# Thread: surface integral help, been stuck for hours

1. ## surface integral help, been stuck for hours

Here is the problem:

Here is my progress:
I have solved the unit normal vector n to be $n = \frac{xi + zk}{3}$ and A dotted with n to be $\frac{5xz}{3}$

So I get to the point where:
$\int{\int{(A dotted with n)dS}} = \frac{1}{3}\int{\int{5xzdS}}$
Now I do not know how to proceed from here. I don't know how to define my upper and lower limits and what dS should be equal to. I have tried to read the book(Schaum's Vector Analysis) but still I cannot seem to get how to approach this, since there has been no example of a horizontally inclined cylinder, only of a vertical one. I have been able to solve a similar problem to this but the cylinder was oriented vertically..
I believe the cylinder in question here has a radius of 3 and runs along the y-axis starting from 0 and ending at 8, and it only exists in the first octant because of its boundaries. Please please help

2. ## Re: surface integral help, been stuck for hours

The unit normal is given by

$\hat{n}=\dfrac{1}{\sqrt{x^2+z^2}}(x,0,z)$

$A=(6z,2x+y,-x)$

$A\cdot \hat{n}=\dfrac{1}{\sqrt{x^2+z^2}}(6 x z,0,-x z)=\dfrac{x z}{\sqrt{x^2+z^2}}(6,0,-1)$

you can probably finish from here.

3. ## Re: surface integral help, been stuck for hours

Originally Posted by romsek
The unit normal is given by

$\hat{n}=\dfrac{1}{\sqrt{x^2+z^2}}(x,0,z)$

$A=(6z,2x+y,-x)$

$A\cdot \hat{n}=\dfrac{1}{\sqrt{x^2+z^2}}(6 x z,0,-x z)=\dfrac{x z}{\sqrt{x^2+z^2}}(6,0,-1)$

you can probably finish from here.
Can i turn that last part from $\dfrac{x z}{\sqrt{x^2+z^2}}(6,0,-1)$ into $\frac{6xz-xz}{\sqrt{x^2+z^2}} = \frac{5xz}{3}$? Like above? and if yes, my issue is actually what to do after this

4. ## Re: surface integral help, been stuck for hours

Originally Posted by catenary
Can i turn that last part from $\dfrac{x z}{\sqrt{x^2+z^2}}(6,0,-1)$ into $\frac{6xz-xz}{\sqrt{x^2+z^2}} = \frac{5xz}{3}$? Like above? and if yes, my issue is actually what to do after this
oops, my mistake, did this before coffee.

$A\cdot \hat{n}$ is of course a scalar, not a vector and is equal to

$A \cdot \hat{n} = \dfrac{1}{\sqrt{x^2+z^2}}(6xz + 0 -xz)=\dfrac{5xz}{\sqrt{x^2+z^2}}$

as you noted $\sqrt{x^2+z^2}=3$ along the surface so this becomes

$A\cdot \hat{n}=\dfrac{5xz}{3}$

From here just do the integral.

$8\displaystyle{\int_0^3}\dfrac{5x\sqrt{9-x^2}}{3}~dx$

the $8$ is from integrating over $y$

5. ## Re: surface integral help, been stuck for hours

Originally Posted by romsek
oops, my mistake, did this before coffee.

$A\cdot \hat{n}$ is of course a scalar, not a vector and is equal to

$A \cdot \hat{n} = \dfrac{1}{\sqrt{x^2+z^2}}(6xz + 0 -xz)=\dfrac{5xz}{\sqrt{x^2+z^2}}$

as you noted $\sqrt{x^2+z^2}=3$ along the surface so this becomes

$A\cdot \hat{n}=\dfrac{5xz}{3}$

From here just do the integral.

