Calc

posted by .

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:
    S=2π∫y(x)√(1+y'(x)^2)dx

    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:
    y'(x)=x²-1/(2x²)
    Now we proceed to evaluate the expression inside the square-root radical:
    √(1+y'(x)²)
    =√(1+(x²-1/(2x²))^2)
    =√(1+((x^4-1)/(2x²)))
    find common denominator and add:
    =√((x^8+2x^4+1)/(4x^4))
    =(x^4+1)/(2x²)

    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
    =(π/6)(x^6/6+2x²-3/(2x²))

    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:
    Aapprox=2πf(1.5)*radic;(1+f'(1.5)^2)
    =2π(0.9)√(1+0.9²)*(x2-x1)
    =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
    =(b-a)/6*(g(a)+4g((a+b)/2)+g(b))
    and in the present case,
    g(x)=2πy(x)sqrt(1+y'(x))
    so
    A=((2-1)/6)*(g(1)+4g(1.5)+g(2))
    =(1/6)(4π/3+4*2.41π+6.73π)
    =9.28, much closer to 9.23 by integration.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus II/III

    A. Find the integral of the following function. Integral of (x√(x+1)) dx. B. Set up and evaluate the integral of (2√x) for the area of the surface generated by revolving the curve about the x-axis from 4 to 9. For part …
  2. Math

    Find the area of the surface generated by revolving the curve y = (x + 3)^1/2, 2 ≤ x ≤ 4 about the x-axis.
  3. Calc

    Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the y-axis. y = cube rt. (x) + 2 Thank you so much!!
  4. 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
  5. Calculus

    The graph of the curve y2=10 x+10 is a parabola, symmetric with respect to the x-axis. Find the area of the surface generated by rotating about the x-axis the part of this curve corresponding to 0 ≤ x ≤ 20.
  6. asdf

    Find a definite integral indicating the area of the surface generated by revolving the curve y = 3√3x ; 0 ≤ y ≤ 4 about the x – axis. But do not evaluate the integral.
  7. calc

    Find the exact area of the surface obtained by rotating the curve about the x-axis. 9x = y2 + 36, 4 ≤ x ≤ 8 I keep on getting 4pi (3^(3/2) +1) so I tried to put the answer in that way and also 24.78pi But it is wrong
  8. calc

    evaluate the surface integral double integral (y dS) z= 2/3 (x^3/2 + y^3/2) 0 ≤ x ≤ 5, 0 ≤ y ≤ 2
  9. Calc

    Evaluate the following integral using three different orders of integration. *triple integral E (xz − y3) dV, where E = (x, y, z) |  −1 ≤ x ≤ 3, 0 ≤ y ≤ 4, 0 ≤ z ≤ 3
  10. Math - Calculus

    Find the area of the surface of revolution generated by revolving the curve y = 3 sqrt (x), 0 <= x <= 4, about the x-axis. Okay, so I've set up the integral like this: 2pi ∫[0,4] (3 sqrt (x))(sqrt(1+(1/4x)))dx Which is coming …

More Similar Questions