Vector Field / Line Integral (Part 2)

Calculus Level 5

The vector field F ⃗ \vec{F} has components ( F x , F y , F z ) = ( x y , y z , z x ) (F_x,F_y,F_z) = (x y , y z , z x) . Consider the integration path C C , which is a semicircle starting at P 1 ⃗ = ( 0 , 0 , 0 ) \vec{P_1} = (0,0,0) and ending at P 2 ⃗ = ( 1 , 2 , 3 ) \vec{P_2} = (1,2,3) (see "details" section). The diameter of the semicircle is D D , and its radius is R R .

P ⃗ = P 1 ⃗ + α u ⃗ + β v ⃗ ( R − α ) 2 + β 2 = R 2 0 ≤ α ≤ D \large{\vec{P} = \vec{P_1} + \alpha \, \vec{u} + \beta \, \vec{v} \\ (R - \alpha)^2 + \beta^2 = R^2 \\ 0 \leq \alpha \leq D}

What is the value of the line integral of the vector field over the path?

∫ C F ⃗ ⋅ d ℓ ⃗ \large{\int_C \vec{F} \cdot \vec{d \ell}}

Details and Assumptions:
1) Vector u ⃗ \vec{u} is a unit-vector in the direction of ( 1 , 2 , 3 ) (1,2,3)
2) Vector v ⃗ \vec{v} is a unit-vector in the direction of ( 1 , 1 , − 1 ) (1,1,-1)
3) The middle point of the curve is M ⃗ = ( ≈ 1.58 , ≈ 2.08 , ≈ 0.42 ) \vec{M} = (\approx 1.58 ,\approx 2.08,\approx 0.42) . Use the positive root for β \beta


The answer is 4.202.

Steven Chase by Steven Chase , 37, USA

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1 solution

The center of the semicircle is at 1 2 ( { 0 , 0 , 0 } + { 1 , 2 , 3 } ) \frac{1}{2} (\{0,0,0\}+\{1,2,3\}) or { 1 2 , 1 , 3 2 } \left\{\frac{1}{2},1,\frac{3}{2}\right\} .

The normalized u ⃗ \vec{u} vector from the center to the start is { − 1 14 , − 2 7 , − 3 14 } \left\{-\frac{1}{\sqrt{14}},-\sqrt{\frac{2}{7}},-\frac{3}{\sqrt{14}}\right\} .

The normalized v ⃗ \vec{v} vector is { 1 3 , 1 3 , − 1 3 } \left\{\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right\} .

p ⃗ ( θ ) \vec{p}(\theta ) is { 7 6 sin ⁡ ( θ ) − cos ⁡ ( θ ) 2 + 1 2 , 7 6 sin ⁡ ( θ ) − cos ⁡ ( θ ) + 1 , − 7 6 sin ⁡ ( θ ) − 1 2 3 cos ⁡ ( θ ) + 3 2 } \left\{\sqrt{\frac{7}{6}} \sin (\theta )-\frac{\cos (\theta )}{2}+\frac{1}{2},\sqrt{\frac{7}{6}} \sin (\theta )-\cos (\theta )+1,-\sqrt{\frac{7}{6}} \sin (\theta )-\frac{1}{2} 3 \cos (\theta )+\frac{3}{2}\right\} .

∂ p ( θ ) ∂ θ \frac{\partial p(\theta )}{\partial \theta } is { sin ⁡ ( θ ) 2 + 7 6 cos ⁡ ( θ ) , sin ⁡ ( θ ) + 7 6 cos ⁡ ( θ ) , 3 sin ⁡ ( θ ) 2 − 7 6 cos ⁡ ( θ ) } \left\{\frac{\sin (\theta )}{2}+\sqrt{\frac{7}{6}} \cos (\theta ),\sin (\theta )+\sqrt{\frac{7}{6}} \cos (\theta ),\frac{3 \sin (\theta )}{2}-\sqrt{\frac{7}{6}} \cos (\theta )\right\} .

f ( x , y , z ) f(x,y,z) is { x y , y z , x z } \{x y,y z,x z\} .

The expression to be integrated, f ⃗ ( p ⃗ ( θ ) ⋅ ∂ p ( θ ) ∂ θ \vec{f}(\vec{p}(\theta )\cdot\frac{\partial p(\theta )}{\partial \theta } expands and simplifies to 1 72 sin ⁡ 2 ( θ 2 ) ( − 34 42 cos ⁡ ( 2 θ ) + 58 42 cos ⁡ ( θ ) + 6 sin ⁡ ( θ ) ( 13 − 97 cos ⁡ ( θ ) ) + 32 42 ) \frac{1}{72} \sin ^2\left(\frac{\theta }{2}\right) \left(-34 \sqrt{42} \cos (2 \theta )+58 \sqrt{42} \cos (\theta )+6 \sin (\theta ) (13-97 \cos (\theta ))+32 \sqrt{42}\right) .

The indefinite integral is 1 864 ( 18 42 θ − 189 42 sin ⁡ ( 2 θ ) + 258 42 sin ⁡ ( θ ) + 34 42 sin ⁡ ( 3 θ ) − 1341 cos ⁡ ( θ ) + 990 cos ⁡ ( 2 θ ) − 291 cos ⁡ ( 3 θ ) ) \frac{1}{864} \left(18 \sqrt{42} \theta -189 \sqrt{42} \sin (2 \theta )+258 \sqrt{42} \sin (\theta )+34 \sqrt{42} \sin (3 \theta )-1341 \cos (\theta )+990 \cos (2 \theta )-291 \cos (3 \theta )\right) .

The result is 1 8 π 7 6 + 34 9 \frac{1}{8} \pi \sqrt{\frac{7}{6}}+\frac{34}{9} or about 4.20194126461.

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