Volume of Solid Revolved Around Y-Axis: Bounds Check

[Saladsamurai]

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

 

[Saladsamurai]

Just wondering about the bounds here..

 

[arildno]

Your bounds&integral look fine to me.
You should get something like $$V=\pi(3+\frac{9}{2}+1-\frac{1}{2})=8\pi$$