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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!!!

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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*}

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  • $\begingroup$ Great thank you. $\endgroup$ – squenshl Nov 14 '19 at 2:08

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