Flux through Surfaces defined by a Function

Now we need to find $\vec{n}$ and dS. If the surface S is defined by a function, z=g(x,y) we can obtain a nice expression for $\vec{n}$ and dS. For the normal vector to S at the point (x₀,y₀,g(x₀,y₀)) we can use the normal vector to the tangent plane. Remember that the tangent plane to g(x,y) at the point (x₀,y₀) is given by:

$$z=g_{x}(x_{0},y_{0})\left( x-x_{0}\right) +g_{y}(x_{0},y_{0})\left( y-y_{0}\right) +g(x_{0},y_{0})$$

or

-g_x(x₀,y₀)x-g_y(x₀,y₀)y+z=c

where c=-g_x(x₀,y₀)x₀-g_y(x₀,y₀)y₀+g(x₀,y₀). A normal vector to the tangent plane is

$$\vec{n}=-g_{x}(x_{0},y_{0})\hat{\imath}-g_{y}(x_{0},y_{0})\hat{j}+\hat{k}$$

If we normalize $\vec{n}$ to get a unit normal vector we get

$$\hat{n}=\frac{-g_{x}(x_{0},y_{0})\hat{\imath}-g_{y}(x_{0},y_... ...t{k}}{\sqrt{g_{x}(x_{0},y_{0})^{2}+g_{y}(x_{0},y_{0})^{2}+1}}$$

So the normal vector as a function of $\left( x,y\right)$ is

$$\hat{n}=\frac{-g_{x}(x,y)\hat{\imath}-g_{y}(x,y)\hat{j}+\hat{k}}{\sqrt{g_{x}(x,y)^{2}+g_{y}(x,y)^{2}+1}}$$

The Area of the j^th patch can be obtained by finding the vectors $\vec{u}$, $\vec{v}$ which define the edges of the patch (one parallel to the xz-plane and one parallel to the yz-plane????). The area of the patch is given by $\left\Vert \vec{u}\times \vec{v}\right\Vert$. If we employ the tangent plane approximation of g(x,y)we can see that $\vec{u}$ is the vector from (x₀,y₀,g(x₀,y₀)) to $(x_{0}+\Delta x,y_{0},g(x_{0},y_{0})+g_{x}(x_{0},y_{0})\Delta x)$ or

$$\vec{u}=\Delta x\hat{\imath}+g_{x}(x_{0},y_{0})\Delta x\hat{k}$$

Similarly $\vec{v}$ is the vector that points from (x₀,y₀,g(x₀,y₀)) to the point $(x_{0},y_{0}+\Delta y,g(x_{0},y_{0})+g_{y}(x_{0},y_{0})\Delta y)$ or

$$\vec{v}=\Delta y\hat{j}+g_{y}(x_{0},y_{0})\Delta y\hat{k}$$

Taking the cross product of $\vec{u}$ and $\vec{v}$ gives

$$\vec{u}\times \vec{v} =\left\vert \begin{array} {lll} \ha... ... \Delta y & g_{y}(x_{0},y_{0})\Delta y \end{array} \right\vert$$

$$=-g_{x}(x_{0},y_{0})\Delta x\Delta y\hat{\imath}-g_{y}(x_{0},y_{0})\Delta x\Delta y\hat{j}+\Delta x\Delta y\hat{k}$$

Finally, the area of the j^th patch is

$$\Delta S_{j}=\left\Vert \vec{u}\times \vec{v}\right\Vert =\s... ...x}(x_{0},y_{0})^{2}+g_{y}(x_{0},y_{0})^{2}+1}\Delta x\Delta y$$

In the limit that $\Delta S_{j}\rightarrow dS$ this becomes

$$dS =\sqrt{g_{x}(x,y)^{2}+g_{y}(x,y)^{2}+1}dxdy$$

$$=\sqrt{g_{x}(x,y)^{2}+g_{y}(x,y)^{2}+1}d$$

When we put these together we get

$$d\vec{A} =\hat{n}dS=\frac{-g_{x}(x,y)\hat{\imath}-g_{y}(x,y)... ...^{2}+g_{y}(x,y)^{2}+1}}\sqrt{g_{x}(x,y)^{2}+g_{y}(x,y)^{2}+1}dA$$

 

$$d\vec{A} =\left( -g_{x}(x,y)\hat{\imath}-g_{y}(x,y)\hat{j}+\hat{k}\right) dA$$ (3)

So we've represented the integral over the surface in terms of an integral over the region R in the xy-plane over which S sits.

$$\int_{S}\vec{F}\cdot d\vec{A}=\int_{S}\vec{F}\cdot \hat{n}dS... ...x,y)\hat{\imath}-g_{y}(x,y)\hat{j}+\hat{k}\right) \right] dA$$

Where R is the ''shadow'' of S on the xy-plane.

Notes:

• We found $d\vec{A}=\hat{n}dS$ for an upward normal vector the formula for the downward normal vector is

  $$d\vec{A}=\left( g_{x}(x,y)\hat{\imath}+g_{y}(x,y)\hat{j}-\hat{k}\right) dA$$

• The area vector $d\vec{A}$ and the infinitesimal piece of area dA are not the same thing, and $\left\Vert d\vec{A}\right\Vert \neq dA$. The are related by $\left( \ref{dA}\right)$ when the surface S is the graph of a function.

Example

1.

Calculate the flux of the vector field $\vec{F}=-z\hat{\imath}+\frac{z^{3}}{y}\hat{j}+x\hat{k}$ through the upper hemisphere of x²+y²+z²=4 that lies above the rectangle $-1\leq x\leq 1$, $\frac{1}{2}\leq y\leq 1$, oriented upward. Since the surface consists of the upper hemisphere we can solve the equation of the sphere for $z=g(x,y)=\sqrt{4-x^{2}-y^{2}}$, so the surface is the graph of a function and we can use $\left( \ref{dA}\right)$ to find $d\vec{A}$.

$$d\vec{A}=\left( \frac{x}{\sqrt{4-x^{2}-y^{2}}}\hat{\imath}+\frac{y}{\sqrt{4-x^{2}-y^{2}}}\hat{j}+\hat{k}\right) dxdy$$

The flux integral is

$$\int_{S}\vec{F}\cdot d\vec{A}=\int_{R}\left( \frac{-xz}{\sqr... ...2}-y^{2}}}+\frac{yz^{3}}{y\sqrt{4-x^{2}-y^{2}}}+x\right) dydx$$

Where R is the rectangle $-1\leq x\leq 1$, $\frac{1}{2}\leq y\leq 1$. Now substitute $z=\sqrt{4-x^{2}-y^{2}}$ to eliminate the z dependence in the integral. The integral reduces to

$$\int_{-1}^{1}\int_{0.5}^{1}\left( -x+4-x^{2}-y^{2}+x\right) dydx$$

$$=\int_{-1}^{1}\int_{0.5}^{1}\left( 4-x^{2}-y^{2}\right) dydx=\frac{37}{12}$$