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-   -   double integral problem #2 (http://mymathforum.com/calculus/342613-double-integral-problem-2-a.html)

zollen November 2nd, 2017 04:00 PM

double integral problem #2
 
Find the area of $\displaystyle x^2 + y^2 + z^2 = a^2 $ that lies above the interior of the circle given in polar coordinates by $\displaystyle r = a \cos \theta $.

zollen November 3rd, 2017 04:24 PM

Here are my solutions in both coordinates. Please let me know if I set correctly. Thanks.

$\displaystyle
\int_{0}^{a} \int_{- \sqrt{ ( \frac {a}{2} )^2 - ( u - \frac {a}{2} )^2 } }^{\sqrt{ ( \frac {a}{2} )^2 - ( u - \frac {a}{2} )^2 }} \frac {a}{ \sqrt{ a^2 - u^2 - v^2 } } ~dv ~du
$

$\displaystyle
\int_{0}^{2 \pi} \int_{0}^{a * cos \theta } \frac {a}{ \sqrt{a^2 - r^2} } ~dr ~d \theta
$

romsek November 3rd, 2017 11:04 PM

bottom one looks correct except it should be $r~dr~d\theta$

romsek November 4th, 2017 12:44 AM

Quote:

Originally Posted by romsek (Post 583534)
bottom one looks correct except it should be $r~dr~d\theta$

actually also the limits on $\theta$ should be $-\dfrac \pi 2 \to \dfrac \pi 2$

romsek November 4th, 2017 12:48 AM

the first one is correct.

I get $(\pi -2) a^2$ for both of them with the two corrections I listed for the 2nd.


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