My Math Forum double integral problem #2

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 November 2nd, 2017, 03:00 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 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 $\displaystyle r = a \cos \theta$.
 November 3rd, 2017, 03:24 PM #2 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 Here are my solutions in both coordinates. Please let me know if I set correctly. Thanks. $\displaystyle \int_{0}^{a} \int_{- \sqrt{ ( \frac {a}{2} )^2 - ( u - \frac {a}{2} )^2 } }^{\sqrt{ ( \frac {a}{2} )^2 - ( u - \frac {a}{2} )^2 }} \frac {a}{ \sqrt{ a^2 - u^2 - v^2 } } ~dv ~du$ $\displaystyle \int_{0}^{2 \pi} \int_{0}^{a * cos \theta } \frac {a}{ \sqrt{a^2 - r^2} } ~dr ~d \theta$ Last edited by zollen; November 3rd, 2017 at 03:28 PM.
 November 3rd, 2017, 10:04 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,468 Thanks: 1342 bottom one looks correct except it should be $r~dr~d\theta$ Thanks from zollen
November 3rd, 2017, 11:44 PM   #4
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Quote:
 Originally Posted by romsek bottom one looks correct except it should be $r~dr~d\theta$
actually also the limits on $\theta$ should be $-\dfrac \pi 2 \to \dfrac \pi 2$

 November 3rd, 2017, 11:48 PM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 2,468 Thanks: 1342 the first one is correct. I get $(\pi -2) a^2$ for both of them with the two corrections I listed for the 2nd. Thanks from zollen

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