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Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis.

y = (x³/6) + (1/2x), 1≤ x ≤ 2

  • Calc-surface of rotation -

    I assume (1/2x) is meant to be 1/(2x), so
    y(x) = (x³/6) + 1/(2x), x1≤ x ≤ x2 ,
    where x1=1, x2=2.

    You are probably aware of the formula for surface of revolution about the x-axis generated by a function y(x) from x=x1 to x=x2:

    We will now calculate each component individually and then do the integration.

    y(x)=x³/6+1/(2x) ....(given)
    By differentiation with respect to x, we get:
    Now we proceed to evaluate the expression inside the square-root radical:
    find common denominator and add:

    Now we are ready to put everything together:

    Area of revolution
    =2π∫(x³/6+1/(2x))*((x^4+1)/(2x²) dx from x1 to x2
    =2π∫((x^4+3)/(6x))*((x^4+1)/(2x²) dx
    =2π∫((x^4+3)(x^4+1)/(12x³)) dx
    =2π∫((x^8+4x^4+3)/(12x³)) dx
    =(π/6)∫((x^5+4x+3/x³)) dx

    Evaluate integral between x1=1 and x2=2 gives 47π/16 (=9.2 approximately).

    As a rough check, we calculate the function at x=(x1+x2)/2=1.5 and estimate the area:
    =7.6, which is not too far from our calculations above.

  • Calc-surface of rotation -

    In this case, a better check is by using simpson's rule for numerical integration, given by:
    ∫g(x) from a to b
    and in the present case,
    =9.28, much closer to 9.23 by integration.

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