Find $\int _ { C } \vec { F } \cdot d \vec { r }$ of the function over the given curve: $\vec { F } ( x , y ) = \langle x , y \rangle$ and the curve C is given by $y = x ^ { 2 } , 0 \leq x \leq 2$
A) 4
B) $8$

C) $10$
D) $12$

Question

Find $\int _ { C } \vec { F } \cdot d \vec { r }$ of the function over the given curve: $\vec { F } ( x , y ) = \langle x , y \rangle$ and the curve C is given by $y = x ^ { 2 } , 0 \leq x \leq 2$
A) 4
B) $8$

C) $10$
D) $12$

Quizplus · Accepted Answer

To solve the line integral $\int _ { C } \vec { F } \cdot d \vec { r }$, we first parameterize the curve $C$ using $x$ as the parameter since $y = x^2$ and $0 \leq x \leq 2$. Thus, the parameterization of $C$ is $\vec{r}(t) = \langle t, t^2 \rangle$ with $t$ ranging from $0$ to $2$. The vector field $\vec{F}(x, y) = \langle x, y \rangle$ becomes $\vec{F}(\vec{r}(t)) = \langle t, t^2 \rangle$. The differential $d\vec{r}$ is the derivative of $\vec{r}(t)$ with respect to $t$, which is $\vec{r}'(t) = \langle 1, 2t \rangle dt$. Therefore, the line integral becomes:$$\int_{0}^{2} \langle t, t^2 \rangle \cdot \langle 1, 2t \rangle dt = \int_{0}^{2} (t + 2t^3) dt = \left[ \frac{1}{2}t^2 + \frac{1}{2}t^4 \right]_{0}^{2} = \frac{1}{2}(2^2) + \frac{1}{2}(2^4) = 2 + 8 = 10.$$

Find ∫CF⃗⋅dr⃗\int _ { C } \vec { F } \cdot d \vec { r }∫C​F⋅dr

Find $\int _ { C } \vec { F } \cdot d \vec { r }$