How to find the area inside $x^2-xy+y^2=1$?

Question

Find the area inside $x^2-xy+y^2=1$. Tip: $x^2-xy+y^2=\big(x-\frac{y}{2}\big)^2+\big(\frac{\sqrt 3y}{2}\big)^2$.

I thought change of variables would be appropriate here. $$ u=x-\frac{y}{2}\\ v=\frac{\sqrt 3y}{2} $$ Then we'd get $u^2+v^2=1$. This is a circle so its area is $\pi$. I guess we just need to multiply it by the Jacobian $J$ which is: $$ \frac{1}{J}=\begin{bmatrix} 1&-0.5\\0&\sqrt 3/2 \end{bmatrix}=\sqrt 3/2\implies J=2/\sqrt 3 $$ Then the final answer would be $2/\sqrt 3\cdot \pi$.

Really not sure if this is correct.

Answer 1

The answer is correct. An easier approach is to realize the shape is an ellipse--the area of which is $\pi a b$. Setting $ x=y$ will lead to a $a = \sqrt{2}$, setting $ x=-y$ will lead to $b = \sqrt{2}/\sqrt{3}$, so $A = 2\pi/\sqrt{3}$.

Answer 2

Another approach is to write the ellipse as a quadratic form:

$$\left[\begin{matrix}x & y\end{matrix}\right]\left[\begin{matrix}1 & -\frac12\\ -\frac12 & 1\end{matrix}\right]\left[\begin{matrix}x\\ y\end{matrix}\right]=1$$

You can calculate the eigenvalues $\lambda_1$ and $\lambda_2$ of this matrix, and the lengths of the semi-major and semi-minor axes $s_1=\frac1{\sqrt{\lambda_1}}$ and $s_2=\frac1{\sqrt{\lambda_2}}$.

The characteristic polynomial is $(1-\lambda)^2-\frac14$, so the eigenvalues are $\lambda_1=\frac12$ and $\lambda_2=\frac32$. That gives us semi-major and semi-minor axes $s_1=\sqrt2$ and $s_2=\frac{\sqrt2}{\sqrt3}$. The area is given by $A=\pi s_1 s_2=\frac{2\pi}{\sqrt3}$, which confirms your calculation.