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Double integral-semi circle

  • #1
if the quest is like this:

Show that double integral over R ( r2 sin(theta)) dr d(theta), where R is the region bounded by the semicircle r=2acos(theta), ABOVE THE INITIAL LINE....


???? theta varies from.....????

finally after 1st integration I got the value as
integral of___ to ___ -[(8a^3)/3] * [((cos^4)theta]....

Am I right till this step?...if not, please correct!


Thanks in advance
AG:approve:
 

Answers and Replies

  • #2
HallsofIvy
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if the quest is like this:

Show that double integral over R ( r2 sin(theta)) dr d(theta), where R is the region bounded by the semicircle r=2acos(theta), ABOVE THE INITIAL LINE....
Show that the double integral [/b]what[/b]? The predicate of your sentence is missing! Do you just mean evaluate the double integral?


???? theta varies from.....????

finally after 1st integration I got the value as
integral of___ to ___ -[(8a^3)/3] * [((cos^4)theta]....

Am I right till this step?...if not, please correct!


Thanks in advance
AG:approve:
That figure is the circle with center at (a, 0) and radius a. The entire circle is swept out as [itex]\theta[/itex] goes from 0 to [itex]2\pi[/itex]. Since r= 2a(cos(0))= 2a, the initial point is (2a, 0) and the "initial line" is the x-axis. "above the initial line" is the upper half of the circle which is swept out as [itex]\theta[/itex] goes from 0 to [itex]\pi[/itex].
You want
[tex]\int_{\theta= 0}^\pi \int_{r= 0}^{2acos(\theta)} r^2 sin(\theta)dr d\theta[/itex]

As for what you have done already, since there is a "[itex]sin(\theta)[/itex]" in your integrand, I don't see how integrating with respect to r could get rid of that!
 

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