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