Posted by **Erica** on Monday, February 7, 2011 at 1:05am.

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 -
**MathMate**, Monday, February 7, 2011 at 3:58pm
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 -
**MathMate**, Monday, February 7, 2011 at 4:08pm
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.

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