<s>[INST] Measure of a subset of $\mathbb{R}^2$ I'd like to know whether the following integral is right or wrong. I have to calculate the measure of $A$, where
$$
A=\left \{ (x,y): \frac{x^2}{3}+y^2\leqslant 1, \, x\geqslant 0, \, -x\leqslant y\leqslant x \right \}
$$
using polar coordinates $\Phi:\left \{ x= \sqrt{3}\rho \cos \theta, y= \rho \sin \theta \right \}$, the polar region is
$B=\left \{ (\rho, \theta): 0 \leqslant \rho \leqslant 1 , -\frac{\pi}{4} \leqslant \theta \leqslant \frac{\pi}{4}  \right \}$, $A=\Phi (B)$ and $|\det J_\Phi|=\sqrt{3} \rho$. Then,
$$
m(A)=2\int_{0}^{\frac{\pi}{4}}\int_{0}^{1}\sqrt{3}\rho \, d\rho d\theta=\int_{0}^{\frac{\pi}{4}}\sqrt{3} \, d \theta=\frac{\sqrt{3}}{4}\pi
$$
The answer would be $\frac{\pi}{\sqrt{3}}$ but I think it's incorrect.
 [/INST] I don't believe $B$ is correctly defined. $\theta$ should run between $-\pi/3$ and $\pi/3$. Notice that for $\theta=\pi/3$, $x=\frac{\sqrt{3}}{2}\rho$ and $y=\frac{\sqrt{3}}{2}\rho$.
Here is another approach to verify the answer. You want to intersect two regions
1)the interior of the ellipse $x^2/3+y^2=1$ 
2)the cone about the positive $x$-axis bounded by $y=x$ and $y=-x$ 
The region is symmetric about the $x$-axis, so we can focus on the first quadrant.
Considering $y$ as the independent variable, the ``top curve'' is the ellipse 
$x=\sqrt{3-3y^2}$ and the bottom curve is $x=y$.
Therefore the area is
\begin{align}
m(A) 
&= 
2\int_{0}^{\sqrt{3}/2} \sqrt{3-3y^2} - y\ dy\\
&=
2\sqrt{3}\int_{0}^{\sqrt{3}/2} \sqrt{1-y^2}\ dy - 2\int_{0}^{\sqrt{3}/2} y\ dy\\
&=
2\sqrt{3}\int_{0}^{\sqrt{3}/2} \sqrt{1-y^2}\ dy - \frac{3}{4}
\end{align}
After a trig substitution $y=\sin(t)$, we have
\begin{align}
m(A) 
&= 
2\sqrt{3}\int_{0}^{\pi/3} \sqrt{1-\sin^2(t)}\cos(t)\ dt - \frac{3}{4}
\\
&= 
2\sqrt{3}\int_{0}^{\pi/3}  \cos^2(t)dt- \frac{3}{4}\\
&=
\left(\frac{\pi}{\sqrt{3}} + \frac{3}{4}  \right) -\frac{3}{4}\\
&=
\frac{\pi}{\sqrt{3}}
\end{align}
</s>