Show that the volume of the …

{tex}v = \frac{1}{3}\pi {r^2}h{/tex} {tex}\left[ {{r^2} = \sqrt {{R^2} - {x^2}} } \right]{/tex}
{tex}V = \frac{1}{2}\pi .\left( {{R^2} - {x^2}} \right).\left( {R + x} \right){/tex}
{tex}\frac{{dy}}{{dx}} = \frac{1}{3}\pi \left[ {\left( {{R^2} - {x^2}} \right)\left( 1 \right) + \left( {R + x} \right)( - 2x)} \right]{/tex}
{tex} = \frac{1}{3}\pi \left[ {\left( {R + x} \right)\left( {R - x} \right) - 2x\left( {R + x} \right)} \right]{/tex}
{tex} = \frac{1}{3}\pi \left( {R + x} \right)\left[ {R - x - 2x} \right]{/tex}
{tex} = \frac{1}{3}\pi \left( {R + x} \right)(R - 3x){/tex}....(1)
Put {tex}\frac{{dv}}{{dr}} = 0{/tex}
R = - x (neglecting)
R = 3x
{tex}\frac{R}{3} = x{/tex}
On again differentiating equation (1)
{tex}\frac{{{d^2}v}}{{d{x^2}}} = \frac{1}{3}\pi \left[ {(R + x)( - 3) + (R - 3x)(1)} \right]{/tex}
={tex}{\left. {\frac{{{d^2}v}}{{d{x^2}}}} \right]_{x = \frac{R}{3}}} = \frac{1}{3}\pi \left[ {\left( {R + \frac{R}{3}} \right)( - 3) + \left( {R - 3.\frac{R}{3}} \right)} \right]{/tex}
{tex}\frac{1}{3}\pi \left[ {\frac{{4R}}{3} \times - 3 + 0} \right]{/tex}
{tex} = \frac{{ - 1}}{3}\pi 4R{/tex}
{tex}\frac{{{d^2}v}}{{d{x^2}}} < 0{/tex} Hence maximum
Now {tex}v = \frac{1}{3}\pi \left[ {\left( {{R^2} - {x^2}} \right)\left( {R + x} \right)} \right]{/tex}{tex}\left[ {x = \frac{R}{3}} \right]{/tex}
{tex}v = \frac{1}{3}\pi \left[ {\left( {{R^2} - {{\left( {\frac{R}{3}} \right)}^2}} \right)\left( {R + \left( {\frac{R}{3}} \right)} \right)} \right]{/tex}
{tex} = \frac{1}{3}\pi \left[ {\frac{{8{R^2}}}{9} \times \frac{{4R}}{3}} \right]{/tex}
{tex}v = \frac{8}{{27}}\left( {\frac{4}{3}} \right)\pi {R^3}{/tex}
{tex}v = \frac{8}{{27}}{/tex} Volume of sphere
Volume of cone {tex} = \frac{8}{{27}}{/tex} of volume of sphere.