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?
[Math] Finding the volume by the shell method.
calculusvolume
Related Solutions
The reason you need two integrals is because over the range $0$ to $4$ in $y$, the height of the shell is $2\sqrt y$ because it extends all the way across the $y$ axis. Over the range $4$ to $9$ in $y$, the height is $\sqrt y -(y-6)$. That changes the integrand, so split it into two integrals.
A sketch is below. The horizontal lines are what are revolved to make the shells. Note that the lower one extends from one side of the parabola to the other. The upper one extends from the line to the parabola. That is why the integral changes
Best Answer
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 $$\int_{-2}^3 2\pi(3-x)((x+6)-x^2)\,dx.$$