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October 31st, 2017, 04:21 PM   #1
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double integral problem

Find the area of z = x^2 - y^2 that lies inside x^2 + y^2 = a^2.

Here is my solution. But I am stuck here because I don't know how to calculate this mess...

$\displaystyle
x = u, y = v, z = u^2-v^2
$

$\displaystyle
\int_{-a}^{a} \int_{- \sqrt{a^2-u^2}}^{ \sqrt{a^2-u^2}} \sqrt{4u^2+4v^2+1} ~dv ~du
$
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October 31st, 2017, 06:40 PM   #2
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you've confused yourself using u's and v's but otherwise what you've done is correct.

however.... when you see $x^2 + y^2 = a^2$ you should immediately be thinking cylindrical coordinates.

here we have

$\displaystyle \int_0^{2\pi}\int_0^a~\sqrt{4r^2 + 1}~r~dr~d\theta$

and this is trivial to integrate.
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