Is the Calculation of the Double Integral Over a Semicircle Correct?

[aspiring_gal]

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

 

[HallsofIvy]

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

That figure is the circle with center at (a, 0) and radius a. The entire circle is swept out as \theta goes from 0 to 2\pi. 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 \theta goes from 0 to \pi.
You want
\int_{\theta= 0}^\pi \int_{r= 0}^{2acos(\theta)} r^2 sin(\theta)dr d\theta[/itex]<br /> <br /> As for what you have done already, since there is a &quot;sin(\theta)&quot; in your integrand, I don&#039;t see how integrating with respect to r could get rid of that!