What is the volume of the solid generated when the region bounded by the graph of $x=\sqrt{y-2}$ and the lines $x=0$ and $y=5$ is revolved about the $y$-axis?

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We can visualize what is going on by graphing the equation. Solving for $y$ allows us to enter it in terms of $x$ in our calculator.

$$x=\sqrt{y-2}$$ $$x^2=y-2$$ $$y=x^2+2$$

Note that the graph includes only the portion in the first quadrant.

The upper limit of integration appears to be $y=5$, and we can find the lower limit by finding when the graph crosses the $y$ axis.

$$0= \sqrt{y-2}$$ $$y=2$$

The volume of the solid can be found by summing the area of each disk:

$$\int\limits_2^5 A_{\text{disk}}   dy$$ $$\int\limits_2^5 \pi r^2   dy$$ $$\int\limits_2^5 \pi (\sqrt{y-2})^2   dy$$ $$\int\limits_2^5 \pi(y-2)   dy$$ $$\approx {14.137}$$