Cylindrical Shell Volumes Problem Use the method of cylindrical shells to find the volume generated by rotating the region bounded by $y=3+2x−x^2$ and $x+y=3$ about the y-axis.
I have already turned $x+y=3$ into $y=3-x$. However I don't know what to do with the polynomial to continue into graphing them and using the cylindrical shell method $dV=2pirht$. I don't understand how to continue this problem. I'd appreciate any help.
 A: Draw a picture.
Algebra shows the two curves meet at $x=0$ and $x=3$. One of the curves is a downward opening parabola that reaches its peak at $x=1$. From $x=0$ to $x=3$ the parabola is above the line $x+y=3$.
Now take a vertical slice of width "$dx$" at $x$. This is at distance $x$ from the $y$-axis. The slice has height $(3+2x-x^2)-(3-x)$. It follows that the volume is
$$\int_{x=0}^3 2\pi x\left((3+2x-x^2)-(3-x)\right)\,dx.$$
A: $3+2x-x^2=(3-x)(1+x)$, so $y=3+2x-x^2$ is a parabola opening down with $x$-intercepts at $-1$ and $3$; meets the line $x+y=3$ at the points $\langle 0,3\rangle$ and $\langle 3,0\rangle$. The region between the two curves therefore lies in the first quadrant, above the straight line and below the parabola. When you slice it into vertical strips, you get a strip at each $x$ in the interval $[0,3]$.
If $0\le x\le 3$, the top end of the strip at $x$ lies on the parabola, so it has $y$-coordinate $3+2x-x^2$. The bottom end of the strip lies on the straight line, so it has $y$-coordinate $3-x$. The length of the strip, and therefore the height of the cylindrical shell that it generates, is
$$\text{top}-\text{bottom}=(3+2x-x^2)-(3-x)=3x-x^2\;.$$
The radius of the shell is the distance from $x$ to the axis of revolution, which in this problem is the $y$-axis; that distance is $|x-0|=x$, since $x\ge 0$. The thickness of the shell is $dx$, so its volume is
$$dV=2\pi x(3x-x^2)\,dx\;.$$
Can you finish it from there?
