# A lil help please?

1. Jul 25, 2004

### Phymath

ok this probley seems simple but i just need to see how to do it, ok well how do u evaluate this...
find the flux of
the vector field...
$$\vec{F}=<x,y,z>$$

throught this surface above the xy-plane..
$$z = 4-x^2-y^2$$

how do u evaluate this with surface integrals method and the divergence therom

$$dS = \sqrt{f_{x}^2+f_{y}^2+1}dA$$ for surface inergrals

thanks very much to anyone who can show me this step by step, i keep getting the wrong answer and i don't know why...

Last edited: Jul 25, 2004
2. Jul 25, 2004

### pnaj

Hi phy_math,

just to make sure I understand you, is this right:

$$\begin{array}{l} \vec F = \vec x = x\hat x + y\hat y + z\hat z \\ f(x,y,z) = z - (4 - x^2 - y^2) = 0\\ \end{array}$$

Since then ...

$$dS = dxdy\sqrt {(\frac{{\partial f}}{{\partial x}})^2 + (\frac{{\partial f}}{{\partial y}})^2 + 1}$$ makes sense

... where you can replace $dA = dxdy$.

To get the total flux of $\vec F$ through the surface, you want the integral $\phi = \int\limits_S {\vec F \bullet \hat ndS}$, where $\hat n = n_x \hat x + n_y \hat y + n_z \hat z$ is the unit normal to the surface.

So, to do this integral, you'll need to calculate the unit normal to the surface. You've already got $\vec F$ and can calculate $dS$ from the above formula.

Have you seen the equation $\nabla f \bullet \hat n = 0$ before?

Also, if you're calculating this flux over the whole of space remember that, by definition, you can write

$$\int\limits_0^\infty {\int\limits_0^\infty {f(x,y)dxdy} } = \begin{array}{*{20}c} {\lim } \\ {X,Y \to \infty } \\ \end{array}\int\limits_0^Y {\int\limits_0^X {f(x,y)dxdy} }$$

That might help as well. See how you get on.

Last edited: Jul 25, 2004
3. Jul 25, 2004

### Corneo

Correct me if I am wrong, I just learned this last week.

By the divergence theorem, $$\int_S \vec F \cdot \vec n \ dS = \int_V \nabla \cdot \vec F \ dV$$

Since $$\vec F = x \vec i + y \vec j + z \vec k$$, then $$\nabla \cdot \vec F = 3$$.

A quick sketch of the surface shows a inverted parabolid. This volume should be easy to find via cylindrical coordinates.

$$flux=\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{4-r^2} 3r \ dz \ dr \ d\theta$$

Side note: Does anyone know if there is a shortcut to writing the "[*tex]...[*/tex]" tags?

4. Jul 25, 2004

### HallsofIvy

Staff Emeritus
By the way, a simpler way to find $\int\limits_S {\vec F \bullet \hat ndS}$
without calculating n and dS separately is this:

The surface is given by z- (4- x2- y2)= 0 or z+ x2+ y2= 4, a constant. If we take F(x,y,z)= z+ x2+ y2 then the surface is a "level surface" of F and so grad F= <2x, 2y, 1> is perpendicular to the surface and n dS is <2x, 2y, 1>dxdy.

Notice that we have "dropped down" to the xy-plane. If the z-component of grad F had not been 1, we would have had to make it 1.

For example if the surface is x2+ y2+ z2= R2 (a sphere) we would take F(x,y,z)= x2+ y2+ z2 so that grad F= <2x, 2y, 2z>. Now we have to decide in what plane we want to do the integration (with a sphere it really doesn't matter!). If we choose to integrate in the xy-plane, then we divide the entire vector by 2z to "normalize" the z-component: <x/z, y/z, 1>dx dy. Of course, we would have to divide the problem into two parts, z positive or z negative, to integrate over the entire sphere.

5. Jul 26, 2004

### Phymath

thanks all, its good to know ur all here to help! ya i did similar things, turns out i was just basicly using the wrong limits for the surface thanks everyone!