# Math Help - Flux help

1. ## Flux help

Hi ,I would appreciate some guidance on this .thanks.
Evaluate the flux of the vector field F=18zi-12j+3yk across that part of the plane 2x+3y+6z=12 which is located in the first octant of the plane 2x+3y+6z=12 .Use the unit normal to the plane that points in the positive z-direction .

So far I have found the normal to the plane ... as 2i+3j+6k/7

I was intending on using the eqn Iintergral F  n ds =
not sure what to do with ds=sqrt(...)
I am not sure from where to where I am integrating ...flux

2. ## Re: Flux help

You mean "that part of the plane 2x+3y+6z=12 which is located in the first octant". The additional "of the plane 2x+ 3y+ 6z= 12" confused me!

When x and y are 0, that becomes 6z= 12 or z= 2 so one corner of the triangle is (0, 0, 2). When y and z are 0, that becomes 2x= 12 or x= 6 so (6, 0, 0) is another corner of the triangle.
And when x and z are 0, that becomes 3y= 12 or y= 4 so (0, 4, 0) is the third corner. Yes, the unit normal to the plane is (2i+ 3j+ 6k)/7 (note the parentheses1)

The "sqrt" in "sqrt(...)dx" is that "7", the length of the vector.

Another way of looking at surface integrals, which I prefer, is this: We can write any curve, in an xyz coordinate system, as x= f(s,t), y= g(s,t), z= h(s, t) where s and t are the parameters.
And we can write that as the vector equation $\vec{r}(s, t)= f(s,t)\vec{i}+ g(s,t)\vec{j}+ h(s,t)\vec{k}$. The two derivatives, $\vec{r}_s= f_s\vec{i}+ g_s\vec{j}+ h_s\vec{k}$ and $\vec{r}_t=f_t\vec{i}+ g_t\vec{j}+ h_t\vec{k}$ are vectors lying in the tangent plane to the surface so that $\vec{r}_s\times \vec{r}_t$ is normal to the surface. Further, we have $\vec{n}dS= \vec{r}_s\times\vec{r}_t dt ds$.

Here, our surface is the plane 2x+ 3y+ 6z= 12 which we can write as z= 2- (1/3)x- (1/2)y and write x= 3s, y= 2t, z= 2- s- t. The "position vector" or a point on the plane would be $\vec{r}= 3s\vec{i}+ 2t\vec{j}+ (2- s- t)\vec{k}$. The derivatives are $\vec{r}_s= 3\vec{i}- vec{k}$ and $\vec{r}_t= 2\vec{j}- \vec{k}$. The cross product of those two vectors (positive z component) is $2\vec{i}+ 3\vec{j}+ 2\vec{k}$ so that $\vec{n}dS= (2\vec{i}+ 3\vec{j}+ 2\vec{k})dsdt$.

The integrand vector function is $\vec{F}= 18z\vec{i}- 12\vec{j}+ 3y\vec{k}$ so $\vec{F}\cdot\vec{n}dS= (36z-36+ 6y)dsdt= (2- s-t- 36+6t)dsdt= (5t-s- 34)dsdt$

Projecting 2x+3y+ 6z= 12 into the xy-plane (z= 0) gives 2x+ 3y= 12 or y= 4- (2/3)x while x runs from 0 to 6. Since s= x and t= y, The integral will be
$\int_0^6\int_0^{4- (2/3)x} (5t- s- 3)dt ds$