1. ## Surface Integral Help

I'm really not sure how to set up the question or solve it, any help would be appreciated.

2. ## Re: Surface Integral Help

You can set this up as three separate integrals over the three separate "smooth" parts of the cylinder:
1) The top which is the disk $x^2+ y^2\le 1$, z= 2
2) The bottom which is the disk $x^2+ y^2\le 1$, z= 0
3) The curved side $x^2+ y^2= 1$, $0\le z\le 1$
Do you know how to find the integral of a vector function over a surface?

Do you know what Gauss's theorem is?

3. ## Re: Surface Integral Help

Hi, thanks for the reply, no I don't know how to find the integral over a surface, or what gauss theorem is unfortunately.

(I'm working on learning it but I have to hand this in tomorrow)

4. ## Re: Surface Integral Help

Hand it in where? Are you saying you were given a problem involving the "Gauss's theorem" in a class where they have never mentioned Gauss's theorem? That would be very strange. Do you not have a Calculus text book?

5. ## Re: Surface Integral Help

No they have mentioned it, I'm just a little behind in class therefore I couldn't do the question.

I've tried to understand it from my book but I'm not getting it at the moment. I need more time, which I don't have for this particular question, hence asking for help here.

6. ## Re: Surface Integral Help

Okay, on the bottom, z= 0, any point is of the form (x, y, 0) which has "position vector" [tex]\vec{r}(x, y)= x\vec{i}+ y\vec{j}[tex]. Differentiating with respect to x and y, we get the "tangent" vectors $\vec{i}$ and $\vec{j}$. The cross product is $\vec{k}$. From that, the "outward (downward) vector differential of area" is $-\vec{k}dxdy$. On z= 0, the vector integrand is $x^2\vec{i}+ z\vec{k}= x^2\vec{i}$. So the integral is $\int\int (x^2\vec{i})\cdot(-\vec{k}dxdy)=0$

On the top, z= 1, the same thing is true except that now the "outward vector differential of area" is upward, $\vec{k}dxdy$ and the vector integrand is $x^2\vec{i}+ 1\vec{k}$. The integral is $\int (x^2\vec{i}+ \vec{k})\cdot(\vec{k}dxy)= \int\int dxdy$.

On the curved surface I would suggest changing to polar coordinates: $x^2+ y^2= 1$ can be written as $cos^2(t)+ sin^2(t)= 1$ so take x= cos(t), y= sin(t).
Then every point, (x, y, z) is of the form (cos(t), sin(t), z) with $\vec{r}(t, z)= cos(t)\vec{i}+ sin(t)\vec{j}+ z\vec{k}$ Again, differentiating with respect to the parameters, we get tangent vectors [tex]\vec{r}_t= -sin(t)\vec{i}+ cos(t)\vec{j}[tex] and $\vec{r}_z= \vec{k}$. The cross product of those two vectors is $cos(t)\vec{i}+ sin(t)\vec{j}$ and the "differential of surface area" is $(cos(t)\vec{i}+ sin(x)\vec{j})dt dz$. With x= cos(t) the integrand is $cos^2(t)\vec{i}+ z\vec{k}$ so that the integral is $\int\int cos^3(t)dt dz$.

You say you are "behind in class" but you also seem to be saying that you do not know any of this- and that concerns me. Surely your text book has a statement of "Gauss's theorem"? Even if it doesn't googling "Gauss's theorem", also known as the "divergence theorem",
$\int\oint \vec{f}\cdot d\vec{S}= \int\int\int \nabla\cdot \vec{f} dV$
That is, the integral of the vector function over the surface is the same as the integral of the divergence of the vector function over the interior.

Here that is $\int\int\int \left(\nabla\cdot (x^2\vec{ix}+ z\vec{k})dxdydz$.

7. ## Re: Surface Integral Help

For the final double integral involving cos^3, shouldn't there be an r^2? and then r = 1 so we are just left with the integral between 2pi and 0 of cos^3(t), which is zero?

8. ## Re: Surface Integral Help

Originally Posted by HallsofIvy
On the top, z= 1, the same thing is true except that now the "outward vector differential of area" is upward, $\vec{k}dxdy$ and the vector integrand is $x^2\vec{i}+ 1\vec{k}$. The integral is $\int (x^2\vec{i}+ \vec{k})\cdot(\vec{k}dxy)= \int\int dxdy$.
Sorry but isn't z = 2 on the top?

Also should it be dxdy not dxy?

$\int (x^2\vec{i}+ \vec{2k})\cdot(\vec{2k}dxdy)= \int\int 2dxdy$.

9. ## Re: Surface Integral Help

Yes. I was thinking it was z from 0 to 1 but it is from 0 to 2.

10. ## Re: Surface Integral Help

Originally Posted by HallsofIvy
Yes. I was thinking it was z from 0 to 1 but it is from 0 to 2.
Ok thanks, also when we are doing the divergence theorem, do we need to integrate in cylindrical polars?

Can't we just say that the integral of the first term is 0, since it's the integral of x over an object that's symmetrical to the reflection along the x axis.

Then that way we are just left with the integral of the volume of the cylinder, pir^2h which is just equal to 2pi?