# Area betweeen curves

## Homework Statement

This is so basic and yet I'm not getting it.

Find the volume of the solid when the region enclosed is revolved arond the y axis.

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

## The Attempt at a Solution

So I first graphed the thing, finding that the enclosed area was between y=1 and y=3. I took the volumne of the equation, or A=pi *$$\sqrt{1 + y}^2$$. I then took the integral, pi * $$\int_{1}^{3} 1 + y dy$$, coming out with 6*pi. My book says this isn't right. Did I do anything wrong? Use the wrong formula?

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AKG
I was graphing with the equation set equal to x, when it should have been set equal to y. This would make $$\int_{-1}^{3} 1 + y dy$$, which, when multiplied by pi, would get 8*pi.