1. ## Flux 2

I need to find the flux of this vector field (in the pic) that goes through this plan (in the pic) and z goes from 0 to 1.
How am I suppose to do that?

2. Originally Posted by asi123
I need to find the flux of this vector field (in the pic) that goes through this plan (in the pic) and z goes from 0 to 1.
How am I suppose to do that?
Do you know that when the surface is defined in the form $z = f(x, y)$ then:

$\int \int_S F \cdot dS = \int \int_S F \cdot n dS = \int \int_{R_{xy}} F \cdot \left( \frac{\partial f}{\partial x} \, i + \frac{\partial f}{\partial y} \, j - k \right) \, dx \, dy$

since $n = \frac{\frac{\partial f}{\partial x} \, i + \frac{\partial f}{\partial y} \, j - k}{\sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1}}$ and $dS = \sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1} \, dx \, dy$.

After making the required substitutions your flux integral becomes $\int \int_{R_{xy}} 2y - 2x + 1 \, dx \, dy$, where $R_{xy}$ is the region of the xy-plane defined by $x^2 + y^2 = 1$. I'd suggest switching to polar coordinates to get this double integral.

Note that the region $R_{xy}$ is found by substituting z = 1 into $z = \sqrt{x^2 + y^2}$.

Of course, you could always switch to cylindrical coordinates at the very start of the problem .......

3. Originally Posted by mr fantastic
Do you know that when the surface is defined in the form $z = f(x, y)$ then:

$\int \int_S F \cdot dS = \int \int_S F \cdot n dS = \int \int_{R_{xy}} F \cdot \left( \frac{\partial f}{\partial x} \, i + \frac{\partial f}{\partial y} \, j - k \right) \, dx \, dy$

since $n = \frac{\frac{\partial f}{\partial x} \, i + \frac{\partial f}{\partial y} \, j - k}{\sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1}}$ and $dS = \sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1} \, dx \, dy$.

After making the required substitutions your flux integral becomes $\int \int_{R_{xy}} 2y - 2x + 1 \, dx \, dy$, where $R_{xy}$ is the region of the xy-plane defined by $x^2 + y^2 = 1$. I'd suggest switching to polar coordinates to get this double integral.

Note that the region $R_{xy}$ is found by substituting z = 1 into $z = \sqrt{x^2 + y^2}$.

Of course, you could always switch to cylindrical coordinates at the very start of the problem .......
Check this out, is that what you meant?

4. Originally Posted by mr fantastic
Do you know that when the surface is defined in the form $z = f(x, y)$ then:

$\int \int_S F \cdot dS = \int \int_S F \cdot n dS = \int \int_{R_{xy}} F \cdot \left( \frac{\partial f}{\partial x} \, i + \frac{\partial f}{\partial y} \, j - k \right) \, dx \, dy$

since $n = \frac{\frac{\partial f}{\partial x} \, i + \frac{\partial f}{\partial y} \, j - k}{\sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1}}$ and $dS = \sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1} \, dx \, dy$.

After making the required substitutions your flux integral becomes $\int \int_{R_{xy}} 2y - 2x + 1 \, dx \, dy$, where $R_{xy}$ is the region of the xy-plane defined by $x^2 + y^2 = 1$. I'd suggest switching to polar coordinates to get this double integral.

Note that the region $R_{xy}$ is found by substituting z = 1 into $z = \sqrt{x^2 + y^2}$.

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.

Code:
$$\int \int_S F \cdot dS$$
And getting $\int \int_S F \cdot dS$, you can type

Code:
$$\iint\limits_S F \cdot dS$$
and get $\iint\limits_S F \cdot dS$, which looks neater.

--Chris

5. Originally Posted by asi123
Check this out, is that what you meant?
Looks fine. Do you know what your integral terminals will be?