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November 2nd, 2017, 04:00 PM   #1
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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 $.
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November 3rd, 2017, 04:24 PM   #2
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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 04:28 PM.
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November 3rd, 2017, 11:04 PM   #3
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bottom one looks correct except it should be $r~dr~d\theta$
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November 4th, 2017, 12:44 AM   #4
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Quote:
Originally Posted by romsek View Post
bottom one looks correct except it should be $r~dr~d\theta$
actually also the limits on $\theta$ should be $-\dfrac \pi 2 \to \dfrac \pi 2$
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November 4th, 2017, 12:48 AM   #5
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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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