There is a right circle cyclinder inside a sphere with radius #8m#. What is the largest surface area of the right circle cyclinder?

1 Answer
Ultrilliam
May 10, 2018

#64 pi (1 + sqrt(5)) approx 650.65 \ m^2#

On basis of assumptions made

Explanation:

Assume that the optimal position for the cylinder is with its centroid at the centroid of the sphere, cylinder is orientated so that its axis is the z-axis, with the edge of the cylinder end caps touching the sphere.

The line from the origin to any point on the edge of the cylinder end caps has length 8m. If the angle between x-axis and that line is #alpha#, then:

  • cylinder has height #h = 2 * 8 sin alpha#

  • cylinder has radius #r = 8 cos alpha#

The surface area of the cylinder is:

#S = 2 pi r h + 2* pi r^2#

# = 2 pi (8 cos alpha) 16 sin alpha + 2* pi (8 cos alpha)^2#

# = 128 pi ( sin 2 alpha + cos^2 alpha) qquad star#

#S_alpha = 128 pi (2 cos 2 alpha - 2 cos alpha sin alpha) = 128 pi (2 cos 2 alpha - sin 2 alpha)#

#S_alpha = 0 implies tan 2 alpha = 2 qquad [alpha approx 32^o]#

#implies sin 2 alpha = 2/sqrt5, qquad cos 2 alpha = 1/sqrt5#

From #star# with another double-angle adjustment:

#S = 128 pi ( sin 2 alpha + 1/2(cos 2 alpha + 1))#

# = 128 pi ( 2/sqrt(5) + 1/2(1/sqrt(5) + 1)) #

# = 64 pi (1 + sqrt(5)) #

# approx 650.65 \ m^2#

Reality check, sphere has surface area # approx 804 \ m^2#

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