17. f(x, y) = <2x, 4y>, c is the arc of the parabola x = y² from (4, -2) to (1, 1)

Explanation:

Step1: Parameterize the curve

Let $y = t$, then $x=t^{2}$. The parameter - interval is determined by the $y$ - values of the endpoints. When $y=-2$, $t = - 2$ and when $y = 1$, $t = 1$. So, $\mathbf{r}(t)=\langle t^{2},t\rangle$, and $\mathbf{r}'(t)=\langle 2t,1\rangle$. Also, $\mathbf{F}(x,y)=\langle 2x,4y\rangle$, substituting $x = t^{2}$ and $y = t$ gives $\mathbf{F}(\mathbf{r}(t))=\langle 2t^{2},4t\rangle$.

Step2: Calculate the line integral

The line integral $\int_{C}\mathbf{F}\cdot d\mathbf{r}=\int_{a}^{b}\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)dt$. [ \begin{align*} \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)&=\langle 2t^{2},4t\rangle\cdot\langle 2t,1\rangle\ &=(2t^{2})\times(2t)+4t\times1\ &=4t^{3}+4t \end{align*} ]

Step3: Evaluate the definite integral

[ \begin{align*} \int_{C}\mathbf{F}\cdot d\mathbf{r}&=\int_{-2}^{1}(4t^{3}+4t)dt\ &=\left[t^{4}+2t^{2}\right]_{-2}^{1}\ &=(1^{4}+2\times1^{2})-((-2)^{4}+2\times(-2)^{2})\ &=(1 + 2)-(16 + 8)\ &=3-24\ &=- 21 \end{align*} ]

Answer:

$-21$