Evaluate C 1/(3+y) ds, whereC is given by the equations x=2t^(3/2) and y=3t ,for0≤t≤1.

Question

Evaluate C 1/(3+y)ds, where C is given by the equations x=2t^(3/2) and y=3t, for 0 ≤ t ≤ 1.

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Answer 1

I think we have to evaluate line integral:∫(C)1/(3 + y)ds;

ds = sqrt((dx)² + (dy)²) = sqrt((2 ·3/2·t^1/2dt)² + (3dt)²) = sqrt(9t + 9)dt = 3sqrt(t + 1)dt.

Now we have ∫(C)1/(3 + y)ds = ∫₀¹1/(3 + 3t)·3sqrt(1 + t)dt = ∫₀¹1/sqrt(1 + t)dt = 2(1 + t)^1/2₀¹ = 2sqrt2 - 2.

Answer: 2sqrt2 - 2