Find $\int _ { C } \vec { F } \cdot d \vec { r }$ of the function over the given curve: $\vec { F } ( x , y ) = \langle 1 , x , 1 \rangle$ and the curve C is given by $\vec { r } ( t ) = \langle \cos ( t ) , \sin ( t ) , t \rangle , 0 \leq t \leq \frac { \pi } { 2 }$
A) $\frac { 3 \pi } { 4 } - 1$
B) $\frac { 3 \pi } { 4 }$
C) $\frac { \pi } { 4 } - 1$
D) $\frac { 3 \pi } { 4 } - \frac { 3 } { 2 }$

Question

Quizplus · Accepted Answer

To solve the line integral $\int _ { C } \vec { F } \cdot d \vec { r }$, we substitute $\vec{F}(x,y) = \langle 1, x, 1 \rangle$ and $\vec{r}(t) = \langle \cos(t), \sin(t), t \rangle$ into the integral. The derivative of $\vec{r}(t)$ with respect to $t$ is $\vec{r}'(t) = \langle -\sin(t), \cos(t), 1 \rangle$. The dot product $\vec{F} \cdot d\vec{r}$ becomes $\langle 1, \cos(t), 1 \rangle \cdot \langle -\sin(t), \cos(t), 1 \rangle = -\sin(t) + \cos^2(t) + 1$. Integrating this from $0$ to $\frac{\pi}{2}$ gives $\int_{0}^{\frac{\pi}{2}} (-\sin(t) + \cos^2(t) + 1) dt$, which simplifies to $\frac{3\pi}{4} - 1$.

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

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