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

 zollen November 2nd, 2017 03:00 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 03:24 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 10:04 PM

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

 romsek November 3rd, 2017 11:44 PM

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 3rd, 2017 11:48 PM

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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