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anonymous

  • 5 years ago

Find the surface area of a sphere of radius r, using calculus.

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  1. myininaya
    • 5 years ago
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    omg i know this

  2. myininaya
    • 5 years ago
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    give me just a sec

  3. anonymous
    • 5 years ago
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    that would be amazing if you could help! :)

  4. myininaya
    • 5 years ago
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    so the equation of a circle having center (0,0) is x^2+y^2+r^2

  5. myininaya
    • 5 years ago
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    oh surface area not volume one sec. let me rething my strategy

  6. anonymous
    • 5 years ago
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    ok. thanks

  7. anonymous
    • 5 years ago
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    Use the formula for the surface area of a general equation revolved around the x-axis. If you start with x^2 + y^2 = r^2, and your function is y, then f(x) = sqrt(r^2-x^2). The formula for the surface area of revolution is, in this case, \[SA = 2 \pi* \int\limits_{-r}^{r} \ \ f(x) * \sqrt{1+f(x)^2}\] Integrate and simplify.

  8. anonymous
    • 5 years ago
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    Sorry, inside the root of the integrand it should be f'(x)^2.

  9. myininaya
    • 5 years ago
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    what is the circumference of sphere

  10. anonymous
    • 5 years ago
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    Thanks QuantumModulus, that helped a lot!!

  11. anonymous
    • 5 years ago
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    Glad to help. :)

  12. myininaya
    • 5 years ago
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    have you done trig substition yet?

  13. myininaya
    • 5 years ago
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    I think you may have to do in one of the steps

  14. myininaya
    • 5 years ago
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    nvm it is a simple integration

  15. myininaya
    • 5 years ago
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    1 Attachment
  16. myininaya
    • 5 years ago
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    ok but i think was suppose to get 4pi*r^2

  17. myininaya
    • 5 years ago
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    let me know what you get?

  18. myininaya
    • 5 years ago
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    ok it is 4*pi*r^2 because the semicirlce is being revolved about the x axis

  19. myininaya
    • 5 years ago
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    Yes! :)

  20. myininaya
    • 5 years ago
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    ignore the extra stuff on that attachment

  21. myininaya
    • 5 years ago
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    I have to go. You can check your work with the attachment just ignore the But part. goodnight

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