Evaluate $\int _ { C } \mathbf { F } \cdot d \mathbf { r }$ for the vector field F and the path C. (Hint: Show that F is conservative, and pick a simpler path.) $\mathbf { F } ( x , y ) = \left( 16 x ^ { 3 } y ^ { 2 } - 4 y \sin x \right) \mathbf { i } + \left( 8 x ^ { 4 } y + 4 \cos x \right) \mathbf { j }$ C: $\mathbf { r } ( t ) = ( 3 - \sin t ) \mathbf { i } + 2 \cos t \mathbf { j } ; \quad 0 \leq t \leq \pi$
A) $- 16 \cos ( 3 )$
B) 54
C) 0
D) 19

Question

Quizplus · Accepted Answer

The vector field is conservative because $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$, where $P(x, y) = 16x^3y^2 - 4y\sin x$ and $Q(x, y) = 8x^4y + 4\cos x$. So, there exists a potential function $f$ such that $\nabla f = \mathbf{F}$.A simpler path from $(3, 2)$ to $(2, -2)$ is the line segment $C_1$. The line segment $C_1$ is parametrized by $\mathbf{r}_1(t) = (3-t)\mathbf{i} + (2-t)\mathbf{j}$ for $0\le t\le 1$. Then,$$\begin{split} \int_C \mathbf{F} \cdot d\mathbf{r} & = \int_{C_1} \mathbf{F} \cdot d\mathbf{r} \\ & = \int_0^1 \mathbf{F}(\mathbf{r}_1(t)) \cdot \frac{d\mathbf{r}_1}{dt} dt \\ & = \int_0^1 \left[ 16(3-t)^3(2-t)^2 - 4(2-t)\sin (3-t) \right] (-1) \\ & \qquad + \left[ 8(3-t)^4(2-t) + 4\cos (3-t) \right] (-1) dt \\ & = \int_0^1 \left[ -16(3-t)^3(2-t)^2 + 4(2-t)\sin (3-t) \right. \\ & \qquad \left. - 8(3-t)^4(2-t) - 4\cos (3-t) \right] dt \\ & = \int_0^1 \left[ -16(3-t)^3(2-t)^2 + 4(2-t)\sin (3-t) \right. \\ & \qquad \left. - 8(3-t)^4(2-t) - 4\cos (3-t) \right] dt \\ & = \left[ 4(2-t)\cos (3-t) - 4(3-t)^4 \right]_0^1 \\ & = 4\cos 2 - 4 \\ & = -16\cos 2. \end{split}$$

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

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