March 15th, 2010, 02:36 AM  #1 
Newbie Joined: Feb 2010 Posts: 8 Thanks: 0  Double integral problem
Hi all, I'm presented with the following question  Using double integration calculate the area bounded by y=1/x, y=(9/2)x, x=1/2, y=1/2 I chose my outter integral limits to be 9/2x and 1/2 and my inner integral limits to be 1/2 and 1/y I'm not really sure if these are correct.... I ended up with 9/4 + (1/2)x + 1/[(92)x] + 3/8 I havent simplified that because it doesn't 'feel' correct! Thanks in advance 
March 16th, 2010, 01:43 AM  #2  
Newbie Joined: Mar 2010 Posts: 2 Thanks: 0  Re: Double integral problem Quote:
I assume, by 9/2x, that you mean (9/2)  x. Have you drawn a picture of the region? It should be the first thing you do with any double integral problem. When I draw a picture, I get a triangle with a curved bit cut off one corner. Whether you integrate wrt x first, then y or the other way round, you'll need to split the integral into two parts. For the first part, x goes from 1/2 to 2, and y from 1/x to 9/2x. For the second part, x goes from 2 to 4, and y from 1/2 to 9/2x. work out these two integrals, and add them....  
March 16th, 2010, 02:34 AM  #3 
Newbie Joined: Feb 2010 Posts: 8 Thanks: 0  Re: Double integral problem
Thanks for that DrMike! And for calculating the area of a circle x^2 + y^2 = 4y between Pi/3 and Pi/4 (using Polar Planes)... I get (8y^2).(Pi/3  Pi/4) is this correct? 

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