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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: enter image description here

thanks again!

Iuli
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2 Answers2

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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.

André Nicolas
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  • Can I solve the intersection point: $\sqrt{x}-\sqrt{a}=\sqrt{a}$, so $\sqrt{x}=2\sqrt{a}$, $x=4a$. $\int^{4a}_{0}{\pi (2\sqrt{a}-\sqrt{x})^2\mbox{dx}}$? is ok? I used the following http://math.hws.edu/~mitchell/Math131S13/tufte-latex/Volume2.pdf, example 6.11 – Iuli Dec 15 '13 at 21:38
  • Make a sketch. The region we are rotating is bounded by the axes and our curve. I am pretty sure that this is intended to describe the vaguely triangular region that stops at $x=a$, so the region below the $x$-axis, above $y=\sqrt{x}-\sqrt{a}$, and to the right of the $y$-axis. Then finding where the curve meets the line $y=\sqrt{a}$ is not useful. – André Nicolas Dec 15 '13 at 21:45
  • Explain me, please, from where $-\pi a$? – Iuli Dec 17 '13 at 11:27
  • Thanks for the typo correction, it is $-\pi a^2$. This is the volume of the cylindrical "hole" radius $\sqrt{a}$, height $a$. We could have used $\int_0^a\pi \left[(2\sqrt{a}-\sqrt{x})^2 -(\sqrt{a})^2\right],dx$, that is, build in the hole into the integral, "washer" style. – André Nicolas Dec 17 '13 at 16:35
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$g_{a}(x)=\sqrt{x}-\sqrt{a}$

$y=\sqrt{a}$.

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

john
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