volume of a solid - rotation $y$-axis

Question

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!

Answer 1

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.

Answer 2

$g_{a}(x)=\sqrt{x}-\sqrt{a}$

$y=\sqrt{a}$.

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