Question 1

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

Accepted Answer

A

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

The answer of Use Stoke's theorem to evaluate  ...

Question 3

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

Accepted Answer

A

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.
A) 
B) 
C) 
D) 
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.
A) 2
B) 
C) 4
D) 3
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 $\boxed{	extbf{(B) }11ea8a3b_b748_8c3a_9fff_ebb09ed0da9d_TB6897_11}$.

Quiz 13: Extension G: Vector Calculus

Use Stokes' Theorem to evaluate where and is the triangle with vertices is oriented counterclockwise as viewed from above.

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

Use Stokes' Theorem to evaluate where is the circle is oriented counterclockwise as viewed from above.

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

Use Stokes' Theorem to evaluate is the part of the paraboloid that lies inside the cylinder oriented upword.