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:

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