Example 1:  We will find the integral that will give the volume of the solid obtained by rotating a region in the first quadrant bounded by $f(y)=4y$ and $g(y)=y^3$ around the $y$-axis.  After looking at the graphs below, we realize that to find the smallest and largest $y$-values of the region, we must find the intersection of $f$ and $g$:  we set the curves equal to each other.  $4y=y^3$ so $y^3-4y=y(y^2-4)=y(y+2)(y-2)=0$.  Since we are in the first quadrant, the curves intersect at $y=0$ and $y=2$.

Solution 2:  $A=\pi(R^2-r^2)$  where $R$ and $r$ can be found by carefully looking at the graph above.  $R$ is the distance between the line $y=4$ and the lower curve.  When the curve is below the axis, its $y$-values are negative, so the distance between the $x$-axis and the curve is $-y=-(x^2-2x)$.  We add this to 4, getting $R=-(x^2-2x)+4$.  When the curve is above the $x$-axis, its $y$-values are positive, and we need to subtract them from 4 to get $R=4-y=4-(x^2-2x)$.  Look carefully at the graph and see that both cases make sense.  Fortunately, in either case we get the same $R$.  Similarly, we see that $r=4-x$ because $r$ is the distance between the line $y=4$ and the line $y=x$. 

We now need our intersection points.  DO: Find these before reading further. 

We set the curves equal to each other:  $x=x^2-2x$ so $x^2-3x=x(x-3)=0$, so our curves intersect at $x=0$ and $x=3$. 

Now we can find our integral: $\displaystyle\int_a^b\pi\left(R^2-r^2\right)\,dx=\pi\int_0^3\left((4-(x^2-2x))^2-(4-x)^2\right)\,dx$, which we can easily compute.
 
NOTICE:  The only integration formula we need to know is that $\int \text{Area}=\text{volume}$.  We compute the area by looking at the graph(s) of the curve(s).