The Surface Area formula for Parametric Surfaces // Vector Calculus

In this video we derive the formula to compute surface area given some surface described parametrically. Thus if you have a parametric description, all you need to do is plug it into this formula. The derivation works by looking at a tiny section of surface area, and approximating this with a little tangential parallelogram whose area can be computed by the length of the cross product of r_u Delta u and r_v Delta b, the partial derivatives of the position vector with respect to the two parameters u and v. Thus the integral is effectively just summing up these little surface areas and becomes a double integral of the length of that cross product. We will see a concrete example of this in the next video in the vector calculus playlit.
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This got uploaded just in time for the corresponding section in my class. Thanks a bunch for these easy to understand videos

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[@DrTrefor]

Thank you!

[@bendavis2234]

Wow, I got that 'aha' moment when you introduced the connection between cross products and areas that I wasn't thinking about for some reason. These equations seem arbitrary at first, but inevitable at the end. Thanks for your helpful videos

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[@filomenamendes5349]

Thank you. This was incredibly helpful.

[@j.o.5957]

I'm assuming this makes sense for a curved area, since r_u X r_v gives direction, while du*dv gives magnitude. The direction would change as the area curves and we're fine. Question to self: what does it mean when the figure is in 2D, but lives in 3D? My guess is that you can have a flat sheet, like a piece of paper, that's curved in some way or another. The thing itself is 2D, yet it exists in 3D. Because it is 2D, we can find a parameterization that uses u and v (and not a third variable), while it still applies to 3d space since u and v are variables in f, g and h. I'm guessing that if we're doing this for something with a volume (like a stack of paper), we'd have to do computations in the 4th dimension with u, v and w.

[@wilurbean]

So, when we're talking about sheets in 3D or surfaces we only need two variables to describe it. We need 3 variables to describe a *volume*. What I mean is that there are unique X and Y points on the plane everywhere. No two points on the x-y plane can have the exact same value. You can't put in (1,1) and get to two different points on the plane. (1, 1) is unique! Yes that's trivial. So when we're looking at volumes or solids [ f(x, y, z) ], we can go to a unique x-y point, but that point has *many* z values which could be described by a function or set (ex 0 < z < 5 meaning z has every value between 0 and 5, and our solid would have a height of 5 at that location on x-y). So for that volume, we need an x and y to get to our point on the plane, and then another function of z to get the points above and below that (x, y). Z *must* be independent to have multiple values at x-y. On the other hand, if the z value is dependent on the x-y point, then any change in x-y changes z. That means that z is a function of an (x, y). Going to an x-y point on the plane generates inputs of x and y that will then go into the function that describes z's height. With this description there can only be a single (or maybe two if you get an x^2 with +/- values) value for z at a given x-y point. If you do this over an an entire region on the x-y plane, you would have a single z value over every x-y point, which forms a surface instead of a solid or volume. You could describe that "z" value as z(x, y), meaning z's value or "height" above the plane is derived from the x-y position, it is dependent. hope that helps

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[@DrTrefor]

That does sound frustrating, but I’m glad at least to have helped a little:)

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i'd definitely support infinite likes:D

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[@DrTrefor]

You’re most welcome!

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[@PriyanshuSingh-uo6dr]

okay so I have a doubt, since at 5:11 you are multiplying r_u and r_v with a scaling factor, but for scaling, shouldn't the vector be a unit vector? and so the cross product will also be unit vector and the magnitude will be 1 ..... it's just doesn't make sense

[@BLVGamingY]

just for visualization purposes, you would take a rectangle in the original 2d space with an area 1, you find its transformed area with the derivatives, then we pretend that the original area was infinitesimal by multiplying by dx and dy

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[@Rjsipad]

is this scaling factor different from the jacobian?

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[@sriraghavt]

Superb, nice refresher. One particular question that has always bothered me is how does the approximation of transformed infinitesimal surface to a parallelogram make sense. Let’s say the surface is contorted protein, will it’s surface infinitesimal also be a parallelogram?

