Meticulously Cutting Fruits

Calculus Level 1

A spherical tomato and a cylindrical portion of a cucumber have the same height and radius. Then they are chopped into slices of equal thickness, as shown above.

Comparing each slice of both kinds, which slice will have more lateral surface area of the peel?

Tomato slice Cucumber slice Each slice of both kinds has the same lateral surface area Uncertain, depending on the level of cut

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1 solution

Relevant wiki: Surface Area of a Sphere

According to Archimedes' Hat-Box Theorem , the lateral surface area of the sphere portion will equal to the lateral surface area of the cylinder with the same height as the portion and the same radius as the whole sphere.

Considering the image above, let θ \theta be the angle the portion's rim makes with the vertical axis. In the image, θ \theta turns out to be a a and b b for the 2 right triangles in the sphere's portion. Then the surface area (A) of the sphere's portion can be calculated as the integration of infinitesimal areas of lateral cylindrical layers.

For each layer, the circumference equals 2 π r ( sin θ ) 2\pi r(\sin \theta) . With an infinitesimal change, d θ \mathrm{d}\theta , the height of each layer is approximately r ( sin ( d θ ) ) r ( d θ ) r(\sin (\mathrm{d}\theta)) \approx r(\mathrm{d}\theta) .

Hence, the surface area can be calculated as:

A = a b ( 2 π r ( sin ( θ ) ) ) ( r ) d θ = 2 π r 2 a b sin ( θ ) d θ = 2 π r 2 [ cos ( θ ) a b ] = ( 2 π r ) r [ cos ( a ) cos ( b ) ] A = \int_a^b (2\pi r(\sin(\theta)))(r) \ \mathrm{d}\theta = 2\pi r^2 \int_a^b\sin(\theta) \ \mathrm{d}\theta = 2\pi r^2 [\left.-\cos(\theta) \right|_a^b] = (2\pi r)r[\cos (a) - \cos (b)]

Since the thickness of each slice h h is the difference between the vertical sides of both triangles, it can calculated as:

h = ( r × cos ( a ) ) ( r × cos ( b ) ) = r [ cos ( a ) cos ( b ) ] h = (r\times \cos (a)) - (r\times \cos (b)) = r[\cos (a) - \cos (b)]

Substituting this term in the previous equation, we will get:

A = ( 2 π r ) r [ ( cos ( a ) cos ( b ) ) ] = 2 π r h A = (2\pi r)r[(\cos (a) - \cos (b))] = 2\pi rh

This is clearly the formula for the cylindrical cucumber slice's lateral surface area with radius r r and h h . As a result, each tomato slice will have the same lateral surface area as the cucumber slice as long as the slice's thickness is the same!

Hence, each slice of both kinds have the same lateral surface area.

Moderator note:

Yea, isn't that theorem amazing! I was really surprised the first time I came across it, and couldn't believe it was for real.

I don't understand why you made it so complicated. Since you're just comparing lateral surface areas, then the lateral surface area of a cylinder (even after chopping it up) is 2 π r h 2\pi r h and the lateral surface area of the sphere (even after chopping it up) is 4 π r 2 4\pi r^2 . With h = 2 r h = 2r , we can show that they are equal.

Pi Han Goh - 5 years, 3 months ago

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I'm not just comparing the whole lateral surface area, but also showing that each slice of a tomato and cucumber cut has equal area of peels as long as the thickness is the same.

Worranat Pakornrat - 5 years, 3 months ago

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The question only asked us to compare the lateral surface areas of the two fruits. Why do you need to show that "each slice of a tomato or cucumber cut has equal area of peels as long as the thickness is the same" as well? It's like asking you how many cats you own, and you answered "I have 6 cats, 3 dogs, 2 hamsters and 5 rabbits," there's a lot of information that I don't need to know or didn't ask for.

Pi Han Goh - 5 years, 3 months ago

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@Pi Han Goh Don't you find it amazing that each slice has got the same lateral surface area? If you're asking why I showed this thing, I would simply answer "I found it amazing." That's all. There's a lot in math that I'd like to explore, and this is merely one of the facts.

Worranat Pakornrat - 5 years, 3 months ago

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@Worranat Pakornrat No, I understand what you're doing but if you're taking the entire height of the sphere/cylinder into account, then there isn't a need to a use that theorem. To make your question more interesting (by forcing readers to apply that theorem), you should ask for one "lateral surface area" of equal height, instead of the total "lateral surface area". With that, we have no choice but to apply that theorem.

By chopping them up and adding up all the lateral surface areas defeats the purpose of using that theorem.

Pi Han Goh - 5 years, 3 months ago

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@Pi Han Goh As a matter of fact, I didn't ask for the total surface area. I did ask for each slice if you notice. (I already bolded it again to emphasize.)

Worranat Pakornrat - 5 years, 3 months ago

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@Worranat Pakornrat You just updated the question after I replied to you. Your previous version did not reflect what you said earlier.

Pi Han Goh - 5 years, 3 months ago

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@Pi Han Goh Well, sorry about that then. Any other suggestion?

Worranat Pakornrat - 5 years, 3 months ago

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@Worranat Pakornrat Well, you should say that you cut a singular equal (but random) thickness for both fruits and then ask to compare the lateral surface area of these the two fruits.

Choice 1: same (right answer)
Choice 2: cylinder > sphere
Choice 3: cylinder < sphere
Choice 4: there is insufficient information.

Even after you made your edit, your image is still misleading and I suggest that you change it to one red and one green instead of 6 red and 6 greens (unless of course if you intentional want to confuse the readers, then it's fine).

Pi Han Goh - 5 years, 3 months ago

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@Pi Han Goh My original fourth option was deleted though. Is there any way to add the choice back in?

Worranat Pakornrat - 5 years, 3 months ago

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@Worranat Pakornrat Just go to the moderation channel and ask Andrew or Calvin or Sandeep or whoever is in charge.

Pi Han Goh - 5 years, 3 months ago

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@Pi Han Goh Thanks. ;)

Worranat Pakornrat - 5 years, 3 months ago

Unfortunately the shape of the red tomato in the picture does not remember a sphere what is confusing. Indeed if one consider other shapes the result would be quite different!

Andreas Wendler - 5 years, 3 months ago

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Sorry, that's my best graphic skills. It's better to stick to the context as the figure "may not be drawn to scale."

Worranat Pakornrat - 5 years, 3 months ago

Thank you for the review, and yes, I couldn't believe it at first either! ;)

Worranat Pakornrat - 5 years, 3 months ago

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