Calculate the flow in the field F along the path C.
-$$\mathbf { F } = \mathrm { y } ^ { 2 } \mathbf { i } + \mathrm { zj } + \mathrm { xk } ; C \text { is the curve } \mathbf { r } ( \mathrm { t } ) = ( 3 + 3 \mathrm { t } ) \mathbf { i } + 2 \mathrm { tj } - 2 \mathrm { tk } , 0 \leq \mathrm { t } \leq 1$$
A) 5
B) -4
C) - 7
D) 28

Question

Quizplus · Accepted Answer

To calculate the flow of $\mathbf{F}$ along the path $C$, we use the line integral of $\mathbf{F}$ along $C$, given by $\int_C \mathbf{F} \cdot d\mathbf{r}$. First, we need to parameterize $\mathbf{F}$ using the given $\mathbf{r}(t)$ and then integrate from $t = 0$ to $t = 1$.Given $\mathbf{F} = y^2\mathbf{i} + z\mathbf{j} + x\mathbf{k}$ and $\mathbf{r}(t) = (3 + 3t)\mathbf{i} + 2t\mathbf{j} - 2t\mathbf{k}$, we have:- $x = 3 + 3t$- $y = 2t$- $z = -2t$Substituting these into $\mathbf{F}$, we get:$\mathbf{F}(t) = (2t)^2\mathbf{i} - 2t\mathbf{j} + (3 + 3t)\mathbf{k}$The derivative of $\mathbf{r}(t)$ with respect to $t$ is:$d\mathbf{r}/dt = 3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}$The dot product $\mathbf{F}(t) \cdot d\mathbf{r}/dt$ is:$((2t)^2 \cdot 3) + (-2t \cdot 2) + ((3 + 3t) \cdot -2)$Simplifying and integrating from $0$ to $1$:$\int_0^1 (12t^2 - 4t - 6 - 6t)dt = \int_0^1 (12t^2 - 10t - 6)dt$This integrates to:$[4t^3 - 5t^2 - 6t]_0^1 = 4(1)^3 - 5(1)^2 - 6(1) - 0 = 4 - 5 - 6 = -7$Therefore, the flow of $\mathbf{F}$ along the path $C$ is $-7$.

Calculate the Flow in the Field F Along the Path