A Double Integral For Calculus I I stumbled upon this questions and really messed around with it for more than half an hour but did not get any where. I checked back and forth and do not think I copied the question incorrectly.  
Question:   Evaluate the double integral:   $\displaystyle \int_{1}^{2} \int_{\frac{1}{y}}^{y} (x^2+y^2)dxdy$ by using the substitution: $x = \frac{u}{v}, y = uv$. 
I drew the "region" bounded by the limits of the integral in both $xy$ and $uv$ coordinates but still don't see the "rectangle" that is supposed to come out. 

Wouldn't this just be $\int_1^2\int_{1/y}^y(x^2+y^2)dxdy=\int_1^2\frac{y^3}{3}+y^3-(\frac{3}{y^3}+y) dy$?
In doing so we're treating the inner integral as simplifying the area in a sense to something of a single variable. If I did the algebra correctly after full evaluation the integral should equal $\frac{263}{24}$.
