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anonymous
 5 years ago
I have to find the Surface Area of the solid that is created by the curve: y=1x^(2)
rotated around the yaxis.
on the interval 0 less than or equal to x less than or equal to 1
anonymous
 5 years ago
I have to find the Surface Area of the solid that is created by the curve: y=1x^(2) rotated around the yaxis. on the interval 0 less than or equal to x less than or equal to 1

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so far I have \[2\pi \int\limits_{0}^{1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{a}^{b} 2pif(x)*\sqrt(1 + (dy/dx)^2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I brought the 2pi out because it's a constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually, let me correct that, the f(x) will just be x because your x value will be the radius of the solid

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you should get: \[2\pi \int\limits_{0}^{1} x \sqrt{1 + 4x^2}dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahhh, that makes sense
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