Given that $f$ is continuous and that the region $D$ (in the first quadrant) is bounded by $xy=1$, $xy=2$, $y=x$ and $y=4x$, I would like to show that $$\iint\limits_D f(xy) dxdy = \ln{2} \int_1^2 f(t)dt$$
The following is my attempted solution:
$$\iint\limits_D f(xy) dxdy = \int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}}\int_{\frac{1}{x}}^{4x}f(xy)dydx+\int_{\frac{1}{\sqrt{2}}}^{1}\int_{\frac{1}{x}}^{\frac{2}{x}}f(xy)dydx+\int_{1}^{\sqrt{2}}\int_{x}^{\frac{2}{x}}f(xy)dydx$$
Using the substitution $t=xy$, I have
$$\iint\limits_D f(xy) dxdy =\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}}\frac{1}{x}\int_{1}^{4x^2}f(t)dtdx+\int_{\frac{1}{\sqrt{2}}}^{1} \frac{1}{x} \int_{1}^{2}f(t)dtdx+\int_{1}^{\sqrt{2}} \frac{1}{x} \int_{x^2}^{2}f(t)dtdx$$
but am unable to go any further. What should be an appropriate next step and are there any errors above?
Best Answer
Hint: Using the change of variables rule in double integrals, we have \begin{gather*} \iint f( x,y) dxdy=\iint f( u,v) dudv\cdotp J\left(\frac{x,y}{u,v}\right)\\ where\ J\left(\frac{x,y}{u,v}\right) \ is\ the\ Jacobian\ of\ x,y\ wrt\ u,v. \end{gather*} For your integral over the area bounded by the curves $xy=1,xy=2,\frac{y}{x} =1,\frac{y}{x} =4$, make a change of variables $u=xy,\ v=\frac{y}{x}$, such that the area of integration becomes the rectangle $( u,v) =( 1,2) \times ( 1,4)$. Can you evaluate the Jacobian and finish it off?