Question 1

Use Stokes' Theorem to evaluate   where   and   is the triangle with vertices     is oriented counterclockwise as viewed from above.&#10;A)  &#10;B)  &#10;C)  &#10;D)  &#10;E)

Accepted Answer

11ea8a3b_b748_b350_9fff_056a804ed125_TB6897_11

Question 2

Use Stoke's theorem to evaluate     C is the curve of intersection of the plane z = x + 9 and the cylinder

Accepted Answer

11ea8a3b_b749_c4dd_9fff_31627e9ef505_TB6897_11

Question 3

Use Stokes' Theorem to evaluate   where     is the circle     is oriented counterclockwise as viewed from above.&#10;A)  &#10;B)  &#10;C)  &#10;D)  &#10;E)

Accepted Answer

11ea8a3b_b749_288a_9fff_23622a78f0d1_TB6897_11

Question 4

Use Stokes' Theorem to evaluate   where     is the curve of intersection of the plane   and the cylinder     is oriented counterclockwise as viewed from above.&#10;A)  &#10;B)  &#10;C)  &#10;D)  &#10;E)

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 $\boxed{	extbf{(A)}}$.

Question 5

Use Stokes' Theorem to evaluate       is the part of the paraboloid   that lies inside the cylinder   oriented upword.&#10;A) 2&#10;B)  &#10;C) 4&#10;D) 3&#10;E) 1

Accepted Answer

We are given that the surface is a part of the paraboloid $z=x^2+y^2$ that lies inside the cylinder $x^2+y^2=1$ and is oriented upwards.

To apply Stokes' Theorem, we need to find the curl of the given vector field $\mathbf{F}(x,y,z) = y\,\mathbf{i} + x\,\mathbf{j} + z\,\mathbf{k}$:

$$	ext{curl}\,\mathbf{F}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ y & x & z\end{vmatrix}=\left(\frac{\partial}{\partial y} - \frac{\partial}{\partial x}ight)\mathbf{i} + \left(\frac{\partial}{\partial z} - \frac{\partial}{\partial y}ight)\mathbf{j} + \left(\frac{\partial}{\partial x} - \frac{\partial}{\partial z}ight)\mathbf{k} = -2\mathbf{i}.$$

Now, we need to find the boundary curve of the surface, which is the intersection of the paraboloid and the cylinder. Since $x^2+y^2=1$ and $z=x^2+y^2$, the boundary curve is the circle $x^2 + y^2 = 1,\ z = 1$. This circle is oriented counterclockwise when viewed from above (i.e. along the positive $z$-axis).

We can now apply Stokes' Theorem:

$$\iint_S 	ext{curl}\,\mathbf{F} \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r}.$$

The left-hand side is simply the flux of $	ext{curl}\,\mathbf{F}$ across the surface $S$. Since the surface is oriented upward and $	ext{curl}\,\mathbf{F} = -2\mathbf{i}$ is in the negative $x$-direction, the flux is $-\iint_S 2\,dS = -2\pi$.

The right-hand side is the line integral of $\mathbf{F}$ around the boundary curve $C$. Since $C$ is a circle of radius 1 centered at the origin, we can parameterize it as $\mathbf{r}(t) = \cos t\,\mathbf{i} + \sin t\,\mathbf{j}$, where $0 \le t \le 2\pi$. Then,

\begin{align*}
\oint_C \mathbf{F} \cdot d\mathbf{r} &= \int_0^{2\pi} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\,dt \
&= \int_0^{2\pi} \left(\sin t,\cos t,1 - \sin^2 tight) \cdot (-\sin t,\cos t,0)\,dt \
&= \int_0^{2\pi} (-\sin^2 t - \cos^2 t)\,dt = -2\pi.
\end{align*}

Since both sides of Stokes' Theorem have the same value, we have confirmed Stokes' Theorem and our answer is $\boxed{	extbf{(B) }11ea8a3b_b748_8c3a_9fff_ebb09ed0da9d_TB6897_11}$.

Deck 13: Extension G: Vector Calculus