Simpson's 1/3 Method, eight subintervals

[francisg3]

The roof of a silo is made by revolving the curve y=10 cos(πx/10) from x=-5m to x=5m about the y-axis. The surface area S, that is obtained by revolving a curve y=f(x) in the domain from 'a' to 'b' around the y-axis can be calculated by

(see attached picture for integral)



I found y'=-π sin(πx/10)

h=(b-a)/N = (5-(-5))/8= 1.25

f(x) = the inside part of the integral

therefore the range is as follows:

x= -5 -3.75 -2.5 -1.25 0 1.25 2.5 3.75 5
y'(x)= 3.14 2.902 2.221 1.202 0 -1.202 -2.221 -2.902 -3.14

f'(x)= -16.48 -11.51 -6.09 -1.95 0 1.95 6.09 11.51 16.48



Using the equation in the second attachment gives me a result of 0. I have done other Simpson's rule problems however there was never an 'x' outside of the square root. Where have I gone wrong?


Thank you.

 

[HallsofIvy]

I don't understand your question. Simpson's method can be used to numerically integrate any function. Why would it matter whether there was an 'x' "outside the square root" or not? You know each x_i and you know each f'(x_i) so it is easy to calculate x_i\sqrt{1+ f'(x_i)}. That, of course, will be the "f(x_i)" in your formula for Simpson's rule.

 

[gus134]

You have to use a=0 and b=5. Since the formula is the same on both sides of the y axis, it cancels out. The 2 in the S formula (before the pi) accounts for the half interval. The answer is 216.3386