calculus II SURFACE AREA Integration

Find the exact area of the surface obtained by rotating the curve about the x-axis.

y = sqrt(1 + 5x) from 1 ≤ x ≤ 7

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  1. The standard formula for the curved surface of revolution about the x-axis for a continuous function f(x) is:
    A=∫2πf(x)sqrt(1+(f'(x))²)dx

    Substitute values
    f(x)=sqrt(1+5x)
    f'(x)=5/(2sqrt(5x+1))
    integrand:
    2πsqrt(5x+1)*sqrt(1+25/(4(5x+1)))
    Limits: 0 to 7

    Integral:
    (π(20x+29)^(3/2))/30
    Evaluate between 0 and 7 to get:

    214 units² approximately.

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  2. Think of the surface as a collection of strips, of radius y, and width a piece of the curve, ds

    A = Int (2π r ds)
    but ds = √(dx^2 + dy^2)
    = √(1 + y'^2) dx

    So, here we have
    y = √(5x+1)
    y' = 5/√(5x+1)

    A = 2π Int(y ds)
    = 2π Int(√(5x+1)*√(1+25/(5x+1)) dx[1,7]
    = 2π Int(√(5x+26))[1,7]
    = 2π * 2/3 * 1/5 (5x+26)^3/2[1,7]
    = 4π/15 (61√61 - 31√31)

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    posted by Steve
  3. Rats. MathMate did it right.
    y' = 5/2√(5x+1)

    Sorry

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    posted by Steve
  4. Except for the limits of integration :-)

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    posted by Steve
  5. Good point!
    I'll let Michael take care of that!

    Michael:
    Limits should be from 1 to 7.

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  6. The answer is 309pi/5 with the limit from 1 to 7

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    posted by Colton

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