Help visualizing a surface.

[qaaq2]

Hello there. I am working on a problem in multivariable calculus. The question is the following:
Given the surface defined by S={(x,y,z) € R2/ 4z=1-(x^2 + y^2 ), z>x+1}
And a vector field F (which i wont write out because its quite long)

Calculate the flux of the vector field along the surface S, which is oriented such that the normal vector of S at point P(0,0,1/4) is (0,0,1).

I am having trouble visualizing this surface. I understand that is is the intersection of a paraboloid with a plane, but i am not quite able to see the surface generated to then see what surfaces i have to "cover up" when applying Gauss/ divergence theorem. Any help would be much appreciated. Thank you.

 

[Dr.Peterson]

Well, as you say, the surface is (part of) a paraboloid, which should not be hard to visualize in itself (particularly if you think in terms of cylindrical coordinates, since x^2 + y^2 = r^2).

If you are asking about the boundary where it is cut off by the plane, that's a little harder to visualize just by looking at the two equations (though I can see that the plane is higher than the vertex of the paraboloid). Just solve the equation [1 - (x^2 + y^2)]/4 = x + 1 to find the boundary. It is a familiar curve.

 

[qaaq2]

Thank you for the reply, Dr. Peterson.

I have eventually been able to see what surface the question is talking about, and have deduced the parametric equation of the circumference resulting from the intersection of plane z=x+1 with paraboloid 4z=1-r^2(cylindrical).

However, i have further issues when it comes to calculating the flux, which I could use help with as i haven't as of yet been able to solve this damn thing. My issue is the following: (NOTE: I apologize for not programming the proper notation but i am unfamiliar with the functions)

Referring to the surface S and vector field F mentioned above, the objective is to calculate the flux of F across S. This is given by the integral of the quantity F•N

Where N is the normal vector to the surface S.

As mentioned, the Surface S is given by all points inside the paraboloid that are above plane z=x+1, with intersection points at x=-1 and x=-3.

As my vector field is a pain in the rear, and it happens to handily be adivergent, Gauss theorem seems the obvious way to go. Because the divergence of the field is zero, is the flux zero?

What I am currently trying to do is to use the parametroc equations of the intersection of the plane and the paraboloid (which comes out to x(t)=(1/2cos t,1/2sint,0) if I am not mistaken) to evaluate a surface integral across the circumference of F•N where N is the normal vector of the surface, obtained by cross multiplying the tangent and the binormal vectors obtained with the derivatives of x(t). However, i am nowhere near being able to evaluating the resulting integral. I'm certain i have the right idea but not the correct application.

Alas, I am sorry for not being able to explain my dilemma any better. Perhaps this is quite a complex issue to solve over a forum. Anyways, i could always use some extra insight. Thank you again.






[MATH][/MATH]

 

[Dr.Peterson]

I think others will be better able to answer the big question, as I haven't done much vector calculus lately; but you should check your curve, because it doesn't look at all right. It should not be in the plane z=0, though it is a circle from the top view.

I don't know what you mean by "with intersection points at x=-1 and x=-3". The intersection of the surface and the plane is a curve, not two points. Also, the surface consists of points on the paraboloid, not "inside" it. These may just be matters of word choice.

As for the divergence theorem, this is not a closed surface. But you can apply it indirectly, which may be what you mean.

 

[qaaq2]

I think others will be better able to answer the big question, as I haven't done much vector calculus lately; but you should check your curve, because it doesn't look at all right. It should not be in the plane z=0, though it is a circle from the top view.

I don't know what you mean by "with intersection points at x=-1 and x=-3". The intersection of the surface and the plane is a curve, not two points. Also, the surface consists of points on the paraboloid, not "inside" it. These may just be matters of word choice.

As for the divergence theorem, this is not a closed surface. But you can apply it indirectly, which may be what you mean.

Yes, you are right. It makes very little sense that my parametric equations for the intersection would be in the plane z=0. When analyzing the intersection of plane with paraboloid (the equation you mention at the end of your first post) I arrive at the conclusion shown here.



I could use some help interpreting this. I understand that this is a "circular" curve, and my initial instinct was to use a simple substitution such that



Leaving a nice and simple circumference, easy to convert to a set of parametrics. However, I understand that this is not the case. This would be the circumference looked at from a top-down-view, but that top down view is not parallel to the Z axis. The question is then, how would i go about calculating the parametric equations of the deformed circumference that results from the intersection?

Finally, why is this not a closed surface? it would seem that what remains from this intersection should be closed, right? Where is it left open? If it is open, but easily closable, then that would be great because i can simply close it manually myself and subtract the flux across the surface i have imagined there to close it from the total integral, leaving me with the correct result. I believe my problem lies in the parametrization of the intersection curve.

