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

Question

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

Quizplus · 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 $\boxed{	extbf{(B) }11ea8a3b_b748_8c3a_9fff_ebb09ed0da9d_TB6897_11}$.

Use Stokes' Theorem to Evaluate Is the Part