what is the volume of the region bound by
y=4/x and y=(x-3)^2 about the y-axis.
I was trying to use this:
2pi*INT[ x(4/x)-x(x-3)^2]
why are you integrating between y= 1 and y= 4? If you draw a graph you will see that parabola extends down to y= 0. I suggest you do this as two parts: integrate what you have from y= 1 to y= 4 then set up a different integral for the part from y= 0 to y= 1.