What is the volume of the solid generated by revolving about x axis, the area bounded by the parabola $x=4y-y^2$ and $y=x$?.
I'm confused with the outer and inner radius. I am a little bit skeptical in using the formula of the volume of solid revolution. I dont know where to substitute those values in the formula.
Is it correct if I'll write $v=\int_{0}^{3}2\pi y(4y-y^2-y)$? I just need confirmation for this so that i can proceed to the next process
Since there are multiple powers of $y$, you need to use the Shell or Collar method. Since you are revolving around the $x$-axis, your functions needs to be in terms of $y$(which they already are), as well as your range. Draw your problem out on paper, and plug everything into the Shell method equation.
You are going to subtract $g(y)$ from $f(y)$ before multiplying by $2\pi$, where $f(y)$ is "on top" on the graph and $g(y)$ is the function "below" $f(y)$.
Edit: $v=\int_{a}^{b}2\pi y(f(y)-g(y))$
The extra $y$ would be your radius from the axis of rotation, but since you are revolving around $y$=0, it can be omitted.