October 31st, 2017, 03:21 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 179 Thanks: 2  double integral problem
Find the area of z = x^2  y^2 that lies inside x^2 + y^2 = a^2. Here is my solution. But I am stuck here because I don't know how to calculate this mess... $\displaystyle x = u, y = v, z = u^2v^2 $ $\displaystyle \int_{a}^{a} \int_{ \sqrt{a^2u^2}}^{ \sqrt{a^2u^2}} \sqrt{4u^2+4v^2+1} ~dv ~du $ 
October 31st, 2017, 05:40 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,009 Thanks: 1042 
you've confused yourself using u's and v's but otherwise what you've done is correct. however.... when you see $x^2 + y^2 = a^2$ you should immediately be thinking cylindrical coordinates. here we have $\displaystyle \int_0^{2\pi}\int_0^a~\sqrt{4r^2 + 1}~r~dr~d\theta$ and this is trivial to integrate. 

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