Let D be the bounded region in the first quadrant of $R^2$ bounded by curves $x^2 + 16y^2 = 16, x^2 + 16y^2 = 1, x = y$ and the positive y−axis. Describe $D$ in the $(u, v)$−plane when $u = x^2 + 16y^2$ and $v =\frac yx,$ and calculate the integral $$\iint_{D} \frac{y}{x} dA$$
So in order to solve the integral I used the new coordinates $1\leq u \leq 16$ and $1\leq v \leq \infty$. Afterwards I calculated the the determinant of the jacobian and got $$ |\det J| = 1+32\left(\frac{y}{x}\right)^2 = 2 + 32v^2$$ Then I did: $$\iint_{D} v (2 + 32v^2) dudv$$ But I got the wrong answer and instead in the answers they divided the determinant of jacobian. $$\iint_{D} \frac{v}{(2 + 32v^2)} dudv$$
Can someone explain why?
Given the change of variables $u=x^2+16y^2$ and $v=y/x$, in your exercise you should consider the Jacobian of the transformation $\Phi$ from $(u,v)$ to $(x,y)$: $$|\det(J_{\Phi}(u,v))|=\Big|\det\Big(\frac{\partial(x,y)}{\partial(u,v)}\Big)\Big|=\begin{vmatrix} \frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial v}& \frac{\partial y}{\partial v} \end{vmatrix}.$$ Instead you computed the determinant of the Jacobian of the inverse transformation $\Phi^{-1}$: $$|\det(J_{\Phi^{-1}}(x,y))|=\Big|\det\Big(\frac{\partial(u,v)}{\partial(x,y)}\Big)\Big|=\begin{vmatrix} \frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y} \end{vmatrix}=\begin{vmatrix} 2x& 32y\\ -y/x^2& 1/x \end{vmatrix}=2+32y^2/x^2$$ which is, due to the chain rule, the reciprocal of $|\det(J_{\Phi}(u,v))|$ (see also Show that the product of the Jacobian and the inverse Jacobian is 1 ). Hence, $$\iint_D\frac{y}{x} dxdy =\iint_{\Phi^{-1}(D)}\frac{y/x}{2+32y^2/x^2} dudv =\int_{1}^{16}\left(\int_{1}^{\infty}\frac{v}{2+32v^2} dv\right)du.$$