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.
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*}