- #1

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## 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!

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- Thread starter Mathn00b!
- Start date

- #1

- 9

- 0

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!

- #2

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- #3

STEMucator

Homework Helper

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

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.

- #4

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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.

- #5

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Got confused..

g = x^{2} + y^{2} + z -1 = 0

n = [itex]\frac{ \nabla g}{|\nabla g|}[/itex]

n =[itex] \frac{ 2x+2y+1}{\sqrt{2x^2+2y^2+1^2}}[/itex]

and dS =[itex]\sqrt{2x^2+2y^2+1^2}[/itex] dxdy

?

g = x

n = [itex]\frac{ \nabla g}{|\nabla g|}[/itex]

n =[itex] \frac{ 2x+2y+1}{\sqrt{2x^2+2y^2+1^2}}[/itex]

and dS =[itex]\sqrt{2x^2+2y^2+1^2}[/itex] dxdy

?

Last edited:

- #6

HallsofIvy

Science Advisor

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This last is wrong. n is ag = x^{2}+ y^{2}+ z -1 = 0

n = [itex]\frac{ \nabla g}{|\nabla g|}[/itex]

n =[itex] \frac{ 2x+2y+1}{\sqrt{2x^{2}+2y^{2}+ 1^{2}}}[/itex]

- #7

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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-x^{2}-y^{2}

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-x

Last edited:

- #8

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

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 [itex]f(x,y) = 1-x^2-y^2 \mbox{ for } x^2+y^2 <= 1[/itex], so use the graph parametrization

[itex] X(x,y) = (y,x, 1 - x^2 - y^2).[/itex]

We shall use the following (very important!) general formula for the upward normal of a surface defined by an equation of the form [itex] z = f(x,y):[/itex]

[itex]N(x,y)=(-f1, -f2,1)[/itex]

For the surface of interest here, it follows that:

[itex]N(x,y) = (-(-2x),-(-2y),1) [/itex]

And we have [itex]F(X(x,y)) = (y,x,1-x^2-y^2)[/itex]

So [itex]F \cdot N = 2xy+2xy+1-x^2 - y^2 [/itex]

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

- #9

STEMucator

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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.

- #10

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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 )##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.

Otherwise your should be the correct one.

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