Volume of 3 intersecting Cylinders

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SUMMARY

The volume of three intersecting cylinders with a radius of 5, intersecting at right angles, is calculated using the formula 8(2 - √2)r³. The proof involves setting up a triple integral in cylindrical coordinates, utilizing symmetry to simplify the calculations. The integration bounds are determined by the geometry of the cylinders, leading to the final volume expression. The detailed derivation includes the integral setup and the evaluation of trigonometric functions.

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Homework Statement



I am trying to find the volume of three intersecting cylinders that intersect at right angles given that the radius is 5.

I have found many sites that state just the answer, but I am after the proof for it showing all of the working. I know how to prove the rule for 2 intersecting cylinders, I can't do it for 3 though.


Homework Equations



The rule I am after is 8(2-sqrt(2))r^3 or any similar form of that.

The Attempt at a Solution



V_3(r,r,r) = 16r^3int_0^(pi/4)int_0^1ssqrt(1-s^2cos^2t)dsdt

A site i have found begins solving with that equation, but I have completely no idea where the numbers come from.
 
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intersection3Cyl.jpg

The above is the looking-down view of the first octant. The next one is slightly cropped and from the z-axis looking out for getting the bounds of integration with the y=x plane included for reference.

intersection3Cyl_crop.jpg

Try setting up the integral now... I'll use cylinders of radius R.

In cylindrical coordinates (r,\theta, z): Using symmetry (multiply by 16), 0\le \theta\le \scriptstyle{\frac{\pi}{4}}
here, z is bounded above by the red cylinder x^2+z^2=R^2 (second image) so 0\le z\le \sqrt{R^2-r^2\cos^2\theta}
and r is bounded by the blue cylinder so 0\le r\le R

so the integral is

16\int_{0}^{\scriptstyle{\frac{\pi}{4}}}\int_{0}^{R}\int_{0}^{\sqrt{R^2-r^2\cos^2\theta}}r\, dzdrd\theta = 16\int_{0}^{\scriptstyle{\frac{\pi}{4}}}\int_{0}^{R}r\sqrt{R^2-r^2\cos^2\theta}\, drd\theta
= -{\scriptstyle{\frac{16}{3}}} R^3\int_{0}^{\scriptstyle{\frac{\pi}{4}}} \frac{\sin^3\theta -1}{\cos^2\theta}\, d\theta = 8R^3\left( 2-\sqrt{2}\right)​
 
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