Volume of liquid needed to fill sphere to height $h$ Find the volume of liquid needed to fill a sphere of radius $R$ to height $h$.
The picture shows $h$ up to maybe a quarter, I am not sure it seems pretty ambiguous. 
No clue what to do here. I just know that the formula I am suppose to memorize is 
$$V = \int \pi(R^2 - x^2)$$
 A: In this answer, I'm not really going to explain how to fit the formula; rather, I'm giving a process that will work for nearly any type of Calc II "application" problem.
The first step in any word problem is to draw a picture (even if you're given one--it still helps for you to draw your own.):

Now, we can turn the drawing into an integral.  If we were to use discrete "slices" of the sphere to compute the volume, we'd have the following:  (Let $r(z)$ be the radius of a slice at a given height $z$, and $\Delta z$ be the height of a given slice.)
$$\sum \left(\pi\cdot\big(r(z)\big)^2\cdot\Delta z\right)$$
This is just summing up the volume of a bunch of really short cylinders.  As $\Delta z \to 0$, we have an integral:
$$\int_0^h\pi\cdot\big(r(z)\big)^2\;\mathrm dz$$
Now, what is $r(z)$?  Well, let's look at a circle of radius $R$, and see what its radius is as a function of $z$:

So, we see that $y = r(z) = \sqrt{R^2 - (z-R)^2}$.  Let's plug this into our integral we have above:
$$\int_0^h\pi\left(\sqrt{R^2 - (z-R)^2}\right)^2\;\mathrm dz$$
$$\int_0^h\pi(R^2 - (z-R)^2)\;\mathrm dz$$
The above integral gives the volume of the water in a sphere, filled to height $h$.  

Note: my answer differs from the other answers because I positioned my sphere with the bottom at the origin, rather than centered at the origin.  In hindsight, the other way is simpler, but I didn't want to have to redo all my graphics. :)  To show they're the same, we perform the subsitution: 
$$u = z-R\implies \mathrm du = \mathrm dz$$
This implies:
$$\int_{-R}^{h-R}\pi(R^2 - u^2)\;\mathrm du$$
A: Hint: Sketch a graph of the curve $y=\sqrt{R^2-x^2}$. It is the upper half of a circle centred at the origin with radius $R$. Now draw a vertical line that intersects the semicircle and shade the part of the circle that is to the right of this line and above the $x$-axis. Imagine rotating this shaded region about the $x$-axis. This will represent the (sideways) liquid in the (sideways) sphere. In order for it to have a (sideways) height of $h$, notice that the equation of the vertical line must be $x=R-h$. Hence, using the disk method, the volume will be given by:
$$
V = \int_{R-h}^R \pi y^2 dx = \int_{R-h}^R \pi \left(\sqrt{R^2-x^2}\right)^2 dx= \pi\int_{R-h}^R (R^2-x^2)~dx
$$
A: Let the sphere be represented by $x^2+y^2+z^2=R^2$.  The $x$ axis is vertical and we start filling it from the bottom, $x=-R$ (to follow your notation).  Think of chopping the filled volume into thin disks.  The disk at a given value of $x$ has a radius of $r(x)=\sqrt {R^2-x^2}$ and a thickness $dx$, so a volume $\pi r^2 dx=\pi(R^2-x^2)dx$  To get the whole volume, we then want $$\int_{-R}^{-R+h}\pi(R^2-x^2)dx$$
Note that your formula is missing limits and the $dx$ that tells us what to integrate with respect to.  You have probably heard that it is better to be able to figure this out than to memorize the formula.
