# Volumes of revolution

Find the volume of the region $$R$$ bounded by the curve $$y=x$$, $$y=1$$, and the $$y$$-axis rotated about $$x=1$$.

We are rotating about a vertical axis so the cross-sectional area will be a function of $$y$$.

So our volume is $$V=\int_{0}^{1} A(y) \; dy$$ but my question is how do I find $$A(y)$$???

Thanks a lot for the help!!!

With simple curves like this, I would immediately translate it to revolving it around the $$x$$ axis. which would translate the problem to $$\pi \int_0^1 (1-(1-x)^2)dx.$$
Although, if you were forced to compute the function in terms of $$y$$ the general notion would be \begin{align*} A(y) &= \pi(\textbf{outer radius})^2 - \pi (\textbf{inner radius})^2 \\ &= \pi(1)^2-\pi (1-y)^2 \\ &= \pi (1-(1-y)^2). \end{align*}