anonymous
  • anonymous
I have to find the Surface Area of the solid that is created by the curve: y=1-x^(2) rotated around the y-axis. on the interval 0 less than or equal to x less than or equal to 1
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
so far I have \[2\pi \int\limits_{0}^{1}\]
anonymous
  • anonymous
\[\int\limits_{a}^{b} 2pif(x)*\sqrt(1 + (dy/dx)^2)\]
anonymous
  • anonymous
I brought the 2pi out because it's a constant

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anonymous
  • anonymous
actually, let me correct that, the f(x) will just be x because your x value will be the radius of the solid
anonymous
  • anonymous
so you should get: \[2\pi \int\limits_{0}^{1} x \sqrt{1 + 4x^2}dx\]
anonymous
  • anonymous
ahhh, that makes sense
anonymous
  • anonymous
Thanks!
anonymous
  • anonymous
no problem

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