document.write( "Question 894206: There is an ice cream cone with a big rounded scoop of ice cream with half the scoop in the cone. The scoop of ice cream has measure of 6, and the cone's height is 8. Please help me to figure out the total volume of the cone and the exposed ice cream!? \n" ); document.write( "

Algebra.Com's Answer #541867 by KMST(5328) $\"\"$ $\"About$

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When I try to put a ball halfway into a cone, I need to deform the bottom half a bit:

\n" ); document.write( "However, once I get the bottom half into the cone, from the outside it looks like this:
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, I can calculate the total volume as the volume of the come plus half the volume of a sphere.
\n" ); document.write( "The circular base of the cone, and the sphere have a diameter of $\"6\"$ units,
\n" ); document.write( "so the radius is $\"3\"$ units.
\n" ); document.write( "The volume of a sphere of radius $\"R\"$ is $\"4pi%2AR%5E3%2F3\"$ .
\n" ); document.write( "The volume of a half of sphere of radius $\"R\"$ is $\"2pi%2AR%5E3%2F3\"$ .
\n" ); document.write( "If the radius is $\"R=3\"$ , that volume is
\n" ); document.write( " $\"2pi%2A3%5E3%2F3=2%2Api%2A27%2F3=18pi\"$
\n" ); document.write( "The volume of a cone of height $\"h\"$ and base radius $\"R\"$ is $\"pi%2AR%5E2%2Ah%2F3\"$ .
\n" ); document.write( "If the radius is $\"R=3\"$ , and the height is $\"h=8\"$ , that volume is
\n" ); document.write( " $\"pi%2A3%5E2%2A8%2F3=pi%2A9%2A8%2F3=24pi\"$ .
\n" ); document.write( "Adding half a sphere plus a cone, the total volume is
\n" ); document.write( " $\"18pi%2B24pi=42pi\"$ \n" ); document.write( "

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