$8\displaystyle{\int_0^3}\dfrac{5x\sqrt{9-x^2}}{3}~dx$

the $8$ is from integrating over $y$
Okay thanks romsek... although I know how to do that integral, what I am trying to find insight about is knowing what integral to execute. I don't know how we can arrive at that last step. Was that actually ${\int_0^8}{\int_0^3}\dfrac{5x\sqrt{9-x^2}}{3}~dxdy$ ? What I am wondering is:
1. how did we know to substitute the z from the equation of the surface
2. how did we know to use dxdy, and i believe this can be integrated in different ways as well, including polar forms, etc?
3. the 0 to 3 and then 0 to 8 as the limits is because the surface has been projected downwards onto the xy plane correct? but if the area covered on the xy plane was not a rectangle then the limits would not all be constants right?

I have tried doing some double integrals in integral calculus but I am still trying to get the essential difference between doing it with vectors, also although I have done some double integrals, it just seems so vast that everytime I seem to learn something about it I forget the whole thing or rather I forget how I analyzed it so I have to keep on cracking my head over this stuff.

Oh my, just now I calculated the integral you gave and it gives me 120 which is not the answer in the book which is 18pi
I tried to divide (A dotted with n) by (n dotted with k) but still no luck because i get 180. So I guess the problem isn't solved yet.. I have no idea what is wrong with this..

6. ## Re: surface integral help, been stuck for hours

Originally Posted by catenary
Okay thanks romsek... although I know how to do that integral, what I am trying to find insight about is knowing what integral to execute. I don't know how we can arrive at that last step. Was that actually ${\int_0^8}{\int_0^3}\dfrac{5x\sqrt{9-x^2}}{3}~dxdy$ ? What I am wondering is:
1. how did we know to substitute the z from the equation of the surface
we're integrating along the surface. Those points are where you evaluate the normal vector and the field A.
2. how did we know to use dxdy, and i believe this can be integrated in different ways as well, including polar forms, etc?
Again that's the surface you're integrating over. It looks like a circle in x and z, and it's constant in y, i.e. it's a cylinder aligned along the y axis. Also you're only integrating a quarter of this cylinder based on the x and z limits.

You could indeed have used cylindrical coordinates, it probably would be beneficial to.
3. the 0 to 3 and then 0 to 8 as the limits is because the surface has been projected downwards onto the xy plane correct? but if the area covered on the xy plane was not a rectangle then the limits would not all be constants right?
I guess the short answer is yes.
I have tried doing some double integrals in integral calculus but I am still trying to get the essential difference between doing it with vectors, also although I have done some double integrals, it just seems so vast that everytime I seem to learn something about it I forget the whole thing or rather I forget how I analyzed it so I have to keep on cracking my head over this stuff.
Can you do just plain surface integrals? I.e. the area of a surface in space? It sounds like you need some practice with those first.

7. ## Re: surface integral help, been stuck for hours

@romsek The value of the definite integral you gave me results in 120. the answer in the book is 18pi. help..?

8. ## Re: surface integral help, been stuck for hours

Originally Posted by catenary
@romsek The value of the definite integral you gave me results in 120. the answer in the book is 18pi. help..?
I'm digging back into this now. I've screwed this all up pretty well. I'll post when I get the right answer. Sorry.

9. ## Re: surface integral help, been stuck for hours

I tried to divide (A dotted with n) by (n dotted with k) before integrating but still no luck because i get 180. So I guess the problem isn't solved yet.. I have no idea what is wrong with this.. I have looked at all the examples in the book...
I suppose it should be possible to solve this and get the same answer using either cartesian or cylindrical coordinates..
I have to sleep, it is late here. I won't be checking, or solving for a few hours

10. ## Re: surface integral help, been stuck for hours

Yes, the book says that if i replace dS using cartEsian coords it has to be with $\frac{dxdy}{ndotk}$
Which in the end gives me a final result of 180. Possibility that the answer on the book is a typo? It's a really old print. Can anyone help verify thanks

11. ## Re: surface integral help, been stuck for hours

Originally Posted by catenary
Yes, the book says that if i replace dS using cartEsian coords it has to be with $\frac{dxdy}{ndotk}$
Which in the end gives me a final result of 180. Possibility that the answer on the book is a typo? It's a really old print. Can anyone help verify thanks
Ok, I think we've been going about this wrong. Applying the divergence theorem gives a quick and easy result.

