Finding the flux of a fluid on a given interval.

[burt]

I've been given this question, and I'm just feeling very stuck. I've tried different things and haven't really gotten anywhere yet. If there is anybody who is able to help me get started, I would really appreciate that.
I realize I am not posting my steps - it is because I haven't gotten anywhere even after much time spent on the problem. However, I need to know how to solve the problem, so I'm trying to learn how. I'm not just looking for an answer, I'm trying to understand the process so I can do it myself.

Compute \(\iint_{\partial Q}F\bullet n\ dS\) using the easiest method available. \(Q\) is bounded by \(z=4-x^2-y^2,\ z=1\), and \(z=0\). \(F=<ze^{\sqrt{x^2+y^2+z^2}},(x^2+y^2+z-4)z\sin(y),-2xze^{\sqrt{x^2+y^2+z^2}}>\)

Any help is appreciated. I do see a pattern - the value given for z has the value given for x in it. In general, the terms do seem to be related, etc. But, I have no clue about what to do with this pattern.

 

[HallsofIvy]

The key to integrating over a surface is that "n dS". n is, of course, the unit (length 1) vector perpendicular to the surface at every point (it might be better written \(\displaystyle \vec{n}\)) while dS is the "differential of surface area". I prefer to write "\(\displaystyle d\vec{S}\)" as the vector that is perpendicular to the surface at every point and whose length is dS. It is, of course, the same thing.

Part of the surface (a paraboloid) is given as \(\displaystyle z= 4- x^2- y^2\). Any point on that surface has "position vector" \(\displaystyle x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (4- x^2- y^2)\vec{k}\). A vector tangent to that surface in the x- direction is \(\displaystyle \vec{i}- 2x\vec{k}\) and a vector tangent to that surface in the y direction is \(\displaystyle \vec{j}- 2y\vec{k}\). The vector ndS (or \(\displaystyle d\vec{S}\)) is the cross product of those:
\(\displaystyle \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & -2x \\ 0 & 1 & -2y \end{array}\right|= 2x\vec{i}+ 2y\vec{j}+ \vec{k}\).

As for \(\displaystyle \vec{F}= (ze^{\sqrt{x^2+ y^2+ z^2}})\vec{i}+ (x^2+ y^2+ z- 4)zsin(y)\vec{j}- (2xze^{x^2+ y^2+ z^2})\vec{k}\)
notice that, on this parabola, the \(\displaystyle \vec{j}\) component is 0! Further \(\displaystyle x^2+ y^2+ z^2= 4- z+ z^2\) so, on this parabola, \(\displaystyle \vec{F}= ze^{\sqrt{4- z+ z^2}}\vec{i}- (2xze^{\sqrt{4- z+ z^2}})\vec{k}\).

So \(\displaystyle \vec{F}\cdot \vec{n}dS= (2xze^{\sqrt{4- z+ z^2}}- 2xze^{\sqrt{4- z+ z^2}})dS= 0dS\)!
The integral over the paraboloid is 0!

That leaves the integral over the flat "top" z= 1. That intersects the parabola where \(\displaystyle z= 1= 4- x^2+ y^2\) or the circle \(\displaystyle x^2+ y^2= 3\) with radius \(\displaystyle \sqrt{3}\). The perpendicular to that is, of course, \(\displaystyle \vec{k}\) and for z= 1, the \(\displaystyle \vec{k}\) component of \(\displaystyle \vec{F}\) is \(\displaystyle -2xe^{x^2+ y^2}\). I I think I would be inclined to do that in polar coordinates. \(\displaystyle x= r cos(\theta)\) and \(\displaystyle x^2+ y^2= r^2\). The limits of integration are r from 0 to \(\displaystyle \sqrt{3}\) and \(\displaystyle \theta\) from 0 to \(\displaystyle 2\pi\). The integral is \(\displaystyle \int_0^{\sqrt{3}}\int_0^{2\pi} -2(r cos(\theta)e^{r^2} rd\theta dr= -2\int_0^{\sqrt{3}}\int_0^{2\pi} r^2e^{r^2}cos(\theta)d\theta dr\).

 

[burt]

notice that, on this parabola, the →jj→\displaystyle \vec{j} component is 0!

How is this so?

