$\text { Find the value of the line integral } \int _ { C } \mathbf { F } \cdot d \mathbf { r } \text {, where } \mathbf { F } ( x , y ) = 4 \mathbf { i } + 2 z \mathbf { j } + 2 y \mathbf { k } \text { and }$$\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + t ^ { 2 } \mathbf { k } , 0 \leq t \leq \pi$
(Hint: If $\mathbf { F }$ is conservative, the integration may be easier on an alternate path.)
A) 4
B) 8
C) 4.
D) $- 4$.
E) $- 8$.

Question

Quizplus · Accepted Answer

The vector field $\mathbf{F} = 4\mathbf{i} + 2z\mathbf{j} + 2y\mathbf{k}$ is not conservative because it does not have a potential function (its curl is non-zero). To find the line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$, we substitute $\mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + t^2 \mathbf{k}$ into $\mathbf{F}$, yielding $\mathbf{F}(\mathbf{r}(t)) = 4\mathbf{i} + 2t^2\mathbf{j} + 2\sin t\mathbf{k}$. The differential $d\mathbf{r}$ is $-\sin t \mathbf{i} + \cos t \mathbf{j} + 2t \mathbf{k} dt$. The dot product $\mathbf{F}(\mathbf{r}(t)) \cdot d\mathbf{r}$ gives $4(-\sin t) + 2t^2(\cos t) + 2\sin t(2t)$ dt. Integrating this expression from $0$ to $\pi$ results in $-8$, indicating that the line integral of the vector field $\mathbf{F}$ along the curve $C$ is $-8$.

r(t)=cos⁡ti+sin⁡tj+t2k,0≤t≤π\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + t ^ { 2 } \mathbf { k } , 0 \leq t \leq \pir(t)=costi+sintj+t2k,0≤t≤π

$\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + t ^ { 2 } \mathbf { k } , 0 \leq t \leq \pi$