- #1
ani9890
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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.
Include at least five decimal places in your answer.
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.
But this is wrong. Please help!
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.
Include at least five decimal places in your answer.
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.
But this is wrong. Please help!