posted by Erica .
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
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.
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
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.
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.