volume of a solid - rotation $y$-axis Let $y=g_{a}(x)=\sqrt{x}-\sqrt{a}$ be a function. The graph of the function $g_a$ together with the coordinate axes bounds a region. Now this region will be rotated about the line $y=\sqrt{a}$.
Determine the volume of this solid. 
What I did. 
I made the intersection with $\cap Oy$ and I obtained point $(0,-\sqrt{a})$. The volume is $$\pi\int^{\sqrt{a}}_{-\sqrt{a}}{x^2\mbox{dy}}.$$ But $x^2=(y+\sqrt{a})^4$. So I have to evaluate $$\pi\int^{\sqrt{a}}_{-\sqrt{a}}{(y+\sqrt{a})^4}\mbox{dy}.$$
Is it ok, or not? 
thanks! 
How can I rotate the region about the line $y=\sqrt{a}$ if the figure looks: 

thanks again!
 A: Make a sketch of the region.
If we are going to use cross-sections (disks, washers) perpendicular to the $x$-axis, then we want to integrate with respect to $x$, from $0$ to $a$. 
The cross-section is a "washer", outer radius $\sqrt{a}-(\sqrt{x}-\sqrt{a})$, and inner radius $\sqrt{a}$. So the volume is 
$$\int_0^a\pi \left(2\sqrt{a}-\sqrt{x}\right)^2\,dx -\pi a^2.$$ 
Another way: We could use cylindrical shells. Consider the horizontal strip "at" $y$, of thickness $dy$. When we rotate it we get a cylindrical shell, where the cylinder has radius $\sqrt{a}-y$ and height $x$.  Thus the volume is
$$\int_{-\sqrt{a}}^0 2\pi x (\sqrt{a}-y)\,dy.$$
Express $x$ in terms of $y$, and integrate. 
A: $g_{a}(x)=\sqrt{x}-\sqrt{a}$ 
$y=\sqrt{a}$.
Volume is evaluated by following expression   since it is rotated by $y=\sqrt{a}$
 $$\pi\int_{x=0}^{4a}{(\sqrt{a}-y)^2\mbox{dx}}.$$ 
And range of x is found by solving $$y$$ $$\text{and}$$ $$ g_a$$ that gives $\sqrt{x}=2\sqrt{a}$ 
But y=0 for x=0 to x= a & $y=\sqrt{x}-\sqrt{a}$ for x= a to x=4a
