# Homework Help: Volume integral

1. Oct 6, 2007

find the volume of the solid resulting when the region enclosed by the curves is revolved around y-axis.

$$x=\sqrt{1+y}$$ x=0 y=3

I am using this integral...

$$V=\int_{-1}^3[\pi(\sqrt{1+y})^2]dy$$

and I am getting the wrong answer.

I think it is just arithmetic, but are my bounds correct?

Thanks,
Casey

Last edited: Oct 6, 2007
2. Oct 6, 2007

You should get something like $$V=\pi(3+\frac{9}{2}+1-\frac{1}{2})=8\pi$$