Two Vaults make a Safe!

Calculus Level 5

Let's call a solid common to two right circular cylinders of equal radii ( r = 6 units ) (r=6 \text{ units}) intersecting at right angles a safe .

(These are the two cylinders which intersect in the manner given above.) (This is the safe .)

Find the volume of this safe .


The answer is 1152.

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1 solution

Otto Bretscher
Apr 27, 2016

Relevant wiki: Volume of Revolution - Problem Solving

Let's define the "safe" by x 2 + z 2 36 x^2+z^2\leq 36 and y 2 + z 2 36 y^2+z^2\leq 36 . Then the horizontal cross section at z = z 0 z=z_0 is a square of area 4 ( 36 z 0 2 ) 4(36-z_0^2) , and the total volume is 6 6 4 ( 36 z 2 ) d z = 1152 \int_{-6}^{6}4(36-z^2)dz=\boxed{1152}

Sir, have a look at this also, I don't know why it's a level 4 and not even in the featured list.

Abhay Tiwari - 5 years, 1 month ago

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hmmm... the link is not working for me

Otto Bretscher - 5 years, 1 month ago

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I hav edited it, try again.

Abhay Tiwari - 5 years, 1 month ago

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@Abhay Tiwari Thanks! That looks like an interesting and tricky problem. I will look at it tonight after work. I don't understand the condition "... and both the revolutions have their center at the origin;" it seems redundant and confusing. Can't we just omit that condition?

Otto Bretscher - 5 years, 1 month ago

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@Otto Bretscher No sir, because of the revolution about the origin and the axis that I have mentioned for each circle, they are forming the same figure as mentioned here in this Question. As they cut each other at 90° which is the condition for steimetz solid.

Abhay Tiwari - 5 years, 1 month ago

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@Abhay Tiwari If you say "Circle B revolves with y = x y=x as the axis", does that not fully determine the revolution? What does it even mean to say that this revolution about a line has "its center at the origin"?

Otto Bretscher - 5 years, 1 month ago

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@Otto Bretscher I understood what you are trying to say here. it is getting redundant to add that part of the statement. But I though to give some help to those who will try to solve but to my surprise no one has gotten it right till now.

Abhay Tiwari - 5 years, 1 month ago

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@Abhay Tiwari It's tricky... ;) As I said, I will give it a try tonight.

Otto Bretscher - 5 years, 1 month ago

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