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Homework problem I recently got:

Find the length of the curve of $y=\sqrt{a^2-x^2}$ from $x=-2$ to $x=1$. Also find the area of the surface of revolution about the $x$-axis.

Having trouble understanding this. I'm not even sure what it would look like. I assume a is supposed to be a constant. Is it just a semi-circle with radius $a$. $X$-intercepts of $(-a,0)$ and $(a,0)$ and a max at $(0,a)$?

I'm a little iffy about how to first find the length.

I do know that I should start with $\int_{-2}^{1} 2\pi y ds$ since I am revolving about the x-axis.

Thanks for anyone who can help me out.

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1 Answer

The graph, as you surmised, of $y=\sqrt{a^2-x^2}$ is the upper half of the semicircle with center at $(0,0)$ and radius $a$. (Square both sides to get $y^2 = a^2-x^2$, or $x^2+y^2=a^2$, the equation of the circle described; since $y\geq 0$, it's the upper half). (Of course, you need $|a|\geq 2$ for your limits to make sense).

The formula for the length of the graph of $y=f(x)$ from $x=r$ to $x=s$ is $$\mathrm{Arc\ Length} = \int_r^s \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \,dx.$$ So here, you have $y=\sqrt{a^2-x^2}$, so take the derivative, square, plug it into the formula, that will give the arc length.

For the area of the surface, I don't know which formula you know. The one I know is that the area is given by $$\mathrm{Area} = \int_r^s 2\pi f(x)\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx.$$

Added. So, since you've essentially worked it out: if $|a|\gt 2$ (so we don't have to worry about the derivative and the curve being defined over the entire interval $[-2,1]$), we have $$\frac{dy}{dx} = -\frac{x}{\sqrt{a^2-x^2}}.$$ So $$ 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{x^2}{a^2-x^2} = \frac{a^2}{a^2-x^2}.$$ So for arc length, we have: \begin{align*} \mathrm{Arc\ Length} &= \int_{-2}^1 \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx \\ &= \int_{-2}^1 \sqrt{\frac{a^2}{a^2-x^2}}\,dx\\ &= \int_{-2}^1 \frac{a}{\sqrt{a^2-x^2}}\,dx\\ &= a\int_{-2}^1\frac{dx}{\sqrt{a^2-x^2}}\\ &= a\arcsin\left(\frac{x}{a}\right)\Biggm|_{-2}^1\\ &= a\arcsin\left(\frac{1}{a}\right) - a\arcsin\left(\frac{-2}{a}\right)\\ &= a\arcsin\left(\frac{1}{a}\right) + a\arcsin\left(\frac{2}{a}\right). \end{align*} For the surface area, you have: \begin{align*} \mathrm{Surface\ Area} &= \int_{-2}^1 2\pi f(x)\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\\ &= 2\pi \int_{-2}^1 \sqrt{a^2-x^2}\left(\frac{a}{\sqrt{a^2-x^2}}\right)\,dx\\ &= 2\pi a \int_{-2}^1 \,dx = 2\pi a(1-(-2)) = 6\pi a. \end{align*} If $a=2$, then the two integrals are improper and you need to use limits, but you'll end up with the same answer in both cases. If $0\lt a\lt 2$, the integrals are not well-defined (but then neither is the problem).

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@Arturo Magidin: Yes, you are correct, my derivative was incorrect. I will work further with this information and report back, thanks. –  Finzz Mar 9 '11 at 5:38
    
(answered my own question) –  Finzz Mar 9 '11 at 5:48
    
@Finzz: No, the arc length formula doesn't have the $2\pi x$ in it. That factor comes from estimating the surface area of a "cylinder" you get by rotating around the $x$ axis. I can give you a derivation of the formulas if you prefer, though I suspect your textbook will have it. –  Arturo Magidin Mar 9 '11 at 5:51
    
Yeah I got mixed up between Area and Arc Length. –  Finzz Mar 9 '11 at 5:53
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@Finzz:Just algebra:$$1 + \frac{x^2}{a^2-x^2} = \frac{a^2-x^2}{a^2-x^2}+\frac{x^2}{a^2-x^2} = \frac{a^2-x^2+x^2}{a^2-x^2} = \frac{a^2}{a^2-x^2}.$$ –  Arturo Magidin Mar 9 '11 at 6:15
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