"Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the curves $y = 3+2x-x^{2}$ and $y+x=3$ about the y-axis. Below is a graph of the bounded region."
I graphed it and it turned out as a cone. area of cone $= 1/3(\pi r^2 h)$
but i don't know what the radius is for the integral equation? but the cylindrical method would imply the use of $(\pi r^2 h)$ ?
Best Answer
It's not a cone, did you lose the parabola? It's a strange rotated displaced parabola (an overturned curved cup with an indentation at the top), with a cone cut out of it at the bottom.
The curves intersect at $x=0$ and $x=3$, so you maximal radius will be $3$. Your volume lies between the strange outer thingy and the inner cone.
You can solve for volumes of surfaces of revolution in more than one way. If you slice the volume into thin disks and integrate over them (best for revolution around $x$ axis, $V=\int \pi y(x)^2\,dx$ where $y(x)$ is the radius of the current disk). However, the method of cylindrical shells works better for revolution around $y$. Instead of thin disks, it sums over cylindrical shells, as suggested by the name:
$$2\pi\int xf(x)\,dx$$
Which uses $f(x)$ as the height of the cylinder and $2\pi x$ as its circumference. This implies the bottom side of the volume sits at $y=0$ (flat bottom on the floor).
If the surface extends between curves $f(x)$ and $g(x)$, the height of the shell in question is the difference $f(x)-g(x)$.
$$2\pi\int x(f(x)-g(x))\,dx$$
Now just plug it in:
$$2\pi \int_0^3 x((3+2x-x^2)-(3-x))\,{\rm d}x$$