integrate $\int_D e^{x^2+3y^2}$ Evaluate $\int_D e^{x^2+3y^2}$, where $D$ is the region bounded in the first quadrant by the lines $y=0, y=x, x^2+3y^2=1$. 
My method is as follows, and I am not sure if it is correct. 
Let $u=x, v=\sqrt{3}y$. Then $D$ becomes bounded by $v=0, \frac{\sqrt{3}}{3}v=u, u^2+v^2=1$. 
$u=r\cos\theta, v=r\sin\theta$, so $\int_D e^{x^2+3y^2} = \int_D e^{r^2}r = \int^{\pi/6}_0e^{r^2}r$
Is this correct? If not, can you tell me where I got this wrong? Thank you. 
 A: Let $u=x$ and $v=\sqrt{3}y$. 
Then the three constraints become $v=0$, $v=\sqrt{3}u$, and $u^2+v^2=1$. Using the equations $x=u$ and $y=\frac{1}{\sqrt{3}}v$, 
we obtain the absolute value of the Jacobian to be: 
\begin{align*}
\Bigg| \frac{\partial(x,y)}{\partial(u,v)} \Bigg| &= 
\Bigg| 
\det
\begin{pmatrix}
x_u & x_v \\ 
y_u & y_v \\ 
\end{pmatrix}
\Bigg| 
= \Bigg| 
\det
\begin{pmatrix}
1 & 0 \\ 
0 & \frac{1}{\sqrt{3}} \\ 
\end{pmatrix}
\Bigg|  
= \frac{1}{\sqrt{3}}. 
\end{align*}
So we have 
\begin{align*}
\iint_{D} e^{x^2+3y^2}dA  
&= \int_{0}^{\frac{1}{2}}\int_{y}^{\sqrt{1-3y^2}} e^{x^2+3y^2}dxdy \\
&= \int_{0}^{\frac{\sqrt{3}}{2}}\int_{\frac{1}{\sqrt{3}}v}^{\sqrt{1-v^2}} e^{u^2+v^2}\Bigg| \frac{\partial(x,y)}{\partial(u,v)} \Bigg|dudv \\ 
&= \int_{0}^{\frac{\sqrt{3}}{2}}\int_{\frac{1}{\sqrt{3}}v}^{\sqrt{1-v^2}} e^{u^2+v^2}\frac{1}{\sqrt{3}} dudv \\ 
&= \frac{1}{\sqrt{3}} \int_{0}^{\frac{\pi}{3}} \int_{0}^{1} e^{r^2} r drd\theta \hspace{4mm}\mbox{ since } u = r \cos \theta\mbox{ and }v = r\sin \theta \\ 
&= \frac{1}{\sqrt{3}}\frac{\pi}{3}   \frac{1}{2}\int_{0}^{1} e^{w} dw \hspace{4mm}\mbox{ since } w=r^2, \mbox{ so }  dw = 2rdr  \\
&=\frac{\pi}{6\sqrt{3}}(e-1). 
\end{align*}