 

[burt]

As for →F=(ze√x2+y2+z2)→i+(x2+y2+z−4)zsin(y)→j−(2xzex2+y2+z2)→kF→=(zex2+y2+z2)i→+(x2+y2+z−4)zsin(y)j→−(2xzex2+y2+z2)k→\displaystyle \vec{F}= (ze^{\sqrt{x^2+ y^2+ z^2}})\vec{i}+ (x^2+ y^2+ z- 4)zsin(y)\vec{j}- (2xze^{x^2+ y^2+ z^2})\vec{k}
notice that, on this parabola, the →jj→\displaystyle \vec{j} component is 0! Further x2+y2+z2=4−z+z2x2+y2+z2=4−z+z2\displaystyle x^2+ y^2+ z^2= 4- z+ z^2 so, on this parabola, →F=ze√4−z+z2→i−(2xze√4−z+z2)→kF→=ze4−z+z2i→−(2xze4−z+z2)k→\displaystyle \vec{F}= ze^{\sqrt{4- z+ z^2}}\vec{i}- (2xze^{\sqrt{4- z+ z^2}})\vec{k}.

So →F⋅→ndS=(2xze√4−z+z2−2xze√4−z+z2)dS=0dSF→⋅n→dS=(2xze4−z+z2−2xze4−z+z2)dS=0dS\displaystyle \vec{F}\cdot \vec{n}dS= (2xze^{\sqrt{4- z+ z^2}}- 2xze^{\sqrt{4- z+ z^2}})dS= 0dS!

Thank you for your work to answer this! I really appreciate it and I'm starting to get it. But, I do not understand the steps you took in this section I just quoted.
Is it possible to explain?

 

[Deleted member 4993]

The key to integrating over a surface is that "n dS". n is, of course, the unit (length 1) vector perpendicular to the surface at every point (it might be better written \(\displaystyle \vec{n}\)) while dS is the "differential of surface area". I prefer to write "\(\displaystyle d\vec{S}\)" as the vector that is perpendicular to the surface at every point and whose length is dS. It is, of course, the same thing.

Part of the surface (a paraboloid) is given as \(\displaystyle z= 4- x^2- y^2\). Any point on that surface has "position vector" \(\displaystyle x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (4- x^2- y^2)\vec{k}\). A vector tangent to that surface in the x- direction is \(\displaystyle \vec{i}- 2x\vec{k}\) and a vector tangent to that surface in the y direction is \(\displaystyle \vec{j}- 2y\vec{k}\). The vector ndS (or \(\displaystyle d\vec{S}\)) is the cross product of those:
\(\displaystyle \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & -2x \\ 0 & 1 & -2y \end{array}\right|= 2x\vec{i}+ 2y\vec{j}+ \vec{k}\).

As for \(\displaystyle \vec{F}= (ze^{\sqrt{x^2+ y^2+ z^2}})\vec{i}+ (x^2+ y^2+ z- 4)zsin(y)\vec{j}- (2xze^{x^2+ y^2+ z^2})\vec{k}\)
notice that, on this parabola, the \(\displaystyle \vec{j}\) component is 0! Further \(\displaystyle x^2+ y^2+ z^2= 4- z+ z^2\) so, on this parabola, \(\displaystyle \vec{F}= ze^{\sqrt{4- z+ z^2}}\vec{i}- (2xze^{\sqrt{4- z+ z^2}})\vec{k}\).

So \(\displaystyle \vec{F}\cdot \vec{n}dS= (2xze^{\sqrt{4- z+ z^2}}- 2xze^{\sqrt{4- z+ z^2}})dS= 0dS\)!
The integral over the paraboloid is 0!

That leaves the integral over the flat "top" z= 1. That intersects the parabola where \(\displaystyle z= 1= 4- x^2+ y^2\) or the circle \(\displaystyle x^2+ y^2= 3\) with radius \(\displaystyle \sqrt{3}\). The perpendicular to that is, of course, \(\displaystyle \vec{k}\) and for z= 1, the \(\displaystyle \vec{k}\) component of \(\displaystyle \vec{F}\) is \(\displaystyle -2xe^{x^2+ y^2}\). I I think I would be inclined to do that in polar coordinates. \(\displaystyle x= r cos(\theta)\) and \(\displaystyle x^2+ y^2= r^2\). The limits of integration are r from 0 to \(\displaystyle \sqrt{3}\) and \(\displaystyle \theta\) from 0 to \(\displaystyle 2\pi\). The integral is \(\displaystyle \int_0^{\sqrt{3}}\int_0^{2\pi} -2(r cos(\theta)e^{r^2} rd\theta dr= -2\int_0^{\sqrt{3}}\int_0^{2\pi} r^2e^{r^2}cos(\theta)d\theta dr\).

Be careful about integrating 0 \(\displaystyle \to 2\pi\)