I was given the following problem:
What is the volume of the solid obtained by rotating the region bound by the curves $x=y^3-y^4$ and $x=0$ around the line $y=-2$
My problem is that I can't figure out what region the question is referring to. Also, how do you do things differently when rotating around something other than an axis?
The region the question refers to must be completely enclosed by curves that are said to be its boundaries. In this example, the only region completely enclosed by the graph of $x=y^3-y^4$ on one side and by the $y$-axis (i.e. the graph of $x=0$) on the other side is that little droplet-shaped region shaded in green here:
Now, as for finding the volume of the solid of revolution, you can choose between the disks/washers method or the cylindrical shells method. See e.g. these nice reviews of these methods: Volume with rings (same as disks/washers) and Volume with cylinders.
In this example, the volume can be more conveniently found using the cylinders method and integrating with respect to $y$. The area of each individual cylindrical shell is $$A(y)=\text{Area of one cylindrical shell}=\text{circumference}\cdot\text{height}=2\pi\cdot\text{radius}\cdot\text{height},$$ where $$\text{height}=(y^3-y^4)-0=y^3-y^4 \quad \text{and} \quad \text{radius}=y-(-2)=y+2.$$ In this method, it's the radius of each cylindrical shell where we use the given axis of revolution, because the radius is the distance between a generic point $y$ and the axis $y=-2$.
Once you've got that, you can integrate the areas of cylindrical shells to find the desired volume: $$V=\int_0^1 A(y)\,dy=\int_0^1 2\pi\cdot(y+2)\cdot(y^3-y^4)\,dy=\cdots$$