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## Homework Statement

Find surface area of solid of revolution obtained by rotating the curve:

y=x

^{2}/40-5lnx

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.

I used:

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?