Finding the volume by the shell method.

Question

A region is bounded by the line $y=x+6$ and the parabola $y=x^2$ rotated about $x=3$. I know the height of the shell: $x+6-x^2$ and the thickness: $dx$. I have trouble finding the radius of the shell since it's rotated about $x=3$. I know that the radius would be $x$ if it was rotated about the y-axis. The thing that is confusing is that I can set the shell to the left or right of the y-axis. The radius could $x+3$ or $3-x$. What am I missing here and how would I determine the radius if the region isn't rotated on the x or y-axis?

Answer 1

You have drawn a picture, and noted that the parabola and the line meet at $x=-2$ and $x=3$.

Fairly conveniently, we are rotating about the line $x=3$.

Let us take a thin vertical strip of width "$dx$" centered at $x$. Draw it, perhaps around $x=1$.

The distance from the strip to the line $x=3$ we are rotating about is $3-x$. For that's how far we have to travel rightwards to get from $x$ to $3$.

The height of the strip is $(x+6)-x^2$. So the volume of the thin shell generated by rotating the strip is approximately $2\pi(3-x)((x+6)-x^2)\,dx$.

"Add up" (integrate) from $x=-2$ to $x=3$. Our volume is $$\int_{-2}^3 2\pi(3-x)((x+6)-x^2)\,dx.$$