Question

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

Answer 1

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.

Answer 2

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.