October 31st, 2017, 03:21 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 193 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: 2,120 Thanks: 1101 
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. 

Tags 
double, integral, problem 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
double integral problem  harley05  Calculus  5  May 18th, 2014 07:51 PM 
Substitution problem (double integral)  unwisetome3  Calculus  0  May 13th, 2014 05:49 AM 
Evaluate Double Integral ( LOG PROBLEM ! )  krampon  Calculus  1  July 1st, 2013 07:21 AM 
Double Integral Problem  Beevo  Calculus  2  November 26th, 2012 02:44 PM 
Double integral problem  j_bloggs  Calculus  2  March 16th, 2010 01:34 AM 