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Surface Area for Implicit & Explicit Surfaces // Vector Calculus
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- เผยแพร่เมื่อ 2 ธ.ค. 2020
- How can we compute the surface area of a surface? In the previous video (see vector calc playlist below) we derived the surface area when the surface was described parametrically. In this video we focus on when the surface is described explicitly and implicitly, and when those formulas work. The formula only workThe formula for the explicit case could be memorized, but it quickly follows from the implicit case so I usually don't bother. Note that I do skip the formal proof of the formula. The basic ingredients of that proof is to come up with a parameterization (via implicit function theorem) and then plug into the surface area formula for parameterized surfaces.
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Hey, I can see that your videos aren't getting much views compared to other youtubers. But the small set of audience you've earned here are probably crazy over your content. If I search for something and one of your videos make up into that list, I'm gonna go for that....no second thoughts
hey, I appreciate that!
I'm crushing this whole playlist in prep for my final and it is a life saver, thank you!
good luck on the final!
Very good. Love that ‘scaling factor’ explanation for the implicit version, very comprehendable for me!
Half way through it and yet again a beautiful lecture delivered. Weldone sir!!!!
Many thanks!
Your voice literally tells how committed you are to your work 😊👍🏻
just another casually incredible video. Keep em comin!
Thanks! Will do!
Doc, you're awesome. Your videos are 3blue1brown and organic chemistry tutor level.
Really beautiful explanation
Great explanations
Towards my finals and this is super duper helpful
Thank you!!
Very 9iC Respected Sir♥️
Respect ❤️ from India sir!!
I like your videos a lot. The examples are great, but i'd like to see you give proofs for all the formulas that you use. For me and I guess for many more, the proofs are the most important thing.
Can simebody help me please? I don't understand why the Gradient is orthigonal to the surface we are describing. Isn't the Gradient a Vektor in the(in this case) x-y plane, pointing into the direktion with the biggest slope? If thats right, it would not be normal to our surface.
I would be habby about explanations.
Because the surface here is defined by the function F(x,y,z) = 0 and we are taking the gradient of this F function. I don't really understand why he failed to define any function here. It really confuses people. Normally I define the z coordinate as function of x,y instead for these kinds of surfaces so if I take the gradient of this function it wouldn't make any sense.
Sir it would be fantastic if u even discuss some problems too..tnqs for the video
So this can be applied when the corners and sides of the surface lines up with the corners and sides of one of the three planes (and no area where the denominator is 0). Pretty cool. Question to self, why is it exactly that formula? I remember you saying it's a stretching factor. nabla f is the gradient to the surface, meaning how much it's changing in that specific point. Call it rate of change. Then the denominator would be rate of change in the p_hat direction. Dot product of u * v = |u| * |v| * cos(theta), whereas theta is the angle between them. Meaning the resulting would be 1/(p_hat * cos(theta)). Unit would be area per p_hat in the direction of the gradient on the plane. So we're essentially left with the area times how much the normal of the plane changes in relation to our p_hat. The area could only grow in that case, considering it would be the least at cos(theta) = 1 (where theta = 0) and greatest as theta gets small (as theta -> pi/2), aka, horizontal to the plane, just as you said.
Thank you for the video.
Why is nabla F dot k equal to -1 ? at 7:24
Is a cylinder an implicit surface? the gradient of g(x,y,z) = x^2+y^2 has gradient where the gradient dotted with k is 0
Trevor, if you have a surface that has minima won't that give regions where the gradient vector passes through zero? Does that mean this method can't be used in these instances? Peter
Why subtitles in indonesian language?
I don't understand eccent sometimes
Professor, shouldn't we have written dxdy instead of dA as we have already considered a double integral. We should have written dA if we had considered a single integral right? But thank you for a superb video anyways.
Both dA and dxdy mean area, it is just in the latter we have made a choice of coordinates to integrate with respect to and in the former we haven’t yet
@@DrTrefor Got it. Thanks prof!
Do you have a reference for the derivation of the implicit formula?
Thanks for such a good Material , love you sir
I have a question in the proof of Gauss law in all textbooks they prove it by integral of electric field of a charge on the area of spherical surface and the result will be of course, that integral [E.n.dA]=q/€ so why this result genarlized on all surface areas however they are sometimes not be spherical?
At 4:30 I don't understand how that flat surface can have ANY gradient. It's only defined in terms of x, y. The x and y components of the gradient are both 0. And the z-component doesn't even exist. Any help here is appreciated.
Why is the gradient vector normal to the curve?
Go search the derivation of tangent plane formula, that will help you
Is the normal vector in the XY plane a UNIT normal vector? Otherwise, why would gradient F dotted with the normal vector in the xy-plane = -1?
great vidoe, thanks professor!
You are welcome!
Thank you so much!
If the denominator a.k.a the squising factor is one ,then how on earth integrating gradient over the region R gives the surface ,can you give me some intuition ?
|∇F| is the area of the target surface while |∇F•k| is the area of the integration surface
Is this finding the surface area of the yellow surface or the white surface?
Grazie.
Can p be just any unit vector?
Exactly. And that projects into the plane defined by that normal.
Dr. Bazett, I really hope it's fine that I write these reflective comments. Let me know if they're not and I'll stop (y)
I enjoy them, and hope others will too:)
Love to hear that!
@@j.o.5957 Yeah they're great.
Woww