Evaluate the surface integral for a vector field

Question

I asked this question among others in another thread (Vector analysis questions). I was given a link to a site (http://mathinsight.org/surface_integral_vector_field_introduction) which I read, and watched youtube videos and read some other sites also. After having done this I am still having trouble understanding and therefore completing the question. The examples I have found seem to have different formats to that of this question also making it hard to attempt. Could anyone help me on how to complete this question.

I can do the line integral for a vector field which looks similar but my attempts at this havent come out well at all.

Answer 1

Start by solving for one variable explicitly.

$$z=f(x,y)=2-\frac{1}{3}x-\frac{1}{2}y$$

You can form a vector which points to any given point on your surface by

$$\vec{r}=<x,y,f(x,y)>$$

Now, if we differentiate the above with respect to $x$ and then with respect to $y$ we will obtain two linearly independent vectors which are tangent to the plane under description. If you are interested why ask and I will try to give more details.

Once, we have the two linearly independent tangent vectors we can take the cross-product of the two vectors to find a vector which is perpendicular to the surface. And then we normalize

$$\hat{n}=\frac{<1,0,\frac{\partial f}{\partial x}>\times<0,1,\frac{\partial f}{\partial y}>}{||{<1,0,\frac{\partial f}{\partial x}>\times<0,1,\frac{\partial f}{\partial y}>}||}$$

Now, once you take an inner product of the unit normal vector with the vector field $\vec{A}$ you will obtain a scalar quantity that you will need to integrate over the appropriate region of the $xy$ plane.

Answer 2

Choose the parameterization $\vec r=x\vec i+y\vec j+(2-\frac{1}{2}x-\frac{1}{3}y)\vec k$.

Then $\vec r_{x}=\vec i-\frac{1}{2}\vec k$ and $\vec r_{y}=\vec j-\frac{1}{3}\vec k$ so

$\tag1\vert \vec r_{x}\times \vec r_{y}\vert =\frac{7}{6}$.

Now the general formula for the integral is

$\tag2\int \int (\vec A\cdot \vec n) \vert \vec r_{x}\times \vec r_{y}\vert dydx$

where $\vec n$ is the unit normal to the surface.

But since the surface is part of the plane $2x+3y+6z=12$ you can read off the normal as $2\vec i+3\vec j+6\vec k$ which normalizes to $\tag3\frac{1}{7}(2\vec i+3\vec j+6\vec k)$

Substituting $(1)$ and $(3)$ into $(2)$ we obtain

$\frac{1}{6}\tag4\int \int (36z+18y-36)dydx$

Now for the limits of integration:

The vector $\vec r$ defines a parameterization in $x$ and $y$ but these vary only over the portion of the surface in the first octant. i.e. $x$ and $y$ vary over the triangle formed by the lines $x=0$, $y=0$ and $2x+3y=12$. Therefore the integral is

$\tag5\frac{1}{6}\int^{6}_{0} \int^{\frac{12-2x}{3}}_{0} (36(\frac{12-2x-3y}{6})+18y-36)dydx=\int^{6}_{0} \int^{\frac{12-2x}{3}}_{0}(6-2x)dydx$

Answer 3

The integration over the surface $S$ defined by $2x+3y+6z=12$, $x\ge 0$, $y\ge 0$, and $z\ge 0$, can be transformed into an integration in the $x-y$ plane making use of the relationship that the differential surface $dS=\dfrac{dx\,dy}{\cos \gamma}$, where $\gamma$ is the angle between the $z$ axis and the normal to the surface.

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NOTE:

This way forward is effectively identical to using a parametric approach in which $x$ and $y$ play the role of parameters $u$ and $v$.

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Proceeding we have

$$\int_S \vec A\cdot \hat n dS =\iint_{S_{xy}}\vec A\cdot \hat n\,\dfrac{dx\,dy}{\cos \gamma}$$

The unit normal $\hat n$ to $S$ is trivial and is given by $\hat n=\dfrac{2\hat x+3\hat y+6\hat z}{7}$, while $\cos \gamma=\hat z\cdot \hat n=\frac67$.

We can then write

$$\begin{align} \vec A\cdot \hat n\,dS&=(\hat x 18z-\hat y12+\hat z 3y)\cdot \left(\dfrac{2\hat x+3\hat y+6\hat z}{7}\right)\,\frac76\,dx\,dy\\\\ &=(12-2x)\,dx\,dy \end{align}$$

Finally, we have

$$\int_S \vec F\cdot \hat n dS=\int_{0}^{3}\int_{0}^{6-2y}(12-2x)\,dx\,dy$$

The evaluation of the integral is left as an exercise for the reader.