Let $\mathbf { F } ( x , y , z ) = x \mathbf { j }$ . Evaluate the line integral $\int _ { C } \mathbf { F } \cdot d \mathbf { r }$ along the elliptical path $\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + \cos t \mathbf { k } , 0 \leq \mathrm { t } \leq 2 \pi$ .
A) $\pi$

B) $\pi \sqrt { 2 }$
C)$\pi \sqrt { 3 }$
D)$2 \pi$
E) $\frac { \sqrt { 2 } } { 2 } \pi$
F) $\frac { \sqrt { 3 } } { 2 } \pi$

G) $\frac { \pi } { 2 }$

H) $\frac { \pi } { 4 }$

Question

Let $\mathbf { F } ( x , y , z ) = x \mathbf { j }$ . Evaluate the line integral $\int _ { C } \mathbf { F } \cdot d \mathbf { r }$ along the elliptical path $\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + \cos t \mathbf { k } , 0 \leq \mathrm { t } \leq 2 \pi$ .
A) $\pi$

B) $\pi \sqrt { 2 }$ 
C)$\pi \sqrt { 3 }$ 
D)$2 \pi$ 
E) $\frac { \sqrt { 2 } } { 2 } \pi$ 
F) $\frac { \sqrt { 3 } } { 2 } \pi$

G) $\frac { \pi } { 2 }$

H) $\frac { \pi } { 4 }$

Quizplus · Accepted Answer

The vector field $\mathbf{F}(x, y, z) = x\mathbf{j}$ means that the force field has a magnitude of $x$ in the direction of $\mathbf{j}$. The line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ along a path $C$ measures the work done by the force field along that path. Given the path $\mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + \cos t \mathbf{k}$, we see that $x$, the coefficient of $\mathbf{i}$, is $\cos t$, and $y$, the coefficient of $\mathbf{j}$, is $\sin t$. Since $\mathbf{F}$ only has a component in the $\mathbf{j}$ direction, we only need to consider the $\sin t$ part of the path for the integral. The integral simplifies to $\int_0^{2\pi} \cos t \cdot \sin t \, dt$, which evaluates to 0 because the integral of $\sin t \cos t$ over a full period $[0, 2\pi]$ is 0 due to the symmetry of the sine and cosine functions. This means there was a mistake in my explanation; let's correct that.Given $\mathbf{F}(x, y, z) = x\mathbf{j}$ and the path $\mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + \cos t \mathbf{k}$, the correct approach to evaluate the line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ involves computing the dot product of $\mathbf{F}$ with the derivative of $\mathbf{r}(t)$ with respect to $t$, and then integrating over the given bounds of $t$.The derivative of $\mathbf{r}(t)$ with respect to $t$ is $-\sin t \mathbf{i} + \cos t \mathbf{j} - \sin t \mathbf{k}$. The dot product of $\mathbf{F}$ with this derivative is $x \cdot \cos t = \cos^2 t$, since $x = \cos t$ and the component of $\mathbf{F}$ in the direction of motion is $x\mathbf{j}$ dot $\cos t \mathbf{j}$.The correct integral to evaluate is $\int_0^{2\pi} \cos^2 t \, dt$. This integral evaluates to $\pi$, because the integral of $\cos^2 t$ over one period is $\frac{1}{2} \times 2\pi = \pi$, taking into account the average value of $\cos^2 t$ over its period.Therefore, the correct answer is $\pi$, indicating a mistake in my initial explanation.

Let F(x,y,z)=xj\mathbf { F } ( x , y , z ) = x \mathbf { j }F(x,y,z)=xj

Let $\mathbf { F } ( x , y , z ) = x \mathbf { j }$