Consider the bounded area by the straights $x=0,\;y=1$, and the parabola $y^2=4y-x$, calculate the volume of the solid of revolution generated when the parabola spins around the $x$ axis.
I think that the volume can be calculated by $$V=\pi\int_{0}^{4}{\left( \sqrt{4-x}+2\right)^{2}dx}+\pi\int_{0}^{3}{4dx}-\pi\int_{0}^{3}{dx}+\pi\int_{3}^{4}{4dx}-\pi\int_{3}^{4}{\left(-\sqrt{4-x}+2\right)^2dx}.$$
Best Answer
Draw a picture. We have a parabola whose axis of symmetry is the line $y=2$, and which opens leftward. The apex of the parabola is at $(4,2)$.
The line $y=1$ divides the part of the parabola to the right of the $y$-axis into two parts, the fat part above the line $y=1$, and a much thinner part below the line $y=1$. It is not clear from the wording which part is being spun. We will assume it is the fat part. Minor modification will take care of things if it is the thinner part.
It is most convenient to use the Method of Cylindrical Shells. Look at a thin horizontal strip, going from height $y$ to height "$y+dy$."
Spin this strip about the $x$-axis. We get a cylindrical shell of thickness "$dy$." The shell has radius $y$ and height $x=4y-y^2$. So the shell has volume approximately equal to $2\pi y(4y-y^2)\,dy$. "Add up" (integrate) from $y=1$ to $y=4$. We get volume $$\int_{y=1}^4 2\pi y(4y-y^2)\,dy.$$
Remark: If we want to integrate with respect to $x$, things get a little more complicated. You did it along those lines, so we write down suitable expressions. The volume from $x=0$ to $x=3$ is equal to $$\int_{x=0}^3 \pi[(2+\sqrt{4-x})^2-1^2]\,dx.$$ To this we must add the volume from $x=3$ to $x=4$, which is $$\int_{x=3}^4 \pi[(2+\sqrt{4-x})^2-(2-\sqrt{4-x})^2]\,dx.$$