Thank you again, Mr. Peterson.

 

[Cubist]

View attachment 19023

You are close, but you made a couple of mistakes on your final, boxed, equation. Why did the x/2 become negative and I can't make out the fraction on the RHS.

You can use this site (click) to visualize 3d curves. And to check a parametric equation you can use the Curve command. Probably best if you don't become too reliant on using this website because you won't have access in an exam!

 

[qaaq2]

[MATH][HR][/HR][/MATH]

You are close, but you made a couple of mistakes on your final, boxed, equation. Why did the x/2 become negative and I can't make out the fraction on the RHS.

You can use this site (click) to visualize 3d curves. And to check a parametric equation you can use the Curve command. Probably best if you don't become too reliant on using this website because you won't have access in an exam!

You are right, that should be (x/2+1)^2, i corrected that mistake shortly after. However, i am still unable to evaluate the integral.

Since the divergence of the vector field is zero, and the surface that i define as {S U surface limited by the curve whose parametric equations are the ones defined by the intersection of paraboloid with plane } is closed, (Assume F is class C1 vector), it follows that the flux of vector field F across the surface is equal to the integral of the divergence of the field extended to the volume limited by the surface (which is 0, because div(F)=0) minus the integral of F•N across the area limited by the intersection curve (boxed equation).

Having gotten this far, i think all i need to do is obtain the parametric equations for the surface whose area is that value, obtain the natural base of the surface by deriving its parametric equations in the form x(u,v) and then evaluating int([F,Xu,Xv] dudv).

Alternatively, i could also use said parametric equations to calculate the normal vector of the surface in the form N= Xu×Xv/|Xu×Xv|, or even thinking about it now since i have two implicit equations that define a curve, in the form F(xyz)=0 and G(xyz)=0, perhaps i could derive a tangent vector by cross multiplying the gradients of each of these functions, and then obtain the normal vector as a vector which is perpendicular to that one. I think i will try that next.

In any case, writing all of this out has helped me realize that my fundamental problem is that my goal is not to make parametric equations for that intersection, rather to make the parametric equations of a surface x(u,v) which is LIMITED by the curve defined by the intersection we have been talking about all along. I am unsure as to how to go about that, but i will fiddle around and see what i can find. Any input from you guys will help me out massively.

Thank you all for your kind replies.

 

[Dr.Peterson]

Picture a bowl turned upside-down. That is the region of the paraboloid that the question is about. It is open at the bottom. The very existence of a boundary curve (the edge of the bowl) is what makes it not a closed surface. I think at the end here you have realized that the edge should not be the focus of attention.

But the plane that cuts off that region forms a lid to the bowl. The paraboloid together with the plane form a closed surface, containing the space inside the "bowl". That is what you can apply the divergence theorem to. Possibly that is what you meant to say, but it is not what you said! This will allow you to integrate over the plane, rather than the paraboloid, and save a little (!) work.

By the way, I would have multiplied your equation by 4 to clear fractions, much earlier!

 

[qaaq2]

Yes, that is what i am meaning to say. I still for the life of me cannot figure out how to solve this question. Granted, it also has to do with the fact that the vector field is quite nasty and performing the actual integration is proving to be difficult. I think there is a certain observation that needs to be made, but i have yet to find it. That, or i just suck at vector calculus. Here is my progress thus far: (starting from the middle of this page)


To keep the drawings simple, i drew only the cross section on the xz plane and made inferences from there. Because Gauss theorem was the idea from the beginning, step 1 was determine the surface of that "Lid" of the bowl, which is a circumference on the XY plane (for a constant z).

Once that has been determined, define surface S1 such that is is the union of given surface S with the lid surface(T), in order to apply divergence theorem to the volume inside. As shown, the flux will therefore be equal to the negative integral of F•N dS over the surface of T.

Next step, calculate the parametric equations of surface T in order to calculate the normal vector needed to compute the integral (At the very bottom of the page). Because the circumference is contained inside the plane z=1+x (in other words, it resides in a plane of constant gradient) i have assigned the variable Z to take the parameter v. (Maybe this is not correct?)

Once i have that, i proceed to attempt to evaluate the flux integral through it's normal definition, establishing the cross vector of Xu×Xv to take the direction of the normal to the surface T:

As you can see, the resulting integral is not very friendly.

Can i say that the binormal vector of the boundary curve will have the direction of the normal vector of the surface, because it is contained inside a plane?. I calculated that vector to be something simple before. Maybe that is the key that unlocks it.

The function here is unusually complicated and drawn out, i believe that is saying that there has to be some key observation to simplify it all. I have been as of yet unable to find it.

