I have , = second quadrant of
I changed to polar coordinates:
Do I evaluate the integral like this?
Last edited by billym; November 13th 2009 at 05:31 AM.
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I think your boundaries are correct.
Don't forget about that extra r when using polar integration.
Should the answer be the same as or am I missing the point here?
Or rather, that is R. Which is basically the upper left portion of a doughnut. But what does the area of integration look like?
Pardon me for being the village idiot, but could somebody please answer this one question for me:
When sketching the area of integration, do I just sketch , or do I have to consider ?
Originally Posted by billym Should the answer be the same as or am I missing the point here? The answer shouldn't have pi in it. .
yeah I realized that last night. Anyway, I've got the right answer now. Thanks dudes.
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