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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 Tags double, integral, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post zollen Calculus 1 October 31st, 2017 05:40 PM harley05 Calculus 5 May 18th, 2014 07:51 PM krampon Calculus 1 July 1st, 2013 07:21 AM Beevo Calculus 2 November 26th, 2012 02:44 PM j_bloggs Calculus 2 March 16th, 2010 01:34 AM

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