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Math Help - Find the surface area generated by revolving the curve

  1. #1
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    Find the surface area generated by revolving the curve

    Need help on finding the surface area generated by revolving the curve

    y=square root of 25-x^2 around the X-axis for -3 ≤ x ≤ 4

    Thank you
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  2. #2
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    Quote Originally Posted by macadamianchica View Post
    Need help on finding the surface area generated by revolving the curve

    y=square root of 25-x^2 around the X-axis for -3 ≤ x ≤ 4

    Thank you

    find it using the definite integral of y in [-3;4]
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  3. #3
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    Quote Originally Posted by macadamianchica View Post
    Need help on finding the surface area generated by revolving the curve

    y=square root of 25-x^2 around the X-axis for -3 ≤ x ≤ 4

    Thank you
    As the curve revolves around the x-axis, each point on graph of the function traces out a circle, with f(x) as radius.

    To calculate the surface area, the circle circumferences are integrated.

    R=f(x),\ -3\le{x}\le{4}

    Hence, the curved surface area is.... 2\pi\int_{-3}^4f(x)dx\ =\ 2\pi\int_{-3}^4\sqrt{25-x^2}dx

    You may or may not want to include the circular side faces.
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  4. #4
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    Quote Originally Posted by macadamianchica View Post
    Need help on finding the surface area generated by revolving the curve

    y=square root of 25-x^2 around the X-axis for -3 ≤ x ≤ 4

    Thank you

    the funct y=sqrt25-x^ is a halfcircle. find the integral by replacing x=5sint t. then u have sqrt of 25-25sin^2 x=sqrt25(1-sin^2 x)=5cost now find the definite integral of y=5cost
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  5. #5
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    This is not the "integration of a function problem".
    It's a 3-dimensional "surface of revolution".
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