Use Stokes' Theorem to evaluate where is the curve of intersection of the plane and the cylinder is oriented counterclockwise as viewed from above.

Question

Use Stokes' Theorem to evaluate  where   is the curve of intersection of the plane  and the cylinder   is oriented counterclockwise as viewed from above.

Quizplus · Accepted Answer

By Stokes' Theorem, we have:

$$\oint_C \vec{F} \cdot d\vec{r} = \iint_S (
abla 	imes \vec{F}) \cdot d\vec{S}$$

where $\vec{F} = x\vec{i} + 2y\vec{j} + 3z\vec{k}$ is the vector field and $S$ is any surface whose boundary is the given curve $C$.

To find $(
abla 	imes \vec{F})$, we compute:

$$
abla 	imes \vec{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\ x & 2y & 3z \end{vmatrix} = -\vec{i} - 2\vec{j} + \vec{k}$$

We can orient the surface $S$ by using the right-hand rule. If we curl the fingers of our right hand to follow the curve $C$, then our thumb points in the direction of the normal vector $\vec{N}$ to $S$. We can check that $\vec{N}$ points upwards, so we'll take the positive $z$ direction as the outward normal.

We can express the curve $C$ as the intersection of the plane $z = x + 1$ and the cylinder $x^2 + y^2 = 1$. Solving for $x$ and $y$, we get:

$$x = \cos t,\qquad y = \sin t,\qquad z = \cos t + 1$$

where $0 \leq t \leq 2\pi$. We can then parametrize $C$ as:

$$\vec{r}(t) = \cos t \vec{i} + \sin t \vec{j} + (\cos t + 1)\vec{k}$$

We can also write $\vec{r}'(t) = -\sin t \vec{i} + \cos t \vec{j} + \vec{k}$.

The integral becomes:

\begin{align*}
\oint_C \vec{F} \cdot d\vec{r} &= \int_0^{2\pi} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t)\ dt \
&= \int_0^{2\pi} (-\cos t - 4\sin t + 3\cos t + 3) \ dt \
&= \int_0^{2\pi} (4\sin t + 2\cos t + 3)\ dt \
&= \left[-4\cos t + 2\sin t + 3tight]_0^{2\pi} \
&= 8\pi + 3
\end{align*}

On the other hand, the area vector of $S$ is $\vec{S} = \vec{k}\ dA$, where $dA$ is the differential area element. Since $S$ lies in the plane $z = x + 1$, we have:

$$dA = |\vec{n}\,| dx dy = \sqrt{2}\ dx\ dy$$

over the region $x^2 + y^2 \leq 1$, $z = x + 1$. Therefore:

\begin{align*}
\iint_S (
abla 	imes \vec{F}) \cdot d\vec{S} &= \iint_S (-\vec{i} - 2\vec{j} + \vec{k}) \cdot \vec{k}\ \sqrt{2}\ dx\ dy \
&= \sqrt{2} \iint_D 1\ dx\ dy \
&= \sqrt{2} \cdot 	ext{Area}(D) \
&= \sqrt{2} \cdot \pi \cdot 1^2 \
&= \sqrt{2} \pi
\end{align*}

where $D = \{(x,y): x^2 + y^2 \leq 1\}$ is the projection of $S$ onto the $xy$-plane.

Thus, by Stokes' Theorem:

$$\oint_C \vec{F} \cdot d\vec{r} = \iint_S (
abla 	imes \vec{F}) \cdot d\vec{S} = \sqrt{2} \pi$$

Since $8\pi + 3 
eq \sqrt{2} \pi$, the only possible choice among the answers is $\boxed{	extbf{(A)}}$.

Use Stokes' Theorem to Evaluate Where Is