Two cylindrical holes drilled through the solid sphere

Calculus Level pending

Two cylindrical holes of radius 5 cm 5\text{ cm} are drilled through perpendicular to each other, along the diametric lines of a solid sphere of radius 13 cm 13\text{ cm} (As shown in the typical figure below). What is the volume of drilled sphere in cm 3 \text{cm}^3 ?


The answer is 5940.353535.

Harish Chandra Rajpoot by Harish Chandra Rajpoot , 30, India

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1 solution

In general, the volume of solid sphere of radius R R having two cylindrical holes of radius r r drilled through perpendicularly along the diametric lines is given by the following formula

V = 4 π 3 ( 2 ( R 2 − r 2 ) 3 / 2 − R 3 ) + 16 3 r 3 V=\frac{4\pi}{3}\left(2(R^2-r^2)^{3/2}-R^3\right)+\frac{16}{3}r^3

setting the value of radius R = 13 c m R=13\ cm & r = 5 c m r=5\ cm , one can find the volume of drilled sphere

V = 4 π 3 ( 2 ( 1 3 2 − 5 2 ) 3 / 2 − 1 3 3 ) + 16 3 ( 5 ) 3 ≈ 5940.353535 c m 3 V=\frac{4\pi}{3}\left(2(13^2-5^2)^{3/2}-13^3\right)+\frac{16}{3}(5)^3\approx 5940.353535\ cm^3

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In general, the volume of solid sphere of radius R R having two cylindrical holes of radius r r drilled through perpendicularly along the diametric lines is given by the following formula

V = 4 π 3 ( 2 ( R 2 − r 2 ) 3 / 2 − R 3 ) + 16 3 r 3 V=\frac{4\pi}{3}\left(2(R^2-r^2)^{3/2}-R^3\right)+\frac{16}{3}r^3

Got a proof?

Pi Han Goh - 4 years, 9 months ago

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one can easily derive the given formula, as the volume left =(volume of sphere of radius R)-2(volume of cylinder of length 2 ( R 2 − r 2 ) 2\sqrt(R^2-r^2) & radius r)-4(volume of hemispherical cap of cylinder)+(volume of intersection of two perpendicular cylinders of radius r)

The volume of intersection can be easily found out to be 16 3 r 3 \frac{16}{3}r^3 using double or triple integration. I have not given the full derivation because it is cumbersome for editing. However if you face any problem in derivation feel free to ask me I will help you.

Harish Chandra Rajpoot - 4 years, 9 months ago

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