Question 192419
<font face="Garamond" size="+2">


Consider a diameter of the sphere perpendicular to the inscribed cylinder's axis.  From the center of the sphere, construct a radius that intersects the circle of intersection between the sphere and one base of the cylinder.  We need to derive a function for the volume of the cylinder in terms of the angle between the above described diameter and the constructed radius.  Let the measure of that angle be <i>x</i>.


Working this problem will be computationally simpler if we consider the sphere to have radius <i>r</i> for the time being.  Given that, the radius of the base of the cylinder is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_c(x) = r\cos{x}]


And the area of the base of the cylinder is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_b(x) = \pi r^2\cos^2{x}]


Half the height of the cylinder is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\sin{x}]


And then the height of the cylinder is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h_c(x) = 2r\sin{x}]


Now we can describe the volume of the cylinder:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_c(x) = \pi \left(r^2\cos^2{x}\right)\left(2r\sin{x}\right)]


Since *[tex \LARGE V_c(0) = 0] and *[tex \LARGE V_c(\frac{\pi}{2}) = 0] and these are the limiting values for <i>x</i>, if we can find an extreme point for the function on the interval *[tex \LARGE 0 < x < {\pi \over 2}], we can be assured that this point is a maximum.


Take the derivative of *[tex \LARGE V_c]:


First factor out the constants:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_c(x) = 2\pi r^3 \left(\cos^2{x}\right)\left(\sin{x}\right)]


Now define *[tex \LARGE f(x) = \cos^2{x}] and *[tex \LARGE g(x) = \sin{x}]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_c(x) = 2\pi r^3f(x)g(x)]


Now use the Product Rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_c^{\,'}(x) = 2\pi r^3 f'(x)g(x) + f(x)g'(x)]


Use the Chain Rule to derive *[tex \LARGE f'(x)]


Let *[tex \LARGE u = \cos{x}]


Then *[tex \LARGE f(x)=y = u^2] and *[tex \LARGE f'(x)=\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=(2\cos{x})(-\sin{x})=-2\cos{x}\sin{x} ]


Also:


*[tex \LARGE g'(x) = \cos{x}]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_c^{\,'}(x) = 2\pi r^3 \left(\left(-2\cos{x}\sin{x}\right)\left( \sin{x} \right)  + \left(\cos^2{x}\right)\left(\cos{x}\right)\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_c^{\,'}(x) = 2\pi r^3 \left(-2\cos{x}\sin^2{x}  + \cos^3{x}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_c^{\,'}(x) = 2\pi r^3\cos{x} \left(-2\sin^2{x}  + \cos^2{x}\right)]


A local extreme (and here, a local maximum) is at the point where the first derivative is equal to zero, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\pi r^3\cos{x} \left(-2\sin^2{x}  + \cos^2{x}\right) = 0]


Applying the zero product rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos{x} = 0]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2\sin^2{x} + cos^2{x}=0]


But *[tex \LARGE \cos{x} = 0 \ \ \Rightarrow\ \ x = \frac{\pi}{2}] which is not in the feasible interval, *[tex \LARGE 0 < x < {\pi \over 2}], hence we exclude this value as an extraneous root.


To solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2\sin^2{x} + cos^2{x}=0]


Use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2{x} = 1 - \sin^2{x}]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2\sin^2{x} + 1 - \sin^2{x} = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3 \sin^2{x} = 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2{x} = \frac{1}{3}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin{x} = \frac{sqrt{3}}{3}]


and, again using *[tex \LARGE \cos^2{x} = 1 - \sin^2{x}] to determine:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2{x} = \frac{2}{3}]


Substituting these values you can use the volume function to calculate the volume of the maximum inscribed cylinder for the general sphere of radius <i>r</i>:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_c(x) = 2\pi r^3 \left(\frac{2}{3}\right)\left(\frac{sqrt{3}}{3}\right)=\frac{4\pi r^3\sqrt{3}}{9}]


Notice that the calculation reduces to simply multiplying the volume of the sphere *[tex \LARGE \left(\frac{4\pi r^3}{3}\right)] by a factor of *[tex \LARGE \frac{sqrt{3}}{3}].


Of course, for your specific sphere of radius 8, you need to substitute for <i>r</i>.  I'll let you do the arithmetic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>