[@DrTrefor]

It's similar to the idea of approximating a single variable function by a tangent line when you zoom in enough. Not EVERY curve works for this (nondiffrentiable ones fail), but if he curve or surface or whatever is nice enough, then it is ok.

[@sriraghavt]

@@DrTrefor Got it. Then a follow up - how do we define "nice" in case of surfaces. We known for single variable it should be differentiable, what is it for a surface?

[@SaiBekit]

@@sriraghavt IIRC, a surface is "nice" or smooth when each point on the surface has a unique tangent plane. Somebody correct me if im wrong, but i think that is equivalent to the point being differentiable.

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[@DrTrefor]

Indeed Fourier transform is coming, probably in about March

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[@DrTrefor]

Happy to help!

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[@moon200070]

Thank you for this video! I have a question about the part in 4:25. I always thought that the derivative with respect to the parameter pointed in the direction of the length with the respective parameter. Why is the partial of "u" in the direction of constant "v" and vice versa?

[@Oberbremser]

I think the following video may help: 67 - The geometric meaning of J kzbin.info/www/bejne/l5nZno2ObrypjLs quick jump to 13:00 - better watch the whole video. Ru is in the direction of constant v like the vector B1B2 in the mentioned video is in the direction of const v what I have understood.

[@hemantamandal6512]

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[@ems__c]

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[@ichigo_nyanko]

How could you generalise this to a, for example, solid? A cross product only exists in 3 dimensions, so is there something mroe general or fundemental here?

[@BREAD-kc4zp]

Is mod of ru cross rv kind of equivalent to jacobian matrix we use when we go from one coordinate system to other?

[@personxy7443]

sir,r_u times delta v dont cover yellow line?

[@bigbluewhale3957]

For the scale factor stretches mentioned at 5:13, why are the scale factors delta u and delta v? Is it to do with the fact that along each of those lines u and v are constant (respectively)?

[@therealnoofle5330]

One year late, but I struggled with it too for a while until I got it so hopefully this helps someone out. Think of a much simpler, less abstract surface like an elliptical paraboloid z=x²+y². For any given point, you can come up with a vector rx and ry such that these two vectors are along the tangent plane of the surface. To give a better mental picture, if you imagine these two vectors stretching out into infinity, the area between these two vectors would essentially be the tangent plane itself. This scaling of the vectors can be described by some coefficient. In the x direction we can describe it with ∆x and in the y direction we can do it with ∆y. Now instead of stretching these vectors out into infinity, imagine them shrinking into an infinitesimally small square. That's essentially your Riemann square to integrate the whole parabola. So now you can imagine u and v being transformation of the xy-plane which will end up distorting the surface, but the basic idea is the same.

[@reardelt]

So there is a difference between surface area and surface integral right? Bdcause the surface integral also has the 4th dimension right?

[@DrTrefor]

Indeed. A surface integral of the function f=1 gives the surface area.

[@fisherofmen489]

whats the point of having scaling factors?

[@camilmoujaber4813]

I need a formula for the surface area of a hyperboloid of height h, between the planes z=1/2h and z=-1/2h. I have calculated this area using numerical integration by summing the lateral area of truncated cones, but I need a formula for the area, if it exists. Can anyone help?

[@KM-om1hm]

Can anyone explain why is that double integral there ? Doesn't double integral means volume under the curve? I'm 11th grader so help me

[@Kappa647]

Its kind of a silly question but.. at 5:05 trefor says quite a bit shorter or longer than.. i cant make out those words.. can someone tell me??

[@jurgenbauer2322]

I think he is saying "left hand edge"

[@Kappa647]

@@jurgenbauer2322 thanks bro

[@hentai6582]

You're saving lives. Thanks man.

[@jacobharris3002]

Would it be possible to use Stokes Theorem to convert surface integrals like this to line integrals? I feel like for some surfaces in three dimensions calculating the line integral would be easier but I've been having a hard time figuring out what the vector field to use. The thing that makes it so difficult is that the surface integral for calculating area is scalar valued whereas the one in Stokes theorem is vector valued. So I somehow have to find a vector field such that 𝐅*𝐧=1.