let A is the region enclosed by $x=1, y=1, y=1-x$. then suppose S is the solid obtained by revolving A about y-axis. Find volume of soild obtained by revolving S about the x-axis.
I've used shell method for revolving once, and I can't imagine how to calculate twice revolving.
I know volume of A is $2\pi \int_0^1 x(1-x)\,dx$.
p.s. if I want to get surface of area S, same method for calculating volume works?
Your region is a right triangle. Revolving it around $y$ gives a cylinder with a cone taken out of it. Your first volume integral is not correct as the height at $x$ is not $1-x$ but $x$-the triangle is above $y=1-x$, not below it. The volume is then $2\pi\int_0^1xydx=2\pi\int_0^1x^2dx$
Having made that shape, to revolve it around the $x$ axis it seems washers are the way to go. For each bit of $dx$ you need to figure out the range in $y$ that the shape covers. My imagination is failing me-I would make model out of paper to help.
To get the surface area, remember you have to integrate arc length of the original surface.