# Find area of surface obtained by rotating the curve, ?

Find area of surface obtained by rotating the curve, URGENT?

Using Simpson's rule n=10, find the area of the surface obtained by rotating the curve
y=x+sqrt(x), 1 less than or equal to x less than or equal to 2, about the x-axis.
Area = ?

I got:

S = = 2π ∫ (x + √x)√[2 + 1/√x + 1/(4x)] dx (from x=1 to 2).
∆x = (2 - 1)/10 = 1/10.
S ≈ (2π)(1/3)(1/10){g(1) + 4g[1 + 1(1/10)] + 2g[1 + 2(1/10)] + 4g[1 + 3(1/10)] + 2g[1 + 4(1/10)] + 4g[1 + 5(1/10)] + 2g[1 + 6(1/10)] + 4g[1 + 7(1/10)] + 2g[1 + 8(1/10)] + 4g[1 + 9(1/10)] + g(2)}
≈ (π/15)[(1 + 1)√(2 + 1 + 1/4) + 4(20.653233395588) + 2(20.119929259013) + (2 + √2)√(2 + 1/√2 + 1/8)]
≈ 27.68876.

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Mark44
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Using Simpson's rule n=10, find the area of the surface obtained by rotating the curve
y=x+sqrt(x), 1 less than or equal to x less than or equal to 2, about the x-axis.
Area = ?

I got:

S = = 2π ∫ (x + √x)√[2 + 1/√x + 1/(4x)] dx (from x=1 to 2).
∆x = (2 - 1)/10 = 1/10.
S ≈ (2π)(1/3)(1/10){g(1) + 4g[1 + 1(1/10)] + 2g[1 + 2(1/10)] + 4g[1 + 3(1/10)] + 2g[1 + 4(1/10)] + 4g[1 + 5(1/10)] + 2g[1 + 6(1/10)] + 4g[1 + 7(1/10)] + 2g[1 + 8(1/10)] + 4g[1 + 9(1/10)] + g(2)}
Why do you have the factor of 1/3 at the beginning of your calculation?
I don't think that should be there.

In your calculations for g(1), g(1.1), etc. does g(x) = (x + √x)√[2 + 1/√x + 1/(4x)]? You didn't say what g was, so I thought I would check.

I don't see anything wrong - your integral is set up correctly and your Simpson's work looks fine, so it would be worthwhile to doublecheck your calculations. The function you're evaluating is pretty complicated, and it would be easy to get incorrect values.

How are you determining that your answer is incorrect? Are you given the answer or is some computer program saying your answer is wrong? I would compare my results to those obtained from wolframalpha and see how close I got.
≈ (π/15)[(1 + 1)√(2 + 1 + 1/4) + 4(20.653233395588) + 2(20.119929259013) + (2 + √2)√(2 + 1/√2 + 1/8)]
≈ 27.68876.