I came across this question in my textbook:
What's the volume generated by rotation of the intersected area between the two curves $Y=X^5$ and $Y=X$ around the axis $X=-1$?
My main problem is the value of r with which I should substitute in the shell formula $dv=2πrhdt$. I believe it should be ($r=x+1$) while my text book substitutes ($r=x$).
Any help would be appreciated!
On
I've done this volume calculation using Pappus's $(2^{nd})$ Centroid Theorem: the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid R, i.e., $2πR$. The volume is simply $V=2\pi RA$.
The figure below shows the area to be revolved about the line $x=-1$. Notice that I've expressed the area in the region $x\in[-1,1]$. It apparent that the centroid $R=1$ and the area is
$$A=2\int_0^1 (x-x^5)\,dx=\frac{2}{3}$$
Hence,
$$V=\frac{4\pi}{3}$$
How does this compare with that of anyone else?
$r$ is the distance from the axis of rotation, and so it is $x+1$ (and $h = x-x^5$). You could use $r=x$ if you compensated for this horizontal shift elsewhere, but I don't know if that's what your book did.