This is a different answer to what you originally posted but the working looks fine to me but you would be better off making sure someone with more recent experience check it too. Check your private messages (PMs) I'll send you one in a sec to give you someone to ask
For part b I think the result is the sum of 2 integrals, as you must integrate over two symmetric but separate regions. I believe the limits for r are from 1/cos(theta) to infinite (and are the same for each integral) and the limits for theta should be -Pi/4 to Pi/4 for one integral and 3Pi/4 to 5Pi/4 for the other.
However, because of the symmetry, both integrals give the same result, so you can just write the answer as 2 times either of them.
2 * Integral (from Theta=-Pi/4 to Pi/4), (from r=1/cos(theta) to infinite)
Last edited by krtxmrtz; Aug 23rd, 2006 at 05:45 AM.
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If I'm not mistaken, this is the integration region for part b.
Lottery is a tax on people who are bad at maths
If only mosquitoes sucked fat instead of blood...
To do is to be (Descartes). To be is to do (Sartre). To be do be do (Sinatra)
This is just a plot showing the integration region. This region "exists" independently of the way you plot it. In fact, using rectangular, polar, elliptic or any other type of coordinates are just procedures for plotting (describe) it.
Lottery is a tax on people who are bad at maths
If only mosquitoes sucked fat instead of blood...
To do is to be (Descartes). To be is to do (Sartre). To be do be do (Sinatra)
is it the shaded region that is to be integrated over or excluded? (i thought excluded...)
I've just realized that's |x|<=1 and I had assumed the other way round, ">=" (sheer absent-mindedness, actually). So definitely the region to integrate over is that constituted by the 4 triangles in between the 2 shaded regions.
The limits for theta are ok but r goes from 0 to 1/cos(theta). The argument about integrating for theta only from -Pi/4 to +Pi/4 and multiplying the result by 2 is still valid.
Sorry if I've been leading you astray for awhile.
Lottery is a tax on people who are bad at maths
If only mosquitoes sucked fat instead of blood...
To do is to be (Descartes). To be is to do (Sartre). To be do be do (Sinatra)
I've just realized that's |x|<=1 and I had assumed the other way round, ">=" (sheer absent-mindedness, actually). So definitely the region to integrate over is that constituted by the 4 triangles in between the 2 shaded regions.
The limits for theta are ok but r goes from 0 to 1/cos(theta). The argument about integrating for theta only from -Pi/4 to +Pi/4 and multiplying the result by 2 is still valid.
Sorry if I've been leading you astray for awhile.
thats fine i think i have got it now, thanks for all your help
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Originally Posted by krtxmrtz
(sheer absent-mindedness, actually). So definitely the region to integrate over is that constituted by the 4 triangles in between the 2 shaded regions.
