[Math] Flux of vector field across surface S

calculusVector Fields

The question I am working on is:

Let S be the part of the plane $1\!x + 3\!y + z = 3$ which lies in the first octant, oriented upwards. Find the flux of the vector field $\mathbf{F} = 3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$ across the surface S.

My work so far:

$\int F dS$

Parametrization: $T: (x, y, 3-3y-x)$

$T_x (1, 0, -1)$

$T_y (0,1,-3)$

$T_x \times T_y (-1,3,1)$

This is the correct orientation (going upwards).

$\int \int (3,2,3) \cdot (-1,3,1)dxdy = -3+6+3 = 6$

$\int\int6dxdy$ with $0 \leq x \leq 3-3y$ and $0 \leq y \leq 1/3$

This isn't yielding me the correct answer. I think my bounds are incorrect but I'm a bit stumped on how to find the correct ones.

Best Answer

Let the parametric equation of the surface be

$${\bf{r}} = {\bf{r}}(u,v)$$

Then the flux across the surface is give by

$$\eqalign{ & \varphi = \int\limits_S {{\bf{F}}({\bf{x}}(u,v)).d{\bf{a}}} \cr & d{\bf{a}} = \frac{{\partial {\bf{r}}}}{{\partial u}} \times \frac{{\partial {\bf{r}}}}{{\partial v}}dudv \cr} $$

Now, consider the following

$$\eqalign{ & {\bf{r}} = x{\bf{i}} + y{\bf{j}} + \left( {3 - x - 3y} \right){\bf{k}} \cr & \frac{{\partial {\bf{r}}}}{{\partial x}} = {\bf{i}} - {\bf{k}},\,\,\,\,\,\frac{{\partial {\bf{r}}}}{{\partial y}} = {\bf{j}} - 3{\bf{k}} \cr & \frac{{\partial {\bf{r}}}}{{\partial x}} \times \frac{{\partial {\bf{r}}}}{{\partial y}} = \left( {{\bf{i}} - {\bf{k}}} \right) \times \left( {{\bf{j}} - 3{\bf{k}}} \right) = {\bf{i}} + 3{\bf{j}} + {\bf{k}} \cr & d{\bf{a}} = \frac{{\partial {\bf{r}}}}{{\partial x}} \times \frac{{\partial {\bf{r}}}}{{\partial y}}dxdy = \left( {{\bf{i}} + 3{\bf{j}} + {\bf{k}}} \right)dxdy \cr & {\bf{F}}({\bf{x}}) = 3{\bf{i}} + 2{\bf{j}} + 3{\bf{k}} \cr & {\bf{F}}({\bf{x}}).d{\bf{a}} = 3 + 6 + 3 = 12\,dxdy\cr} $$

and hence

$$\eqalign{ & \varphi = \int_{y = 0}^1 {\left( {\int_{x = 0}^{3 - 3y} {12dx} } \right)dy} = \int_{y = 0}^1 {12(3 - 3y)dy} = 36\int_{y = 0}^1 {(1 - y)dy} \cr & \,\,\,\,\, = 36\left. {\left( {y - \frac{{{y^2}}}{2}} \right)} \right|_0^1 = 18 \cr} $$


Your Mistakes

1) The normal was wrong.

2) The bounds was wrong.

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