Evaluate $\int _ { C } \mathbf { F } \cdot d \mathbf { r }$ for the vector field $\mathbf { F }$ and the path $C$. (Hint: Show that $\mathbf { F }$ is conservative, and pick a simpler path.)
$$\mathbf { F } ( x , y ) = \left( 18 x ^ { 2 } y ^ { 2 } - 5 y \sin x \right) \mathbf { i } + \left( 12 x ^ { 3 } y + 5 \cos x \right) \mathbf { j }$$
$C : \mathbf { r } ( t ) = ( - 2 - \sin t ) \mathbf { i } + 2 \cos t \mathbf { j } ; \quad 0 \leq t \leq \pi$
A) 24
B) 47
C) 0
D) $- 20 \cos ( - 2 )$

Question

Quizplus · Accepted Answer

To show $\mathbf{F}$ is conservative, we check if $\nabla \times \mathbf{F} = 0$. The curl of $\mathbf{F}$ in 2D is given by $\frac{\partial}{\partial x}(12x^3y + 5\cos x) - \frac{\partial}{\partial y}(18x^2y^2 - 5y\sin x) = 36x^3 + 5\sin x - (36x^2y - 5\sin x) = 0$, confirming $\mathbf{F}$ is conservative. For conservative fields, the integral depends only on the endpoints. The path $C$ starts at $(-2- \sin(0), 2\cos(0)) = (-3, 2)$ and ends at $(-2-\sin(\pi), 2\cos(\pi)) = (-3, -2)$. A potential function $\Phi$ for $\mathbf{F}$ satisfies $\nabla\Phi = \mathbf{F}$, leading to $\Phi(x, y) = 6x^3y^2 - 5y\cos x + g(y)$ and $\Phi(x, y) = 6x^3y^2 + 5\sin x + h(x)$. Matching terms, $g(y)$ and $h(x)$ are constants with respect to their variables, so $\Phi(x, y) = 6x^3y^2 - 5y\cos x$. Evaluating $\Phi$ at the endpoints gives $\Phi(-3, -2) - \Phi(-3, 2) = [6(-3)^3(-2)^2 - 5(-2)\cos(-3)] - [6(-3)^3(2)^2 - 5(2)\cos(-3)] = -20\cos(-3)$.

Evaluate ∫CF⋅dr\int _ { C } \mathbf { F } \cdot d \mathbf { r }∫C​F⋅dr

Evaluate $\int _ { C } \mathbf { F } \cdot d \mathbf { r }$