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This is incorrect. The problem with the answer starts here:
"Also, the point will lie on the plane which divides the top and bottom such that the volumes are equal."
This would only be true if the cg's for both halves were the same distance from that plane. But they aren't.
The solution is pi*int( t*(a2-t2)dt,0 to a)/(volume of hemisphere=2/3*pi*a3)
pi's cancel.
cg at ([a2t2/2-t4/4],0,a)/(2/3*a3)
= (a4/4)/(2/3*a3)
= 3/8 a from the base.
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Whoops! I did not mean to wipe out the original answer, and am a bit surprised that I apparently have the power to do that. Sorry!
The original answer found the distance from the base to a plane that divided the hemisphere into two equal halves, at t= ~0.347a. That might be useful information but is not the cg.
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What is a steradian?
A steradian is the solid angle subtended at the centre of a sphere by radius r by a portion of the surface of the sphere which has area r2.
What is a solid half circle called?
hemisphere
If you know the radius of a cross section 8 inches from the center of a circle how do you calculate the radius at the center?
The radius of a circle is constant regardless of where it is, so if the radius of the circle at a cross section is 8 inches, then the radius of the circle is 8 inches. However, if you mean you know the length of the chord 8 inches from the centre of the circle, then you can create an isosceles triangle by joining the ends of the chord to the centre of the circle. If you now drop a perpendicular from the centre of the circle where the radii drawn in meet, it will bisect the chord and form two congruent right triangles which have one leg 8 inches long, one leg half the length of the chord and a hypotenuse the radius of the circle. You can now use Pythagoras to find the unknown radius from the two lengths you do know: radius² = (8 in²) + (chord_length/2)² There is more to this problem which you have not stated as it is very unusual to use the phrase "CROSS-SECTION 8 inches from the centre OF A CIRCLE"; cross-sections are used with 3 dimensional objects whereas a circle is a 2 dimensional object. I would guess that you have some solid (say a sphere) where a circular cross-section has been created 8 inches from the centre of the object and you want to know its radius. In this case you can use Pythagoras as above: radius_object² = (8 in)² + (radius_circle)² by considering a plane that slices through the centre of the original object and bisects the circle created in the cross-section.
What solid figure has two faces but no vertices or edges?
A hemisphere
A solid figure which has both a flat and a curved surface?
A cylinder or a cone. Or a hemisphere.
