SURFACE AREA of a SPHERE SECTION

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[@tayzonday]

I’m just so amazed to see an impeccable chalk-board used in 2023- not a marker board and not a digital drawing pad.

[@Sugarman96]

Damn Tay, you keep showing up in the most pleasantly surprising places.

[@sarithasaritha.t.r147]

Chocolate rain

[@luaiderar6600]

No way It's really you!

[@user-ct1ns6zw4z]

makes me want to get one

[@kkanden]

omg, you mentioning spivak's calculus brought back so much calc 3 trauma lol. my calc 3 prof kept referencing the book and saying "you'll find this in spivak" no matter the theorem or definition we stumbled on. hopefully one day i'll be able to get down and read through it because i have heard otherwise great things about it!

[@bscutajar]

There's also a similar fact that if you take the same setup, and subtract the cylinder from the middle of the sphere (to be left with an olive shaped ring) the RADIUS of the sphere doesn't matter for the volume. As the radius grows larger you get a bigger but thinner ring with constant volume.

[@Noam_.Menashe]

I didn't know the answer, but I guessed that just like a sphere has the same area as a cylinder without bases with height 2r, this will just be the area of a cylinder with height H.

[@bndrcr82a08e349g]

That is, does the area of that figure correspond to the lateral surface of a cylinder of equal height and radius equal to that of the sphere?

[@General12th]

Hi Dr. Penn! This problem reminds me of the napkin ring video VSauce made several years ago. Is it just me, or is Dr. Penn's voice a little off this video? I don't know if he was sick or choked up, or if the mic was a little funny, or if the audio got changed during editing, but it sounds like there's something.

[@ddognine]

My calculus undergrad text was Thomas/Finney 7th ed. I looked it up and on page 340 is the exact same problem but posed as cutting a slice from a spherical loaf of bread. And interestingly it uses what I assume is the customary convention of a slice of length h. It would be interesting to see the solution in polar and spherical coordinates.

[@tylerduncan5908]

To answer your question in the early part of the video about whether the volume is dependent on where a given slice is taken, the answer is that the choice of h₀ matters. We can see this by choosing a Δh that is decently less than r and h₀ as the tip of the sphere. we can see that by replacing h₀ with h₀+k, we remove a slice of thickness k from the tip of the sphere, and replace it with a slice of thickness k between h₁ and h₁+k But since any given cross section of a sphere is larger as it approaches the equator, we can show that k₀< k₁ To close a quick loophole, we would not be allowed to do this with a shape who's surface is not both continous, and also has dr/dx>0 I also believe that (mostly on intuition) that if you were to define a function of the thickness slice such that the volume between is constant, any polynomial term would have to be degree 3, since you would integrating with respect to y over a quadratic a quadratic both in x and y

[@Wielorybkek]

interesting outcome!

[@OhGreatSwami]

Hi Dr. Penn….are you saying the surface area of a band around the equator of height h (i.e. planes at y = -h/2 and y = + h/2) is the same as the surface of the polar cap of height h (i.e. planes at y = +(r-h) and y = +r)? Interesting result.

[@ouni2022]

Thanks! 🙂

[@sinecurve9999]

Ah yes. Zach Star mentioned this in his video a few weeks back.

[@powercables1611]

What happens if the planes are non parallel so that one side touch each other while the other is h units apart? That seems very difficult

[@gp-ht7ug]

Probably you have to split the surface into pieces and apply the method Michael showed

[@Krekkertje]

Well you could calculate the surface area of the two pieces that are being 'cut off' using the method in this video and then subtract those from the total area of the sphere. All you would need would be the distance from the center of the sphere to those planes. Of course this would only work if the planes do not intersect within the sphere.

[@dalehall7138]

An interesting fact is that this result was proven by Archimedes, several years before the development of calculus.

[@x12_79]

I love you

[@AnkhArcRod]

The answer can be deduced from the fact that no radius has been specified. Therefore, the answer should be independent of R. Thus, we consider a specific case of sphere with radius h/2 whose entire surface is between parallel planes that are h apart. Thus the answer is pi*h^2. Incidentally, a similar problem that I had learned in my childhood before I knew about calculus was: What is the volume of remaining portion of a sphere in which a hole is drilled through center and the height of the hole in remaining portion is h? The solution can be arrived at in the same fashion.

[@bjornfeuerbacher5514]

But the answer _does_ depend on the radius: It's 2 pi r h.

[@Chiavaccio]

👏👏👍

[@Jack_Callcott_AU]

At 2:32 we tell the professor. Hey mike, "A cone with its top cut off" is called a frustum. Apologies for the pedantry❕

[@General12th]

This is a circular frustum!

[@davidgerber9317]

I thought the technical term was "Lampshade".:)

[@Jack_Callcott_AU]

@@General12th That's correct. There are frustums with square bases, and other shapes❕

[@manucitomx]

Every time Spivak comes up, I smile. Thank you, professor.

[@Vladimir_Pavlov]

It can be added that although the area of the calculated part of the surface of the sphere does not depend on how the planes are located relative to the center of the sphere, the volume of the part of the ball between these parallel sections varies from this factor. And reaches a maximum equal to V= π*h*(r^2 - h^2/12) with a symmetrical arrangement of these sections relative to the center of the ball (sphere), that is, when they are at a distance h/ 2 from the center. Keep this in mind when you are offered to cut a piece of watermelon for yourself with a fixed distance between the sections.

[@mikeschieffer2644]

So this method seems to find what I was taught was called the lateral surface area. What is our answer if we include the areas of the two circular faces?

[@madcapprof]

"cone with the top cut off"=frustum.

[@RigoVids]

Integrate baby

[@bilalabbad7954]

Your solutions are amazing

[@erazorheader]

In this case the answer is just obvious through the conformal transformation of a cylinder of radius r and height h onto corresponding part of sphere.

[@gp-ht7ug]

Love these videos!