I'm really not sure how to set up the question or solve it, any help would be appreciated.
Thanks in Advance.
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 $\displaystyle x^2+ y^2\le 1$, z= 2
2) The bottom which is the disk $\displaystyle x^2+ y^2\le 1$, z= 0
3) The curved side $\displaystyle x^2+ y^2= 1$, $\displaystyle 0\le z\le 1$
Do you know how to find the integral of a vector function over a surface?
Do you know what Gauss's theorem is?
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.
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 $\displaystyle \vec{i}$ and $\displaystyle \vec{j}$. The cross product is $\displaystyle \vec{k}$. From that, the "outward (downward) vector differential of area" is $\displaystyle -\vec{k}dxdy$. On z= 0, the vector integrand is $\displaystyle x^2\vec{i}+ z\vec{k}= x^2\vec{i}$. So the integral is $\displaystyle \int\int (x^2\vec{i})\cdot(-\vec{k}dxdy)=0$
On the top, z= 1, the same thing is true except that now the "outward vector differential of area" is upward, $\displaystyle \vec{k}dxdy$ and the vector integrand is $\displaystyle x^2\vec{i}+ 1\vec{k}$. The integral is $\displaystyle \int (x^2\vec{i}+ \vec{k})\cdot(\vec{k}dxy)= \int\int dxdy$.
On the curved surface I would suggest changing to polar coordinates: $\displaystyle x^2+ y^2= 1$ can be written as $\displaystyle cos^2(t)+ sin^2(t)= 1$ so take x= cos(t), y= sin(t).
Then every point, (x, y, z) is of the form (cos(t), sin(t), z) with $\displaystyle \vec{r}(t, z)= cos(t)\vec{i}+ sin(t)\vec{j}+ z\vec{k}$ 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 $\displaystyle \vec{r}_z= \vec{k}$. The cross product of those two vectors is $\displaystyle cos(t)\vec{i}+ sin(t)\vec{j}$ and the "differential of surface area" is $\displaystyle (cos(t)\vec{i}+ sin(x)\vec{j})dt dz$. With x= cos(t) the integrand is $\displaystyle cos^2(t)\vec{i}+ z\vec{k}$ so that the integral is $\displaystyle \int\int cos^3(t)dt dz$.
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",
$\displaystyle \int\oint \vec{f}\cdot d\vec{S}= \int\int\int \nabla\cdot \vec{f} dV$
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 $\displaystyle \int\int\int \left(\nabla\cdot (x^2\vec{ix}+ z\vec{k})dxdydz$.
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?