Originally Posted by
mr fantastic Do you know that when the surface is defined in the form $\displaystyle z = f(x, y)$ then:
$\displaystyle \int \int_S F \cdot dS = \int \int_S F \cdot n dS = \int \int_{R_{xy}} F \cdot \left( \frac{\partial f}{\partial x} \, i + \frac{\partial f}{\partial y} \, j - k \right) \, dx \, dy$
since $\displaystyle n = \frac{\frac{\partial f}{\partial x} \, i + \frac{\partial f}{\partial y} \, j - k}{\sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1}}$ and $\displaystyle dS = \sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1} \, dx \, dy$.
After making the required substitutions your flux integral becomes $\displaystyle \int \int_{R_{xy}} 2y - 2x + 1 \, dx \, dy$, where $\displaystyle R_{xy}$ is the region of the xy-plane defined by $\displaystyle x^2 + y^2 = 1$. I'd suggest switching to polar coordinates to get this double integral.
Note that the region $\displaystyle R_{xy}$ is found by substituting z = 1 into $\displaystyle z = \sqrt{x^2 + y^2}$.
Of course, you could always switch to cylindrical coordinates at the very start of the problem .......
mr. fantastic, I'd like to suggest something to improve your notation.
Instead of typing
Code:
[tex]\int \int_S F \cdot dS[/tex]
And getting $\displaystyle \int \int_S F \cdot dS$, you can type
Code:
[tex]\iint\limits_S F \cdot dS[/tex]
and get $\displaystyle \iint\limits_S F \cdot dS$, which looks neater.
--Chris