A container encloses the volume defined by rotating the curve x=z^{3/2} about the z axis, where 0<z<2 and x and z are measured in metres.

Step 1
So first graph the line $z^{\frac{3}{2}}$. Now we want to visualize spinning this around the z axis. On the domain [0:2] this is goig nto look like "/" and you can find a graph of this on something like wolfram alpha or Sage or a calculator.
We also have that $0<z<2$. We want to find a solid of revolution. It is slightly ambiguous here whether or not we want the part inside or outside (especially because you are using $x=f(z)$ rather than the standard $f(x)=y$. This is a solid of revolution, so you are going to want to find the area squared between the maximum height and the line and then multiply by $\mathrm{\pi <!--\; \pi \; -->}$. The reason for this is that we are taking little bits of the function and rotating them around. So we are basically making circles that are of different radius (the radius is the function we are talking about) next to each other and adding them up to find the volume. Thus, we need the areas (multiplied by a dz of course) to sum up to make an area. $\mathrm{\pi <!--\; \pi \; -->}\ast <!--\; \ast \; -->{\int <!--\; \int \; -->}_0^2{z}^{3}dz$. This integral shouldn't be too hard to do or evaluate. note that $z^3/2^2=z^3$.
Step 2
If you need help visualizing it, try graphing using a calculator. Only look at the domain given. Then imagine spinning the line that you get and filling it int, like with water. You should end up with a cross between a cup and a cone.
The way solids of revolution work is that you have some function and you want to spin it around an axis and then you want the volume. This can be thought of as summing a bunch of area's together. So we have what is kind of like a cylinder, except that each circle gets bigger, so more like a cone. Since this is calculus, we use integrals and if the dimensions bother you realize that we are taking $\mathrm{\pi <!--\; \pi \; -->}\ast <!--\; \ast \; -->r^2\ast <!--\; \ast \; -->dx$ which is the volume of a cone. As dx goes to 0, the number of pieces goes up and our approx approaches the actual volume. So that's why you want a sum (integral) of the radius (the function because it is literally the distance from the axis in question) squared.