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Apply Stokes theorem to prove that $\int_{c} ydx+zdy+xdz =-2\sqrt{2}\pi a^2$
Where C is the curve given by $x^2+y^2+z^2-2ax-2ay=0, x+y=2a$ ; and begins at the point (2a,0,0) and it goes first below the z plane.

My work

Figure

I move clockwise on the boundary of the sphere cut by the plane since i want the surface to be on the left side upon movement on the curve inline with stoke's theorem.

The normal on the curve is $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$

Curl of F $y\hat{i} + z\hat{j} + x\hat{k}$ is $-\hat{i} -\hat{j} -\hat{k}$

Now taking surface of the sphere as $S_1$ and surface of the great circle as $S_2$ we have using gauss divergence theorem

$$ \int_{S_1 + S_2} (\nabla \times F) \hat{n} dS = 0 \\ \int_{S_1} (\nabla \times F) \hat{n} dS = - \int_{S_2} (\nabla \times F) \hat{n} dS \\ \int_{S_1} (\nabla \times F) \hat{n} dS = - \int_{S_2} (-\hat{i} -\hat{j} -\hat{k}) * \frac{\hat{i} + \hat{j}}{\sqrt{2}} dS \\ \int_{S_1} (\nabla \times F) \hat{n} dS = \int_{S_2} \sqrt{2} dS \\ \int_{S_1} (\nabla \times F) \hat{n} dS = 2 \sqrt{2} \pi a^2 $$

The above calculation is off by a factor of - sign

Also I am unable to understand how does goes below the z axis affects the calculation.

J. W. Tanner
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dreamboat
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  • The minor problems first: by "$z$ plane" you presumably mean the $xy$-plane. By the "curve" to which $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$ is the normal you presumably mean the disk that is in the plane $x+y=2a$ and caps both hemispheres of $x^2+y^2+z^2-2ax-2ay=0,.$ Now my advice: Make your life simpler and study how the sign of the flux of $\nabla\times F$ through the unit disk around the origin and in the $xy$-plane is related to the orientation of the line integral of $F$ around the disk's boundary. – Kurt G. Jan 07 '24 at 21:30
  • I know that keeping the target surface on the left / imagining that the surface is a bottle cap and moving it towards the direction of outward normal gives the direction of the movement on curve which here is clockwise on the curve / disc enclosing this surface @KurtG. – dreamboat Jan 08 '24 at 05:12
  • If you know that, what is $S_2$ and why the minus sign in front of the third integral equation? – Kurt G. Jan 08 '24 at 05:37
  • Please see S2 here https://i.stack.imgur.com/zeVul.jpg – dreamboat Jan 08 '24 at 05:38
  • The minus sign is there because the total over S1 and S2 is zero. Since our aim is to find for S1 we have to take negative over S2 – dreamboat Jan 08 '24 at 05:39
  • The parametrization of your curve that starts at $(2a,0,0).$ and goes below the $xy$ plane does not look compatible with thst picture. This is the heart of the problem. – Kurt G. Jan 08 '24 at 05:57
  • Let us continue this discussion in chat. – dreamboat Jan 08 '24 at 06:01
  • Yesterday it did not go fast enough. Now we have an answer and you are silent. Any thoughts? – Kurt G. Jan 09 '24 at 17:16
  • My apologies @KurtG. I did not want to rush discussion after seeing your answer since i wanted to think a bit. Thank you for taking the time to add a detailed answer. Pls see my comment on your answer – dreamboat Jan 10 '24 at 04:36

1 Answers1

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Your surface $$ x^2+y^2+z^2-2ax-2ay=0 $$ is a sphere around $(a,a,0)$ with radius $\sqrt{2}\,a\,.$ The plane $$ x+y=2a $$ intersects that sphere in a great circle (of radius $\sqrt{2}\,a$). The unit normal to that plane is $$\tag{$\color{red}{red}$ arrow} \frac{\hat{i}+\hat{j}}{\sqrt{2}}\,. $$ The curve that starts at $(2a,0,0)$ and goes below the $xy$-plane is oriented counterclockwise when we look from outside down to the $\color{blue}{blue}$ disk $S_2$ that caps the hemispheres. It is correct that the flux of $\nabla \times F=(-1,-1,-1)$ through the hemisphere $S_1$ plus the $\color{green}{green}$ disk $S_2$ is zero but this seems to only confuse you. What you have to calculate is just (by Stokes, and without extra minus sign) $$ \int_{\partial S_2}F\cdot\,ds=\int_{S_2}(\nabla\times F)\cdot\frac{\hat{i}+\hat{j}}{\sqrt{2}}\,dS $$ because that has the right orientation indicated by the $\color{orange}{orange}$ counterclockwise arrows.

The labels in the picture assume $a=1\,:$

enter image description here

Kurt G.
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  • Thank you for this detailed explanation @kurt.

    The problem forces a direction to traverse on the curve for us, given the direction we have a normal (N1) by right hand rule and the normal given by gradient is the same as normal (N1) and hence we are in the positive orientation.

    Now if there was no direction enforced we would have to think of the direction of traversal on the curve opposite to what we are taking now since we would want to keep the spherical curve on the left, which would orient normal opposite to N1 giving us negative of current answer. Is this line of thought correct

     – dreamboat Jan 10 '24 at 04:35
  • @dreamboat Thanks for coming back. I cannot tell if your line of thought is correct because I don't understand it. Key is the right hand rule that links the normal with the orientation of the curve. The latter is given in OP ("goes below the $xy$ plane"). As I show in the answer this kills the extra and wrong minus sign you had in your attempt. – Kurt G. Jan 10 '24 at 08:46
  • If I remove and begins at the point (2a,0,0) and it goes first below the z plane. and add Assume surface S for stoke theorem is the green surface in the above picture, then I will have to take clockwise path on the curve correct.? – dreamboat Jan 10 '24 at 15:37
  • Let me put it this way: by the Stokes theorem the line integral around the counter clockwise path is identical to the integral of $\nabla \times F$ over any surface that has the same orientation as the blue disk. This applies to the green hemisphere with inward pointing normal, or to (not in the picture) the opposite hemisphere above the blue disk with outward pointing normal. The same fact can be formulated as follows: when we flip the normal of the green hemisphere to be outward pointing and add it to the surface integral of the opposite hemisphere above, then their sum ... – Kurt G. Jan 10 '24 at 17:18
  • will be zero (same value opposite sign). In short: neither the line integral, nor the surface integral will have automatically the correct sign unless you give it an orientation after knowing what you are doing. – Kurt G. Jan 10 '24 at 17:19
  • Small clarification: should have written "...that has the same orientation as the blue disk and the same boundary." – Kurt G. Jan 10 '24 at 17:28
  • Understood @Kurt. Thank you again for taking out the time to explain and solve my doubts. :) Also I wanted to ask you what software did you use to create the above diagram? – dreamboat Jan 12 '24 at 10:24
  • Great that we have one more mathematician who now understands this well. The graph was created with Desmos. Unlike the 2d version the 3d software does not seem to allow to put in labels. Then I copied the image into Paint and added the labels manually. – Kurt G. Jan 12 '24 at 10:32