sharks said:
OK, i think it's simpler if i just show my work.
Well, that's what you should do from the start!
(a) ##z=4-x^2-y^2## gives ##\phi (x,y,z)=z-4+x^2+y^2## as the surface is oriented upwards, so z has to be positive.
##∇\phi = 2x\hat i +2y\hat j + \hat k## and ##|∇\phi| = \sqrt{4x^2 i +4y^2 + 1}##
There should be no "i" inside that square root, of course.
##\hat n = 1/ \sqrt{4x^2 i +4y^2 + 1}.(2x\hat i +2y\hat j + \hat k)##
##\vec F= 4x \hat i+4y \hat j +3 \hat k##
$$\iint_S \vec F.\hat n\,.dS=\iint_S (4x \hat i+4y \hat j +3 \hat k).\frac{1}{ \sqrt{4x^2 i +4y^2 + 1}}.(2x\hat i +2y\hat j + \hat k) \,.dS$$
Projecting the surface S onto the xy-plane:
##z_x=2x## and ##z_y=2y##
$$\iint_S \vec F.\hat n \,.dS= \iint_R (4x \hat i+4y \hat j +3 \hat k).\frac{1}{ \sqrt{4x^2 i +4y^2 + 1}}.(2x\hat i +2y\hat j + \hat k). \sqrt{4x^2+4y^2+1}\,.dxdy
\\=\iint_R (8x^2+8y^2+3)\,.dxdy$$Parametrization of the region R:$$
Okay, all of that's correct but I really hate that way of representing the differential of surface area! Did you notice that you had to calculate the length of \nabla \phi
twice and then they canceled?
Simpler, in my opinion, is this. If a surface is given by the "position vector" \vec{r}(s, t)= x(s,t)\vec{i}+ y(s,t)\vec{j}+ z(s,t)\vec{k}, with parameters s and t, then the partial derifvatives, \vec{r}_s= x_s\vec{i}+ y_s\vec{j}+ z_s\vec{k} and \vec{r}_t= x_t\vec{i}+ y_t\vec{j}+ z_s\vec{k} are vectors in the tangent plane to the surface at every point. Their cross product, \vec{r}_s\times\vec{r}_t, is normal to the surface and gives the "vector differential of surface area": \vec{n} dS= (\vec{r}_s\times\vec{r}_t)dsdt. (And, by the say, the "differential of surface area is given by the length of that vector: dS= |\vec{r}_s\times\vec{r}_t| dsdt.
In this case, z= 4- x^2- y^2 so the "position vector" of any point on the surface, using x and y as parameters, is \vec{r}= x\vec{i}+ y\vec{j}+ (4- x^2- y^2)\vec{k}. So \vec{r}_x= \vec{i}- 2x\vec{k} and \vec{r}_t= \vec{j}- 2y\vec{k}. Their cross product is 2x\vec{i}+ 2y\vec{j}+ \vec{k} and so d\vec{S}= (2x\vec{i}+ 2y\vec{j}+ \vec{k})dxdy which gives exactly what you have- but is much easier don't you think?
\pi}_0 \int^2_0 (8r^2+3)\,.rdrd\theta=76\pi$$
(b) Using Stoke's theorem: $$\iint_S ∇\times \vec F. \hat n .dS$$ but first calculating the curl of F. I get the answer 0. This is a bit surprising as it voids any other calculations involving the product with ##\hat n## and subsequently calculating the surface area, so here is my work:
That's true, of course, but it does NOT use "Stoke's theorem". That is a direct calculation (which happens to be remarkably easy).
Stoke's theorem says that
\int \nabla\times \vec{F}\cdot d\vec{S}= \oint \vec{F}\cdot d\vec{\sigma}
where the integral on the right is the the integral around the boundary of the surface, here, the circle x^2+ y^2= 4, z= 0.
We can take as parameterization x= 2cos(\theta), y= 2 sin(\theta), so that d\vec{\sigma}= dx\vec{i}+ dy\vec{j}+ 0\vec{k}= (-2sin(\theta)\vec{i}+ 2cos(\theta)\vec{j})d\theta and the integral becomes
\int_0^{2\pi} (8 cos(\theta)\vec{i}+ 8sin(\theta)\vec{j}+ 3\vec{k})\cdot (-2sin(\theta)\vec{i}+ 2 cos(\theta)\vec{j})d\theta
What is that?
$$∇\times \vec F=\begin{vmatrix}\hat i & \hat j & \hat k \\ \partial /\partial x & \partial /\partial y & \partial /\partial z \\ 4x & 4y & 3\end{vmatrix}=0$$
Therefore, $$\iint_S ∇\times \vec F. \hat n .dS=0$$
(c)$$\iiint_V div\, \vec F .dV$$
##div\, \vec F=4+4=8##
Describing the volume in terms of cylindrical coordinates:
For ##r## and ##θ## fixed, ##z## varies from ##z=0## to ##z=4-r^2##
For ##\theta## fixed, r varies from r=0 to r=2
θ varies from 0 to ##2\pi##
Therefore, $$\iiint_V div\, \vec F .dV=\int^{2\pi}_0 \int^2_0 \int^{4-r^2}_0 8\,.rdzdrd\theta=64\pi$$
Is my work correct?
