1. ## Surface Integral

Calculate $\iint (x+y+z) dS$
If S is part of the plane 2x + y + z =1

Ok I think i have done this but want to double check, here's my working:
Sub in z = 1 - 2x - y to eliminate Z
$ds = \frac{dA}{cos(\theta)} = \frac{dx dy}{\vec{n}\vec{k}}$
$\vec{N} = \nabla g{x,y,z}$
$\vec{n} = \vec{N}/\abs{\vec{N}}$

Do all the subbing, i get
$\iint \sqrt{6}(-x+1) dxdy$

Limits are between [1-2x..0] for y and [1..0] for x.

Thus i obtain $\sqrt{1/6}$

Is this correct?

2. Originally Posted by imagemania
If S is part of the plane 2x + y + z =1
I assume you mean in the first ocatant for the plane

If that is the cases the x limit should go from zero to one half.

3. Ohh of course it's 2x, thanks

4. Originally Posted by imagemania
Ok I've come across a similar question which I am having troubles with, this time its the method
A vector field exists so that:
$\vec{F(r)}=f(r)\vec{r}$
where
$r = \sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$

C is the path on the sphere which is ${x}^{2}+{y}^{2}+{z}^{2}=1$

Prove that $\int \vec{F}.d\vec{r} = 0$

Now im not entirely sure how to best approach this, i can easily see:

$\int \vec{F}.d\vec{r} = \int f(r)\vec{r}.d\vec{r}$

Though i am unsure how to encorporate the path to deduce zero.

I guess the main thing throwing me here is there f(r) i have r, and i could use sphereical coordinates, but I do not know f(r), just part of it (the r). Do i assume f(r) = r?? Then convert into sphereical coordinates, work out normal etc ?
Use Stokes Theorem and note that

$\text{Curl} f(r)\mathbf{r}=\nabla f(r) \times \mathbf{r}+f(r)\nabla \times \mathbf{r}= \nabla f(r) \times \mathbf{r}$

5. Ugh Oh yeah, forgot about that theory! Thanks

6. Ok had another look, and it would seem straight foward ie do a matrix between the derivatives and $\vec{r}$. The issue is do we know $\vec{r}$. I know r as in f(r). Stokes theorem:
$\oint \vec{F}.d\vec{r} = \int \nabla \times(f(r)\vec{r}.d\vec{s})$

As you quite rightly said we can simply look at the cross product, but if i imply use r despite it not being the vector r, will the path C not be used anywhere in this formula (Which if its not would be logical, considering it would show its independent of path) perhaps on my sheet it should say:
$\vec{r} = \sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$
$r = \sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$
??

7. Originally Posted by imagemania
Ok had another look, and it would seem straight foward ie do a matrix between the derivatives and $\vec{r}$. The issue is do we know $\vec{r}$. I know r as in f(r). Stokes theorem:
$\oint \vec{F}.d\vec{r} = \int \nabla \times(f(r)\vec{r}.d\vec{s})$

As you quite rightly said we can simply look at the cross product, but if i imply use r despite it not being the vector r, will the path C not be used anywhere in this formula (Which if its not would be logical, considering it would show its independent of path) perhaps on my sheet it should say:
$\vec{r} = \sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$
$r = \sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$
??
The above does not make sense the supposed "vector" does not have any components!

What you have is the position vector and by defintion it is

$\mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$

8. Ok so i do the cross product between that and nabla, but then is the surface
${x}^{2}+{y}^{2}+{z}^{2}=1$
and r,
$r = \sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$

Used at all in this question :s

9. Originally Posted by imagemania
Ok so i do the cross product between that and nabla, but then is the surface
${x}^{2}+{y}^{2}+{z}^{2}=1$
and r,
$r = \sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$

Used at all in this question :s
By Stokes theorem you have that

$\int_{C}f(r)\mathbf{r}\cdot d \mathbf{s}=\iint_{S}(\nabla \times (f(r)\mathbf{r}))\cdot d\mathbf{S}=\iint_{S}(\nabla f(r) \times (\mathbf{r}))\cdot \mathbf{n}dS$

The normal vector to the sphere is some multiple of the position vector

This gives as the integrand

$(\nabla f(r) \times (\mathbf{r}))\cdot \mathbf{r}$

but these two vectors are perpendicular so their dot product is zero.