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Calculus Level 5

There are two circles:

$\text{Circle-A}$ with equation $(x-10)^{2}+(y-10)^{2}=4$
$\text{Circle-B}$ with the equation $(x+10)^{2}+(y-10)^{2}=4$ .

$\text{Circle-A}$ revolves with $y=-x$ as the axis while $\text{Circle-B}$ revolves with $y=x$ as the axis.

What is the total volume of the solid structure formed by the revolution of two circles?

Clarification and Caveat: The Structure will have more than one Steinmetz solid , so be carefull while entering your answer. And correct your answer to three decimal places. Assume the ring shaped structure formed by each circle to be a perfect cylinder.

This problem is original.

The answer is 2147.903.

1 solution

Abhay Tiwari
Apr 29, 2016

Since both the circles have the same radii= $2$ and are at equal distance from the origin (i.e.)= $10\sqrt{2}$

Both will have the same volume of revolution .

Calculating for one revolution=Area of the circle $\times$ the length traversed by its center $(2\times\pi\times10\sqrt{2})$

We get $= \pi\times2^{2}\times (2\times\pi\times10\sqrt{2})=1116.618272$

The volume due to revolution of other circle will be same, so total volume formed= $2\times1116.618272=2233.236544$

Now, there will be $two$ Steinmetz solid(Bi cylinder) , one at $Z>0$ and other one at $Z<0$ .

Formula for Volume of a $Steinmetz$ $Bi$ $Cylinder$ (only in case of the intersecting cylinders have equal radii)= $\frac{16}{3}\times r^{3}=\frac{16}{3}\times 2^{3}=\frac{128}{3}$

Volume of two $Steinmetz$ $Bi$ $Cylinder$ = $2\times\frac{128}{3}=\frac{256}{3}$

Subtracting this from total volume, we get = $2233.236544-\frac{256}{3}=\boxed{2147.903}$

Hi Abhay, The tori (the volumes created by revolution) are not cylinders, so their intersection will not quite be a bicylinder, right? (Although it will be fairly close in shape to one.) The straight lines on cylinders make for easy integrals in that case; I'd love to know how one can determine the volume of the intersections in the curvier case in your problem.

Mark C - 5 years, 1 month ago

Sir, if the radius of the circle is very small and the radius of revolution is too large as compared to the radius of the circle, then it would be possible to apply the method that I applied. I made the problem by assuming that the rings formed are just like cylinders bent into the shape of ring. I don't use AUTOCAD, or some other software similar to it, it's just pure imagination from my side. And I know that you are correct, they will not be perfect cylinders, and then it becomes difficult for me to determine the actual volume. So, I guess I should mention there to assume the structure formed by each circle to be a perfect cylinder. Thanks a lot!

Abhay Tiwari - 5 years, 1 month ago

Oh my god! Nice solution....

Ashish Menon - 5 years, 1 month ago

Ha ha, Thanks a Lot. :)

Abhay Tiwari - 5 years, 1 month ago

Love Calculus?, Try this!

The answer is 2147.903.

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