# Math Help - Surface Integral

1. ## Surface Integral

Find the flux of F across S. For closed surfaces, use the positive orientation.

F(x,y,z) = xy i +yz j + zx k

S is the part of the paraboloid z = 4 - x^2 - y^2 that lies above the square 0 <= x <= 1 and 0 <= y <= 1 and has upward orientation.

The part that's confusing me is finding an appropriate paramaterization.

2. Originally Posted by Math Major
Find the flux of F across S. For closed surfaces, use the positive orientation.

F(x,y,z) = xy i +yz j + zx k

S is the part of the paraboloid z = 4 - x^2 - y^2 that lies above the square 0 <= x <= 1 and 0 <= y <= 1 and has upward orientation.

The part that's confusing me is finding an appropriate paramaterization.
If we write the surface as a function of 3 variable we get

$g(x,y,z)=x^2+y^2+z-4 \implies \nabla g = 2x \vec i+ 2y \vec j + \vec k$

$|\nabla g|=\sqrt{1+4x^2+4y^2}$

and $dS=\sqrt{1+\left( \frac{\partial z}{\partial x} \right)^2+ \left( \frac{\partial z}{\partial y} \right)^2}dA=\sqrt{1+4x^2+4y^2}dA$

Then we get...

$\iint \vec F \cdot d\vec S = \iint_D F(x,y,z) \cdot \frac{\nabla g}{|\nabla g|}dS=\iint F(x,y,z)\cdot \nabla g dA$

$\int_{0}^{1}\int_{0}^{1}(2x^2y+2y^2z+xz)dxdy = \int_{0}^{1}\int_{0}^{1}(2x^2y+2y^2(4 - x^2 - y^2)+x(4 - x^2 - y^2))dxdy$

3. If you need to use a parametrization of the surface you can use the natural one

$\begin{pmatrix}x\\y\\4-x^2-y^2\end{pmatrix}$

$\overrightarrow{dS} = \frac{\partial}{\partial x}\begin{pmatrix}x\\y\\4-x^2-y^2\end{pmatrix} X \frac{\partial}{\partial y}\begin{pmatrix}x\\y\\4-x^2-y^2\end{pmatrix}\:dx\:dy$

$\overrightarrow{dS} = \begin{pmatrix}2x\\2y\\1\end{pmatrix}$

$\overrightarrow{F}.\overrightarrow{dS} = \left(2x^2y + 2y^2(4-x^2-y^2) + x(4-x^2-y^2)\right)\:dx\:dy$

which fortunately gives the same result as TheEmptySet

4. Thanks to you both for the speedy replies!

Originally Posted by running-gag
If you need to use a parametrization of the surface you can use the natural one

$\begin{pmatrix}x\\y\\4-x^2-y^2\end{pmatrix}$

$\overrightarrow{dS} = \frac{\partial}{\partial x}\begin{pmatrix}x\\y\\4-x^2-y^2\end{pmatrix} X \frac{\partial}{\partial y}\begin{pmatrix}x\\y\\4-x^2-y^2\end{pmatrix}\:dx\:dy$

$\overrightarrow{dS} = \begin{pmatrix}2x\\2y\\1\end{pmatrix}$

$\overrightarrow{F}.\overrightarrow{dS} = \left(2x^2y + 2y^2(4-x^2-y^2) + x(4-x^2-y^2)\right)\:dx\:dy$

which fortunately gives the same result as TheEmptySet
When parameterizing, I can always just let x = u, y = v, and then find z in terms of u and v, correct?

5. Originally Posted by Math Major
Thanks to you both for the speedy replies!

When parameterizing, I can always just let x = u, y = v, and then find z in terms of u and v, correct?
Yes that's it

Due to the fact that x=u and y=v, I let x and y instead of using u and v