Double integral with polar coordinates

• Apr 19th 2011, 07:59 PM
tangibleLime
Double integral with polar coordinates
Evaluate the given integral by changing to polar coordinates.

...where D is the region bounded by the semicircle http://www.webassign.net/latexImages...a5dd9384b0.gif and the y-axis.

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I do all my scrap work on a tablet, so here's what I came up with. I did all of the integration in WolframAlpha just to make sure it was correct (did it manually the first three times). What am I doing wrong?

http://mikederoche.com/temp/Untitled.png

Any help or hints would be appreciated!

Thanks.
• Apr 19th 2011, 08:15 PM
TaylorM0192
The limits of integration are incorrect. Theta should be from -pi/2 to pi/2 as the region of integration is bounded by x = sqrt(49-y^2) and x = 0 (i.e. the right hemicircle of radius 7).

EDIT:

Separating the integration with respect to theta from r, we obtain immediately theta evaluated from -pi/2 to pi/2 which is equal to pi. Evaluating the integration with respect to r, not forgetting the Jacobian factor r, we obtain after the substitution u = -r^2, change of limits (including reversal due to the introduced negative sign): (pi)(1/2)(e^u) | (u1 = -49, u2 = 0) which gives (pi/2)(1-e^-49).
• Apr 20th 2011, 03:23 PM
SammyS
Your work looks fine to me.

The limits of integration are not quite right as pointed out by TaylorM0192.

The integration over r looks fine to me.
• Apr 20th 2011, 03:44 PM
tangibleLime
Thanks, both of you, for your assistance! From your advice I used all of my previous work but just changed the limits of integration on theta to (-pi/2) to (pi/2) and it worked like a charm.

Cheers!