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 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 9/2-x 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/[(9-2)x] + 3/8 I havent simplified that because it doesn't 'feel' correct! Thanks in advance
March 16th, 2010, 01:43 AM   #2
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Re: Double integral problem

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 Originally Posted by j_bloggs 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 9/2-x 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/[(9-2)x] + 3/8 I havent simplified that because it doesn't 'feel' correct! Thanks in advance

I assume, by 9/2-x, 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 9/2-x.
For the second part, x goes from 2 to 4, and y from 1/2 to 9/2-x.

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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