Question 1085534
<pre>

He's correct for {{{x^2 + y^2 = 25}}}.

He used 25 instead of 16.

{{{drawing(500,500,-4.5,4.5,-4.5,4.5,
graph(500,500,-4.5,4.5,-4.5,4.5), circle(0,0,4),
line(2.5,-3.122398999,2.5,3.122398999),
line(2.4,-3.122398999,2.4,3.122398999),
locate(2.4,-3.12,(matrix(1,3,x,",",-sqrt(16-x^2)))),
locate(2.4,3.6,(matrix(1,3,x,",","" + sqrt(16-x^2)))),
green(line(2.5,3.122398999,.5,4.4),

line(2.5,-3.122398999,.5,4.4-2*3.122398999),
line(.5,4.4-2*3.122398999,.5,4.4),

line(2.4,3.122398999,.5,4.4),

line(2.4,-3.122398999,.5,4.4-2*3.122398999-.1),
line(.4,4.4-2*3.122398999,.4,4.4-.1)
)

 )}}}

The differential of volume is a thin square slab,
(think of a thin square linoleum floor tile) whose
thickness is dx thick. Think of the green slab being
a square (I know it doesn't look like a square, but
pretend it is a square whose edge is on the thin black
strip and imagine it sticking straight up out of the 
paper straight up toward you, perpendicular to the 
xy-plane, not slanted as it looks like here.)

The length of the edge of the slab is 

{{{("" + sqrt(16-x^2))-(-sqrt(16-x^2))}}} = {{{sqrt(16-x^2)+sqrt(16-x^2)}}} ={{{2sqrt(16-x^2)}}}

The vertical height of the slab is the same, since it's square,
and its thickness is dx, so the differential of volume is

{{{(2sqrt(16-x^2))^2*dx}}} or {{{4(16-x^2)dx}}}

The slabs go from where x=-4 to +4

So

{{{V=int(4(16-x^2),dx,-4,4)}}} = {{{4*int((16-x^2),dx,-4,4)}}} =  

Finish that by breaking it into two integrals and you'll
end up with {{{1024/3}}}.

Edwin</pre>