$\oint A\cdot \hat{n}=\int_V \nabla \cdot A~dV$

$\nabla\cdot A = 0+1+0=1$

and thus we can just find the volume of the quarter cylinder, i.e.

$V_{cyl}=8\dfrac {\pi (3)^2}{4}=18\pi$

and multiply it by 1

$\oint A\cdot \hat{n}=1\cdot 18\pi = 18\pi$

Otherwise we have to do surface integrals on 5 surfaces and add them all up, which while doable is a lot more work than above.

Are you allowed to apply the Divergence theorem yet?

12. ## Re: surface integral help, been stuck for hours

We haven't discussed volume integrals yet in our vector analysis class, which means maybe I was really expected to intuit that I need to get 5 surfaces and then add them? But isn't it the problem states that I have to find ${\int}{\int_S}A\cdot \hat{n} dS$... which indicates a surface integral? By the way thanks for sticking with me on this.. Yeah we have already discussed divergence, gradient, and curl
I know that the divergence of A is nabla dot A but I don't know what it has to do with volume, I always thought the value of divergence just meant that the particles are moving away from or towards a certain point. I do know how nabla dot A is calculated and why that became 0 + 1 + 0 = 1. Also I know how you got the volume of the quarter cylinder it was (pi*(r^2)*h)/4. But why would multiplying the volume by the divergence give you the value of the surface integral? and why is everything above in our previous posts wrong and this one correct?

13. ## Re: surface integral help, been stuck for hours

Originally Posted by catenary
We haven't discussed volume integrals yet in our vector analysis class, which means maybe I was really expected to intuit that I need to get 5 surfaces and then add them? But isn't it the problem states that I have to find ${\int}{\int_S}A\cdot \hat{n} dS$... which indicates a surface integral? By the way thanks for sticking with me on this.. Yeah we have already discussed divergence, gradient, and curl
I know that the divergence of A is nabla dot A but I don't know what it has to do with volume, I always thought the value of divergence just meant that the particles are moving away from or towards a certain point. I do know how nabla dot A is calculated and why that became 0 + 1 + 0 = 1. Also I know how you got the volume of the quarter cylinder it was (pi*(r^2)*h)/4. But why would multiplying the volume by the divergence give you the value of the surface integral? and why is everything above in our previous posts wrong and this one correct?
Basically what you are asking is why the Divergence Theorem works. There are plenty of online and other sources that can explain that better than I can.

The reason that this flux integral is so tightly coupled to the volume here is because the divergence of your field is a constant. If this weren't the case you would have contribution from the field divergence more than just a multiplicative constant.

As for intuiting 5 surface integrals, well... , the problem did say the total flux, and there are 5 surfaces, so of course you need to look at 5 surface integrals.

As far as why we were previously incorrect it's mainly because we didn't consider all 5 surfaces. I may have also had the formula for the surface integral wrong as well as I don't think I properly accounted for $dS$.

It would be an instructive exercise to go ahead and compute the flux integral for all 5 surfaces and verify that you get the same answer as when you use the divergence theorem.

14. ## Re: surface integral help, been stuck for hours

As for intuiting 5 surface integrals, well... , the problem did say the total flux, and there are 5 surfaces, so of course you need to look at 5 surface integrals.
'divergence' and 'divergence theorem' are two different things?
the word flux has never been uttered in my vectors class yet... anyway ive got class tomorrow so i can update here if any new insight. really thanks a lot man.

15. ## Re: surface integral help, been stuck for hours

Originally Posted by catenary
the word flux has never been uttered in my vectors class yet...
It's another word for the "amount" of a field that crosses a surface.

A surface integral of a field is called a flux integral.

Page 1 of 2 12 Last