Vector Calc 8-3-2020 Part 2

Calculus Level 3

Consider the vector field:

F ⃗ = ( F x , F y ) = ( x 2 y , x + y ) \vec{F} = (F_x, F_y) = (x^2 y, x + y)

Evaluate the line integral of the vector field over the path from ( x , y ) = ( − 1 , 1 ) (x,y) = (-1,1) to ( x , y ) = ( 1 , 1 ) (x,y) = (1,1) over the curve y = x 2 y = x^2 .

∫ F ⃗ ⋅ d ℓ ⃗ \int \vec{F} \cdot \vec{d \ell}


The answer is 1.733.

Steven Chase by Steven Chase , 37, USA

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1 solution

Hosam Hajjir
Aug 3, 2020

We have to parametrize the curve we're integrating along, so the natural choice is r ⃗ ( t ) = ( t , t 2 ) \vec{r}(t) = (t, t^2) , then our integral is,

∫ C F ⃗ ⋅ d r ⃗ = ∫ t = − 1 t = 1 F ( t ) ⃗ ⋅ d r ⃗ ( t ) d t d t = ∫ − 1 1 ( t 4 , t + t 2 ) ⋅ ( 1 , 2 t ) d t = ∫ − 1 1 ( t 4 + 2 t 2 + 2 t 3 ) d t = 2 5 + 4 3 = 26 15 = 1.733 \large \displaystyle \int_C \vec{F} \cdot \vec{dr} = \int_{t = -1 }^{t = 1} \vec{F(t)} \cdot \dfrac{d\vec{r}(t)}{dt} dt = \int_{-1}^{1} (t^4, t+ t^2) \cdot (1, 2t) dt = \int_{-1}^{1} (t^4 + 2 t^2 + 2 t^3) dt = \dfrac{2}{5} + \dfrac{4}{3} = \dfrac{26}{15} = \boxed{1.733}

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