1. ## Stokes' Theorem

Let C be the intersection of the sphere $x^2 +y^2 +z^2 = 2$ and the plane z=1, oriented counter-clockwise as viewed from high above the z-axis. Use Stokes' Theorem to evaluate.

$\int_{c} (-y^3 +z) dx + (x^3 +2y) dy + (y - x) dz$

Is my solution right?

Set $\vec F=(-y^3 +z) \vec i + (x^3 +2y) \vec j + (y - x) \vec k$
$curl(\vec F)= \vec i + \vec 2j + (3x^2 + 3y^2) \vec k$

By Stokes' Theorem, the line integral is equal to the surface integral
$\int \int_{\Sigma} (\nabla \times F) dS$

$\int \int_{\Sigma} (\nabla \times F) dS = \int \int_{\Sigma}' (\nabla \times F) \cdot \vec n ds
\int \int_{\Sigma}' (\nabla \times F) \cdot \vec k ds$

Sigma': ${ (x,y,z): x^2 +y^2 \leq 1, z=1}$ flat disk
so $(x, y, z) \rightarrow (cos \theta, sin \theta, 1)$

so
$\int \int_{\Sigma} ' 3x^2 +3y^2 ds$

$3 \int_{0}^{2 \pi} \int_{0}^{\sqrt{2}} r dr d\theta = 6\pi$

2. Originally Posted by yzc717
Let C be the intersection of the sphere $x^2 +y^2 +z^2 = 2$ and the plane z=1, oriented counter-clockwise as viewed from high above the z-axis. Use Stokes' Theorem to evaluate.

$\int_{c} (-y^3 +z) dx + (x^3 +2y) dy + (y - x) dz$

Is my solution right?

Set $\vec F=(-y^3 +z) \vec i + (x^3 +2y) \vec j + (y - x) \vec k$
$curl(\vec F)= \vec i + \vec 2j + (3x^2 + 3y^2) \vec k$

By Stokes' Theorem, the line integral is equal to the surface integral
$\int \int_{\Sigma} (\nabla \times F) dS$

$\int \int_{\Sigma} (\nabla \times F) dS = \int \int_{\Sigma}' (\nabla \times F) \cdot \vec n ds
\int \int_{\Sigma}' (\nabla \times F) \cdot \vec k ds$

Sigma': ${ (x,y,z): x^2 +y^2 \leq 1, z=1}$ flat disk
so $(x, y, z) \rightarrow (cos \theta, sin \theta, 1)$

so
$\int \int_{\Sigma} ' 3x^2 +3y^2 ds$
No. To integrate on the surface (disk), $x= rcos(\theta)$, $y= r sin(\theta)$ and you must integrate over both r and $\theta$, with r going from 0 to 1 and $\theta$ going from 0 to $2\pi$.

$3\int_{r=0}^1\int{\theta= 0}^{2\pi} (r^2cos^2(\theta)+ r^2sin^2(\theta)) rd\theta dr$
$= 3\int{r= 0}^1\int{\theta= 0}^{2\pi} r^3 d\theta dr$
$= 6\pi \int_{r= 0}^1 r^3 d\theta= \frac{3}{2}\pi$

$3 \int_{0}^{2 \pi} \int_{0}^{\sqrt{2}} r dr d\theta = 6\pi$

3. The integral to be found is

$\oint_C(-y^3+z)\,dx+(x^3+2y)\,dy+(y-x)\,dz=\int_{\Sigma}\mbox{curl}\,\mathbf{F}\cdot\m athbf{n}\,d\sigma.$

In our case,

\begin{aligned}
z&=\sqrt{2-x^2-y^2}\;\;\;\;\;\;\;\;\;\;\mbox{(Surface)}\\
\mathbf{F}&=\langle-y^3+z,x^3+2y,y-x\rangle \\
\mathbf{n}&=\left<-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1\right>\cdot\left(\sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}\right)^{-1}
\end{aligned}

Since

$\int_{\Sigma}\mbox{curl}\,\mathbf{F}\cdot\mathbf{n }\,d\sigma=\int_D\mbox{curl}\,\mathbf{F}\cdot\math bf{n}\,\sec\gamma\,dA,$

where $\gamma$ is the angle made by the upper unit normal of the surface with the $z$-axis, and

$\sec\gamma=\sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1},$

our integral becomes

$\int_D\mbox{curl}\,\mathbf{F}\cdot\left<-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1\right>\,dA.$

We can now evaluate it with the mnemonic rule

$\mbox{curl}\,\mathbf{F}=\nabla\times\mathbf{F}=
\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
{\partial}/{\partial x} & {\partial}/{\partial y} & {\partial}/{\partial z} \\
-y^3+z & x^3+2y & y-x
\end{array}\right|.$

4. Originally Posted by Scott H
Is below way also right?

Originally Posted by HallsofIvy
No. To integrate on the surface (disk), $x= rcos(\theta)$, $y= r sin(\theta)$ and you must integrate over both r and $\theta$, with r going from 0 to 1 and $\theta$ going from 0 to $2\pi$.

$3\int_{r=0}^1\int{\theta= 0}^{2\pi} (r^2cos^2(\theta)+ r^2sin^2(\theta)) rd\theta dr$
$= 3\int{r= 0}^1\int{\theta= 0}^{2\pi} r^3 d\theta dr$
$= 6\pi \int_{r= 0}^1 r^3 d\theta= \frac{3}{2}\pi$

5. Originally Posted by Scott H
our integral becomes

$\int_D\mbox{curl}\,\mathbf{F}\cdot\left<-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1\right>\,dA.$
if I were to do this in your way, what is the D refer to? the unit disk? then go to polar coordinates?

6. Yes.

When $z=1$, the equation for the surface $x^2+y^2+z^2=2$ becomes

\begin{aligned}
x^2+y^2+1&=2\\
x^2+y^2&=1,
\end{aligned}

defining a circle of radius $1$. Our limits for $\mbox{curl}\,\mathbf{F}\cdot\left<-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1\right>$ in polar coordinates will therefore be

$\int_0^{2\pi}\int_0^1.$

I believe that HallsofIvy was just giving the integral

$\int_D 3x^2+3y^2\,dA,$

which was not what I calculated for the surface integral in question. In fact, what you have tried to use is called Stoke's theorem in the plane, and while you have applied it correctly, it only works for planar surfaces. Here, we must use the general form of Stoke's theorem as the surface $z=\sqrt{2-x^2-y^2}$ is curved. In this case, there is a useful theorem that

$\int_{\Sigma} \mbox{curl}\,\mathbf{F}\cdot\mathbf{n}\,d\sigma=\i nt_R \mbox{curl}\,\mathbf{F}\cdot\left<-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1\right>\,dA,$

which we just proved by multiplying the unit normal $\mathbf{n}$ by $\sec\gamma$ in calculating the surface integral.