How to determine the area of a rotated ellipse? The ellipse $6x^2+4xy+5y^2+8x+8y+1=0$ is neither expressed in terms of $x$; like $y=\pm\sqrt{a^2-x^2}$, nor in terms of $y$; like $x=\pm\sqrt{a^2-y^2}$.
Separation of $x$ (or $y%$) may be impossible.
I was thinking to let $x=r\cos(\theta)$, and $y=r\sin(\theta)$, and then integrating, but finding the limits of integration [with respect to $\theta$] is difficult, or may be impossible.
How to determine the area of the ellipse shown below?

 A: Consider the ellipse $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$. It is well known that (1) this conic is an ellipse iff $AB - H^2 > 0$, so that we can assume $A > 0$; and (2) assuming $A > 0,$ the ellipse has real points iff $L > 0$, where
$$L = AF^2 + BG^2 + CH^2 - ABC - 2FGH.$$
To find area of the ellipse, one can apply an affine transformation that doesn't change the area. Such a transformation is $x = \lambda y + z$ where $\lambda$ is any constant.
This gives 
$$(A\lambda^2 + 2H\lambda + B)y^2 + 2(A\lambda + H)yz + Az^2 +2(G\lambda + F)y + 2Gz + C = 0.$$
Get rid of $yz$ by choosing $\lambda = -H/A.$ Then
$$(AB - H^2)y^2 + 2(FA - GH)y + A^2z^2 + 2GAz + AC = 0,$$
that is
$$(AB - H^2)\left(y - {{(AF - GH)}\over{AB - H^2}}\right)^2 + (Az + G)^2 - {{(AF - GH)^2}\over{AB - H^2}} - G^2 + AC = 0.$$
Moving the centre of the ellipse to the origin, this can be put as
$$(AB - H^2){y_1}^2 + A^2{z_1}^2 - K = 0,$$
where
$$K = {1\over{(AB - H^2)}}\left[ (AF - GH)^2 + (AB - H^2)(G^2 - AC)\right],$$
which reduces to $K = LA/(AB - H^2)$ (with $L$ as above). Hence the semi-axes of the transformed ellipse are
$a = \sqrt{K/(AB - H^2)}$ and $b = \sqrt{K}/A,$ and the required area is $\pi ab = \pi L/(AB - H^2)^{3/2}.$
In particular, the ellipse in the question has area $86\pi/26^{3/2}.$
A: Write the equation as:
$$\begin{pmatrix}x & y & 1\end{pmatrix}
\left(\begin{array}{ccc|c} \\&Q& &P \\ \\ \hline &P^T& &R \end{array}\right)
\begin{pmatrix}x \\ y \\ 1\end{pmatrix}
=\begin{pmatrix}x & y & 1\end{pmatrix}
\left(\begin{array}{cc|c}6 & 2 & 4 \\ 2 & 5 & 4 \\ \hline 4&4&1 \end{array}\right)
\begin{pmatrix}x \\ y \\ 1\end{pmatrix} = 0$$
Set the gradient equal to zero to find the translation (same as solving $Qx=-P$):
$$\nabla(6x^2+4xy+5y^2+8x+8y+1)=0 \Rightarrow \begin{cases}12x+4y+8=0\\4x+10y+8=0\end{cases} \Rightarrow x=-\frac 6{13}, y=-\frac 8{13}$$
So the translation vector $t=-\frac 1{13}\begin{pmatrix}6\\8\end{pmatrix}$.
Divide the equation by $-(P\cdot t + R)$ to normalize it. The resulting $Q'$ is:
$$Q'=-\frac{1}{P\cdot t + R}Q$$
Its determinant is the product of the eigenvalues, which are $\frac 1{a^2}$ and $\frac 1{b^2}$.
So:
$$\det Q' = \frac{1}{a^2b^2}$$
The area of an ellipse is $\pi a b$ so that:
$$\text{Area} = \pi a b = \frac{\pi}{\sqrt{\det Q'}} = \frac{\pi\cdot |P\cdot t+R|}{\sqrt{|\det Q|}}
=\frac{\pi\cdot |-\frac 1{13}(4\cdot 6 + 4\cdot 8)+1|}{\sqrt{|6\cdot 5-2\cdot 2|}}
=\frac{43\pi}{13\sqrt{26}}$$
The general formula is:
$$\text{Area}=\frac{\pi|P^TQ^{-1}P-R|}{\sqrt{|\det Q|}}$$
A: The simplest way to compute the area is that of using the formulas given here to find semi-axes $a$ and $b$ and then area $=\pi ab$.
If you insist on using an integral, you can of course solve for $y$ to find two solutions 
$$
y_\text{up/down}={1\over5}\left(
-2x-4\pm\sqrt{-26x^2-24x+3}
\right)
$$
and then compute the area as
$$
\int_{x_\text{left}}^{x_\text{right}}(y_{up}-y_{down})\,dx,
$$
where $x_\text{left/right}$ are the values of $x$ giving a vanishing square root.
A: For an ellipse rotated by the angle $\theta$, this should work.
$$I (\theta) = Ixx Cos^2(\theta) + Iyy Sin^2(\theta)$$
