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Let $[a, b] \subset \Bbb R, a, b \in \Bbb R, a < b$, and let $f: [a, b] \rightarrow [0, \infty)$ be a continuous function. Define:

$K := \{(x, y, z) \in [a, b] \times \Bbb R^2 : y^2 + z^2 \le (f(x))^2 \} \subset \Bbb R^3$.

Now, I wonder what this object does look like, because it's hard for me to draw a picture inside of my head. I'm quite sure that this is a ball. $y^2 + z^2 \le (f(x))^2$ looks like the equation of a circle. But normally, one would choose an "arbitrary" radius $r \ge 0$, but the radius $(f(x))^2$ depends on $x \in [a, b]$ here. Does this look any different from a normal ball?

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You are right when you say that $y^2 + z^2 \leq (f(x))^2$ resembles the equation of a circle, because it really is.

Notice how $(x, y, z) \in [a, b] \times \Bbb{R}$ means that $x$ goes from $a$ to $b$ and notice that in the expression $y^2 + z^2 \leq (f(x))^2$, $x$ only appears on the right side. That means $x$ will be setting the radius of a circle, when $x$ is fixed, through $f(x)$.

To picture the resulting solid, imagine you are in a fixed $x$ coordinate, somewhere between $a$ and $b$. Call it $x_0$ and let $r_0 = f(x_0)$. Then, at $x_0$, we have $y^2 + z^2 \leq r_0^2$ and thus we have a circle, perpendicular to the $x$-axis, which has radius $r_0$. If you walk just a little bit more further along the $x$-axis, changing $x$ to $x_1$, we now have a new $r_1 = f(x_1)$ and at $x_1$ you have a new circle perpendicular to the $x$-axis with radius $x_1$. And so on and so forth.

That means your solid, when cut perpendicularly to the $x$-axis, is just a circle. When you make $x$ go from $a$ to $b$, you create solid that is built around the $x$-axis and that has circular sections, when cut perpendicular to the $x$-axis.

I used Mathematica to draw 4 such $K$, with $x \in [1, 4]$:

$f(x) = 5$:

y^2 + z^2 <= 5, 1 <= x <= 4

$f(x) = 2x$:

y^2 + z^2 <= 2 x, 1 <= x <= 4

$f(x) = 2(x^2 - 3)$:

y^2 + z^2 <= 2 (x^2 - 3), 1 <= x <= 4

and $f(x) = |20\sin(x)| + 2$, this one with $x \in [1, 40]$:

enter image description here

so you can pretty much draw a wacky continuous line on the $xOz$ plane (or the $xOy$ plane) and then rotate it around the $x$-axis, filling in the interior.

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    So when looking at it from the front, would it be similar to a, well, dartboard? – 2017-01-07
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    @Julian pretty much :P Depending on $f(x)$ you may be only able to see a single circle (like in my 3 examples). But you can draw yourself such a $K$ where you would see a "dartboard". I will actually add another example with a more interesting shape. – 2017-01-07
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    @Julian the fourth one is a bit more funny looking, but not quite that dartboard. Nonetheless, in the end I tell you just how you can build your dartboard. – 2017-01-07
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    Ha, thank you very much! :-) – 2017-01-07
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    You might have an idea for my current problem too: https://math.stackexchange.com/questions/2087467/for-k-x-y-z-in-a-b-times-bbb-r2-y2-z2-le-fx2-pro – 2017-01-07
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For each $x$, $y^2+z^2\leq (f(x))^2$ describes a disc with radius $f(x)$, but that does in general (see below) not make $K$ into a ball.

Try to imagine what happens for various simple functions $f$.

If $f$ is constant, you get a disc with the same radius for each $x$, that makes $K$ look like a ...?

If $f(x)=x$, you get something looking like ...?

...

And if $a=-b$ and $f(x)=\sqrt{1-\frac{x}{b}^2}$, $K$ actually does become a ball (and you can find $f$'s that does the same for other values of $a$).