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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=1-x^(2) rotated around the y-axis. on the interval 0 less than or equal to x less than or equal to 1

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  1. anonymous
    • 5 years ago
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    so far I have \[2\pi \int\limits_{0}^{1}\]

  2. anonymous
    • 5 years ago
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    \[\int\limits_{a}^{b} 2pif(x)*\sqrt(1 + (dy/dx)^2)\]

  3. anonymous
    • 5 years ago
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    I brought the 2pi out because it's a constant

  4. anonymous
    • 5 years ago
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    actually, let me correct that, the f(x) will just be x because your x value will be the radius of the solid

  5. anonymous
    • 5 years ago
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    so you should get: \[2\pi \int\limits_{0}^{1} x \sqrt{1 + 4x^2}dx\]

  6. anonymous
    • 5 years ago
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    ahhh, that makes sense

  7. anonymous
    • 5 years ago
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    Thanks!

  8. anonymous
    • 5 years ago
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    no problem

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