^2) +36) dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when I simplify, I am getting negative numbers, or no right answer at all

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\pi \int\limits_{1}^{3}((36/x4)−(72/x2))dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\pi \int\limits_{1}^{3}(36/x^4) + (72/x^2) dx\]

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1\(\pi\int_1^3 72x^{2}36x^{4}\,dx\) \(\pi(72x^{1}+12x^{3}\mid_1^3)\) Did you get that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, I think maybe I did not integrate properly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Everything you put up so far was right ilovemath

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think the answer is 328/9 pi, so I will check it really quick

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks for helping! The answer is correct!
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