Flux Integral Evaluation Problem: Check My Workings and Get Advice

[Mathn00b!]

Homework Statement

Hello,
So I've got to evaluate a flux integral, but I get 0. Could someone please, check my workings and give me advice.
Please find my workings and the problem attached here as an image.
Thank you very much!

 

[LCKurtz]

I will give you that your handwriting is pretty nice. Still it would be better to type your work. Is that third component of your flux vector a $z$ or a $2$? Some mathematical issues are: Is your normal directed upward like it is supposed to be? In fact, is it even a vector? That's another reason to type your question, so we can edit it. Also you can't say $x^2+y^2=1$ for the normal. That is only true on the boundary of the surface but not on the surface itself. And check your partial derivatives.

 

[STEMucator]

You can parametrize the surface if you want, let:

$$\vec r(x, y) = x \hat i + y \hat j + z \hat k = x \hat i + y \hat j + (1 - x^2 - y^2) \hat k$$

Find $\vec r_x(x, y)$ and $\vec r_y(x, y)$. Use those to obtain $\vec r_x \times \vec r_y$. Now you have a theorem which states:

$$\iint_S \vec F(x, y, z) \cdot d \vec S = \iint_D \vec F(\vec r(x,y)) \cdot (\vec r_x \times \vec r_y) \space dA$$

Where the region $D$ is given by the plane $z = 1 - x^2 - y^2$ for $0 \leq z \leq 1$.

Plotting the plane for the points $(x, 0, 0), (0, y, 0)$, and $(0, 0, z)$ will give you the limits easily, and a nice graph to look at.

 

[LCKurtz]

You can parametrize the surface if you want, let:

$$\vec r(x, y) = x \hat i + y \hat j + z \hat k = x \hat i + y \hat j + (1 - x^2 - y^2) \hat k$$

Find $\vec r_x(x, y)$ and $\vec r_y(x, y)$. Use those to obtain $\vec r_x \times \vec r_y$. Now you have a theorem which states:

$$\iint_S \vec F(x, y, z) \cdot d \vec S = \color{red}{\pm}\iint_D \vec F(\vec r(x,y)) \cdot (\vec r_x \times \vec r_y) \space dA$$

Where the region $D$ is given by the plane $z = 1 - x^2 - y^2$ for $0 \leq z \leq 1$.

Plotting the plane for the points $(x, 0, 0), (0, y, 0)$, and $(0, 0, z)$ will give you the limits easily, and a nice graph to look at.

The $\pm$ sign is necessary because one has to decide whether $\vec r_x \times \vec r_y$ points in the required direction to agree with the given orientation. The OP didn't check that.

 

[Mathn00b]

Got confused..

g = x² + y² + z -1 = 0
n = \frac{ \nabla g}{|\nabla g|}
n =\frac{ 2x+2y+1}{\sqrt{2x^2+2y^2+1^2}}
and dS =\sqrt{2x^2+2y^2+1^2} dxdy
?

 

[HallsofIvy]

g = x² + y² + z -1 = 0
n = \frac{ \nabla g}{|\nabla g|}
n =\frac{ 2x+2y+1}{\sqrt{2x²+2y² + 1²}}

This last is wrong. n is a vector not a scalar. What is \nabla g?

 

[Mathn00b]

g(x,y,z) = z - f(x,y) = 0 according to my lecture notes.
I tried out the latex output, since the preview didn't work out. I've made other edits.
I am quite confused, I solved other questions using the method with g, but this one, don't know if it should be solved with rdrdθ or just dxdy . The confusing thing is that here I have z = 1-x²-y²

 

[gl0ck]

You can parametrize the surface if you want, let:

$$\vec r(x, y) = x \hat i + y \hat j + z \hat k = x \hat i + y \hat j + (1 - x^2 - y^2) \hat k$$

Find $\vec r_x(x, y)$ and $\vec r_y(x, y)$. Use those to obtain $\vec r_x \times \vec r_y$. Now you have a theorem which states:

$$\iint_S \vec F(x, y, z) \cdot d \vec S = \iint_D \vec F(\vec r(x,y)) \cdot (\vec r_x \times \vec r_y) \space dA$$

Where the region $D$ is given by the plane $z = 1 - x^2 - y^2$ for $0 \leq z \leq 1$.

Plotting the plane for the points $(x, 0, 0), (0, y, 0)$, and $(0, 0, z)$ will give you the limits easily, and a nice graph to look at.

Hello, I've got exam on the same things, so please give some feedback on my workings

This surface is the graph of the function f(x,y) = 1-x^2-y^2 \mbox{ for } x^2+y^2 <= 1, so use the graph parametrization
X(x,y) = (y,x, 1 - x^2 - y^2).
We shall use the following (very important!) general formula for the upward normal of a surface defined by an equation of the form z = f(x,y):
N(x,y)=(-f1, -f2,1)

For the surface of interest here, it follows that:
N(x,y) = (-(-2x),-(-2y),1)

And we have F(X(x,y)) = (y,x,1-x^2-y^2)

So F \cdot N = 2xy+2xy+1-x^2 - y^2

So we have \ \int_0^{2\pi} \int_0^1 (4cos( \theta )sin( \theta ) + 1 - r^2) rdrd \theta

 

[STEMucator]

Sorry for that a bit earlier where I mentioned $z = 1 - x^2 - y^2$ is a plane, it's a paraboloid.

As for the actual question, I worked it out to be:

Sorry for the squeeze near the bottom I was writing a bit too quickly and it got cut off.

 

[gl0ck]

Sorry for that a bit earlier where I mentioned $z = 1 - x^2 - y^2$ is a plane, it's a paraboloid.

As for the actual question, I worked it out to be:View attachment 83290

Sorry for the squeeze near the bottom I was writing a bit too quickly and it got cut off.

Yeah, I got the same, I just played it smart, but I think your solution is correct. I substituted r = x + y in the integral $1 - x^2 - y^2$ with $1 - ( x^2 + y^2)$ to get $1 - r^2$ and also for x and y since r = 1 I think I saw somewhere to replace $rcos( \theta )$ with $1cos( \theta )$
Otherwise your should be the correct one.