Find the length of the curve from x=-2 to x=1 of $y=\sqrt{a^2-x^2}$ 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.
 A: 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).
