Computing the Surface Area of a surface parametrically // Example 2 // Vector Calculus

We compute the surface area of a cylinder between two planes. We will use the parametric integral formula for surface area that we computed in the previous video in my vector calculus playlist (see below). The first step is to come up with a parameterization. Because of the cylinder it makes sense to use something based out of cylindrical coordinates. As this has a fixed radius, we use theta and z as our parameters.

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[augustodantedecarlizolet]

I'll tell all my friends here at Unicamp about you, and my math teachers as well, because they have a lot to learn with you too.
Share knowlegd is priceless, and youre doing it better then everyone. Really hope you know how wonderfull it is.
You're making the world better, thank you.

[Hi_Brien]

Woulda been useful yesterday XD, shoulda procrastinated on that homework assignment

[j.o.]

Ahhh, I didn't get it at first, but then I realized (when you said there were three parameters in cylindrical coordinates), that I could just let r = 2. That's felt really darn good. I'm kinda beginning to realize how to chose parameterizations (referring to my comment on a previous video, seems like I'm closing in on accomplishing that goal). Question to self: what if, instead of a plane, it was a cone, starting at z = 16 and going down on all sides? Ah, that's ez, we'd just have a circle and an easy time computing it. But what if there was a complex shape that created spikes on the side? In that case, I'm pretty sure we could still use cylindrical coordinates and just change the upper bound to something more complex than just 16-2(r*cos(theta)) (given that the complex shape have no overlapping's in the z-axis. Oh, that begs the question: what if it was a complex shape with multiple areas crossing the z-axis? How do we know which z to use? I'm not quite sure actually, I'll let that one brew in my brain

[Dani-ppgy]

What is the difference between the cross product for surface integrals and the jacobian for triple and double integrals? I mean the difference in concept, they actually mean the same thing right? In fact, the Jacobian and that cross product is the same ... Thanks for your videos

[sumitkumarsahoo]

thank you, sir, great job, you are helping me too much

[folsomboytbone]

Dr. Bazett, the surface for which you produced the area is ... the wall area of the cylinder? the ellipse at the top? the entire surface area of the cylinder, including the circle at the base and the ellipse at the top? Thank you.

[anujmishra]

Here i tried to evaluate it before playing the video, i was tempted to use Z Parameterization as [16-4cos@ ], wondering how to get second variable, now after playing for 5 min i was wondering about what was my wrong parametric equation really shows ?what was its messy integration of magnitude of wrong equation [<2cos@, 2sin@, 16-4cos@ > ] , 0 <@<2pie will show .

[boscolebosco]

So the result of 64 pi is the same as if we were to take the line integral of the given plane over the constraint circle of radius 4 in the xy plane parameterized with theta. It’s just the cylinder wall.

[kevvv]

The surface area we get is every side of the cylinder or just the top of the cylinder?

[jamwheeler]

so the answer you found is just the surface area of the cylinder wall correct? if you wanted the top and the base of the cylinder as well you would have to do 2 more double integrals with variables of theta and r instead of z. is my understanding correct here?

[Amoney]

Quick question Dr, why is the last line when you have integral 2(16-4cos(theta)) dtheta not evaluate to 32 being a constant which you can pull out but wouldn’t the -4costheta with respect to dtheta be -4sintheta plugging in 2pi and 0 it would just be 32*0=0?

[adam_bh]

I dont understand how the ans you got is correct. Given a cylinder with radius 2 and height 16, has a surface area on the sides equivalent to the answer you got 64pi, but if you are looking for the surface area of of that same cylinder with a little less it would only be logical that you answer should have a lower surface area

[ebrahiemalmansori]

@Dr.Trefor hey can we do r(r, @) instead of z ? i tried it and I got a different answer

[buqrepb]

What would it be for a hexa-hexaflexagon or mor general n-hexaflexagon?

[jorgel.j.jimenezg.]

Hola, tus videos son geniales, los veo a menudo desde mi celular y, como público desde allí, te recomiendo que aumentes el tamaño de la letra porfa