Here is the exercise:
Find the volumes of the solids obtained if the plane regions R are rotated about (a) the x-axis and (b) the y-axis.
R is the finite region bounded by $y = x$ and $x = 4y – y^2$.
This is how I plot the situation:
I try to find the volume of the red graph from $x = 0$ to $x = 3$ rotated about the x-axis using the shell method. This is because I'm planning on subtracting this area from the volume of the blue graph. However, I get a negative number, which is obviously wrong. Can somebody please tell me where I go wrong?
You have not set up the integral correctly. You are taking shell length as $(3 - (4y-y^2))$ but that is between vertical line $x = 3$ and the parabola. Also between $y = 1$ and $y = 3$, $(3 - (4y-y^2))$ is negative as the line $x = 3$ is to the left of the parabola. That is why your integral gives you negative value.
As you are finding volume of the shaded region (see below diagram) rotated around x-axis and using shell method, note that the shell length is $(4y - y^2 - y)$ as the region is bound on the left by the line $x = y$ and on the right by the parabola $x = 4y - y^2$.
So, $ ~ V = \displaystyle \int_0^3 2 \pi y \cdot (3y - y^2) ~ dy = \frac{27 \pi}{2}$