Anyways, it'd be great to head your thoughts on this if you have any.




[MATH][MATH][SUB][/SUB][/MATH][/MATH]

 

[Dr.Peterson]

I haven't looked too closely, but I expected to see the normal of the surface that forms the "lid" ([MATH]N_T[/MATH]?)stated simply and early. Perhaps what you are not noticing is that, since it is a region of a plane, the normal vector is just the normal to that plane, which is a constant that you could determine before doing anything else.

I'm also concerned that you still seem to be focusing attention on the curve where the paraboloid and the plane intersect, rather than on the surfaces of which that curve is the boundary. I may just be misreading what you say.

It will help me read this if you define what S and T are. What you write under "Parameters of T" confuses me, as does "[MATH]S_1\cap S_2 \text{ forms } S[/MATH]".

 

[qaaq2]

If you still think I am concerned with the curve that defines the boundary of the surface, then i definitely believe this is my problem.

T is the name of the bottom surface, the Lid of the bowl. All points such that (x+2)^2 +y^2<_1

My intention is to create a set of parametric equations that describe the bottom surface only(T). That is what i have attempted to do there under "parametrics of T", but it doesnt seem like ive got it right. I have established that the divergence through the top part of the paraboloid is zero, i just need to calculate the flux across the surface T (lid), which will give me the negative value of the flux across the surface S according to divergence theorem.

The reason i feel like i have to use Gauss theorem is because evaluating the regular surface integral requires quite a few more calculations. Perhaps it is as you say, and since the normal vector is constant across that plane, i just need to calculate the normal vector of plane z=x-1 and then simply write F•N.

However, from theory i know that int(F•N)ds = int (F•(Xu×Xv)) du dv for any surface expressed in parametric equations in the form x=x(u,v), where the derivatives of this vector with respect to u and v, Xu and Xv, form the natural base of the surface.

Therefore, it still seems i would need to calculate the parametric equations that define the surface T. I think that's what i really need help with.

Thank you.

 

[Dr.Peterson]

Do you not know how to find the normal to a plane, given the equation? It's very easy.

Calculus III - Equations of Planes

In this section we will derive the vector and scalar equation of a plane. We also show how to write the equation of a plane from three points that lie in the plane.

tutorial.math.lamar.edu

As for the parametric equations, can you explain the work you did, so I can try to start with what you know? In my mind, you already have an equation for the plane, and just need to choose the limits of integration, in whatever coordinate system you choose. (I'd start with the ordinary Cartesian coordinates, until I saw a reason to do otherwise. That is, it would have the form (u, v, f(u,v)), where z = f(x, y).)

 

[qaaq2]

Eureka! It seems it is as we both expected, and the trick has to do with the constant normal vector to the plane. I was blind but now i can see, and yes in the end it is an easy problem once you have realized this. I wrote up a cleaner solution:



I do know how to calculate (in this case, just see) the normal vector of a plane (as just the gradient of the implicit equation that defines it). However, I was contemplating that very same vector in an unnecessarily complicated way. I was visualizing the normal vector to the plane as the binormal vector to the curve which makes up its boundary.

I know the focus should not be here, on this curve. That was the root of all the previous complications, but my argument was that the binormal vector of that curve will have the same direction as the normal vector to the lid T, and since that lid is contained inside a plane, the normal vector will be constant and equal to the normal vector of that plane. However, this vector is immediately seen from the implicit equation of the plane, removing the problem of calculating it entirely (which is why i needed the parametric equations in the first place). Silly of me to not see this, but it happens.

Once that vector is seen (1,0,-1), then the dot product with that nasty vector field cancels out leaving a very simple integral, which i transferred to polar coordinates to evaluate even more simply.

So, mystery solved. Your input really helped me out, Dr. Peterson. Thank you kindly for your time.

Sincerely
-
Alex.

 

[Dr.Peterson]

Yes, you've got it. And I imagine the struggle was worth it.

Lesson: Look for the easy way first, rather than being confined to the "official" way. Math is about making things easier, not just following rules.

 

[qaaq2]

Yes. You are very right, it seems following theory in this case just gets you killed. I think this problem has just been made to showcase the utility of Gauss theorem in certain cases, given that if i were to evaluate the integral normally, is practically impossible to integrate the function that would result from the dot product of the vector field with the normal to the paraboloidal surface.


I did at least reckognize that Gauss was the way to go here, and it makes this problem not only doable but quite easy once you realize that you already have the normal vector that you need.

I must say I am enjoying this course in vector calculus. Hopefully I will pass my exams soon! Thanks for the help.



[MATH][MATH][/MATH][/MATH]