1. stoke's thrm

Let $\vec{F} = y\vec{i} + z\vec{j} + x\vec{k}$ and C be the boundary of $z = 5 - (x^2 + y^2)$ for z ≥ 3, oriented upward. Find $\int\limits_c\vec{F} · d\vec{r}$.

2. Originally Posted by TheRekz
Let $\vec{F} = y\vec{i} + z\vec{j} + x\vec{k}$ and C be the boundary of $z = 5 - (x^2 + y^2)$ for z ≥ 3, oriented upward. Find $\int\limits_c\vec{F} · d\vec{r}$.
So to use stokes theorem we need to find the curl of the vector

$\begin{vmatrix}
\vec i & \vec j & \vec k \\
\frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\
y & z & x
\end{vmatrix} = (0-1)\vec i-(1-0)\vec j+(0-1)\vec k=-\vec i -\vec j -\vec k$

Now we need the normal vector to the surfe

$F(x,y,z) = x^2+y^2+z-5$

$\nabla F = 2x \vec i + 2y \vec j + \vec k$

$|\nabla F| = \sqrt{4x^2 + 4y^2 + 1}$

$dS=\sqrt{1+(z_x)^2+(z_y)^2}= \sqrt{1+4x^2+4y^2}dA$

$\oint \vec F \cdot d\vec r = \iint_S \nabla \times \vec F\cdot d \vec S= \iint_S \nabla \times \vec F\cdot \frac{\nabla \vec F}{|\nabla F|}\sqrt{1+(z_x)^2+(z_y)^2}dA$

$-\iint_S(2x+2y+1)dA$

We need to integrate over the circle $x^2+y^2=(\sqrt{2})^2$

Just change to polar coordinates to finish

3. here's what I got:

$- \int_0^{2\pi}\int_0^{\sqrt(2)} (2\sqrt(2)cos\theta + 2\sqrt(2)sin\theta + 1) r dr d\theta$

is this correct?

4. Originally Posted by TheRekz
here's what I got:

$- \int_0^{2\pi}\int_0^{\sqrt(2)} (2\sqrt(2)cos\theta + 2\sqrt(2)sin\theta + 1) r dr d\theta$

is this correct?
Note quite you jumped the gun on the $\sqrt{2}$

$- \int_0^{2\pi}\int_0^{\sqrt(2)} (2rcos\theta + 2r\sin\theta + 1) r dr d\theta$

This will do it

5. I got it! thanks

6. Originally Posted by TheRekz
I got $-2\pi\sqrt(2)$ but it marks the answer as wrong

$- \int_0^{2\pi} \frac{4\sqrt(2)cos(\theta)}{3} + \frac{4\sqrt(2)sin(\theta)}{3} + \sqrt(2) d\theta$
$- \frac{4\sqrt(2)sin(\theta)}{3} + \frac{4\sqrt(2)cos(\theta)}{3} + \sqrt(2)\theta$ evaluated from 0 to 2pi
and I got the above answer

$- \int_0^{2\pi}\int_0^{\sqrt(2)} (2r^2cos\theta + 2r^2\sin\theta + r) dr d\theta=-\int_{0}^{2\pi} \frac{2}{3}r^3\cos(\theta)+\frac{2}{3}r^3\sin(\the ta)+\frac{1}{2}r^2 \bigg|_{0}^{\sqrt{2}}d\theta$

$=-\int_{0}^{2\pi} \frac{4\sqrt{2}}{3}\cos(\theta)+\frac{4\sqrt{2}}{3 }\sin(\theta)+1 d\theta =$

$-\left(\frac{4\sqrt{2}}{3}\sin(\theta)-\frac{4\sqrt{2}}{3}\cos(\theta)+\theta \bigg|_{0}^{2\pi}\right)=-2\pi$

7. what is the normal vector to this surface:

the surface z = 9 - x^2 for 0 ≤ x ≤ 3 and -6 ≤ y ≤ 6, oriented upward

8. Originally Posted by TheRekz
what is the normal vector to this surface:

the surface z = 9 - x^2 for 0 ≤ x ≤ 3 and -6 ≤ y ≤ 6, oriented upward
write the function in the form

$F(x,y,z)=0$

$z=9-x^2 \iff x^2+z-9=0$

This gives

$F(x,y,z)=x^2+z-9$

$\nabla F = \frac{\partial F}{\partial x} \vec i+ \frac{\partial F}{\partial y} \vec j+ \frac{\partial F}{\partial z} \vec k =2x\vec i +\vec k$

9. so if $\vec{F} = xy \vec{i} + yz\vec{j} + xz\vec{k}$ and what is dS here as what you did with the question above?

10. Originally Posted by TheRekz
so if $\vec{F} = xy \vec{i} + yz\vec{j} + xz\vec{k}$ and what is dS here as what you did with the question above?
$d\vec S=\vec n \cdot dS$

Where n is a normal unit vector from the surface

$\vec n =\frac{\nabla F}{|\nabla F|}$

This will always give you a normal unti vector.

Note that IF z can be isolated on your surface then ie

$z=g(x,y)$

then $|\nabla F|=dS=\sqrt{1+(z_x)^2+(z_y)^2}$

and then

$\iint_S F \cdot d\vec S = \iint F \cdot \vec n dA$

11. so dS above is equal to 2xi + k dA ??

12. ok after doing all that I got:

$\iint_S (-2xy - x) d\vec{A}$

I need the integration surface now... having a hard time to set it