Here's a more general version of user11667's answer that yields the discriminant formula in Chris Taylor's answer. In particular, I'd like to explain how the fact that a coordinate transform as seen in user11667's answer exists can be used to get an area formula of the form $\frac{2\pi}{\sqrt{4ac-b^2}}$ (in this case, $\frac{2\pi}{\sqrt{3}}$ for the whole ellipse, and $\frac{\pi}{\sqrt{3}}$ for the part above the $x$-axis).
If you have an ellipse described by the equation $ax^2 + bxy + cy^2 = 1$, the Principal Axis Theorem asserts that you can always change coordinates to diagonalize the quadratic form $ax^2 + bxy + cy^2$. That is, there are always variables $u$ and $v$ that are linear combinations of $x$ and $y$ such that substitution yields the equation $u^2 + v^2 = 1$. The area of the ellipse is then given dividing $\pi$ by the determinant of the linear transformation, i.e., the extra factor you see when changing variables for double integrals.
The coordinate change is given by a change of bases to the principal axes of the quadratic form, and you can find a description of an explicit algorithm in the Wikipedia article linked in the previous paragraph. The upshot is roughly as follows: We define the matrix $A = \binom{\,a \quad b/2}{b/2 \quad c\,}$, so the form $ax^2 + bxy + cy^2$ can be written as the matrix product $(x \quad y) A \binom{x}{y}$. The Principal Axis Theorem asserts that we can find a matrix $P$ satisfying $P^T \binom{10}{01}P = A$. Therefore, we can make a coordinate change $\binom{u}{v} = P\binom{x}{y}$ to get the equality $ax^2 + bxy + cy^2 = (x \quad y) A \binom{x}{y} = (u \quad v)\binom{10}{01} \binom{u}{v} = u^2 + v^2.$
The area enclosed by the ellipse is then $\frac{1}{\det(P)}$ times the area enclosed by a unit circle, that is, $\frac{\pi}{\det(P)}$.
In the case of the problem at hand, the symmetry between $x$ and $y$ implies the principal axes are on the lines $y = \pm x$, so you don't even need to execute the full algorithm to find user11667's variable change. Furthermore, you don't even need to compute the explicit coordinate change to find the area, because $\det(A) = \det(P^T)\det(P) = \det(P)^2$. That is, we need only divide by the square root of $\det A = ac-\frac{b^2}{4}$. The area of the ellipse is then $\frac{pi}{\sqrt{ac-b^2/4}} = \frac{2\pi}{\sqrt{4ac-b^2}}$.
This method works in higher dimensions: if you have an $n$-dimensional ellipsoid centered at the origin, defined by $q(\vec{x}) = 1$ for $q$ a (positive definite) quadratic form in $n$ variables, you can convert that form into a symmetric matrix $A$. The volume enclosed by the ellipsoid is then given by dividing the volume of the unit $n$-ball by $\sqrt{\det(A)}$. (If you want to know a formula for the volume of the unit $n$-ball I have a short derivation on MathOverflow.)