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In my Calculus 3 course, we were told that the formula for a sphere is as follows:

$(x-h)^2+(y-k)^2+(z-l)^2 = r^2$ where the center is $C(h, k, l)$ and $P(x, y, z)$ is some valid point on the edge of the sphere that is $r$ distance from the center. This formula makes sense.

We were told that an extension of this formula is one for a ball:

$(x-h)^2+(y-k)^2+(z-l)^2 \leq r^2$

At first, I thought the $\leq$ represents the notion that one can squeeze a ball and force its radius to become smaller than it originally was.

But then I realized that squeezing part of the ball naturally displaces the air that occupied that space, so while the radius in the pinched portion of the ball is greater than $r$, shouldn't the various radii in the other portions of the ball be slightly larger than the normal radius of the ball to account for that displaced air, in order for the volume of the ball to remain the same? If my assumption is correct, why do we use $\leq$ for the formula of a ball?

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    The sphere is just the outer surface a distance $r$ from $(h,k,l)$. The ball is the whole solid and so includes the interior with points with a distance $0$ through to $r$ from $(h,k,l)$ – 2017-01-06
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    Air? Is there air in $\Bbb R^3$? – 2017-01-06
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    Everything less than the radius is in the ball. – 2017-01-06
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    @Henry Ooooh. Wow, I was overthinking this wayyy too much lol :) Thanks for the clarification, guys. – 2017-01-06
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    The inequality represents the fact that all points inside the sphere are included as well. So while the sphere is "hollow", the ball is "solid". I don't get where this concept of air is coming from. – 2017-01-06
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    @NambiarM. I was thinking of a real-life ball inflated with air. – 2017-01-06

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If you just had an $=$ sign, it would be the the surface of the sphere, a shell.

If you had an $\le$ sign, it would include the inside stuff. Imagine all of the shells equal to or less than your shell mushed together. You'd get a ball.