Find surface area of solid of revolution obtained by rotating the curve:
from x=5 to x=7, rotated about x=-4
The Attempt at a Solution
The problem is I know how to do this if I rotated it about x-axis/y-axis, but I have no idea how to do it if the axis is shifted. I did it pretending it was rotated about the y-axis (since x=-4 is parallel to the y-axis), and my answer came out wrong.
SA=∫2∏x ds, where ds=√(1+[f'(x)]2) dx
So I found the derivative of the function, everything was fine. I squared and square-rooted, found the integral to be:
∏/10 ∫ x^2+100 dx (evaluated from 5 to 7)
but i didn't use the piece of information "axis of rotation is x=-4" anywhere in my solution. I don't really know how to go about using it, please help?