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I am having a little trouble figuring out how to integrate this problem.

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis.

$y=3(2 − x)$

$y=0$

$x=0$

This is how I think we should do it, but I am still not 100 percent sure $$V=\pi\int_0^1 (3(2 − x))^2\,dy$$

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1 Answer 1

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If $y=3(2-x)$ then $x=\frac{6-y}{3}$

Next we would the bounds for the integration. $y=0$ is given. The other is whatever $y$ value makes $x=0$. When $x=0$, $y=6$

So$$V=\pi\int_0^6 ({\frac{6-y}{3}})^2\,dy$$

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