How to calculate the volume of revolution around $x=3$ for the region bounded by x, y axis and $\sqrt{x}+\sqrt{y}=1$?

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So I intend to do it in both shell and disk ways.

Let's first use shell method (formula: $\int 2\pi x(f(x)-g(x))dx$):

Since $x=3$ is the major axis, and $y=(1-\sqrt{x})^2$, we have $V=\int_0^1 2\pi (3-x)((1-\sqrt{x})^2)dx$

Next, disk method (formula: $\int \pi (f^2(x)-g^2(x))dx)$: we have $x=(1-\sqrt{y})^2$$V=\int_0^1 \pi ((3-0)^2-(3-(1-\sqrt{y})^2))dy$

Did I write my equations wrong?