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I am a bit stuck over here I need to find the area between

$x^2+y^2\leqslant4$ and $z=\frac{1}{8}(x^2-y^2)$

I have tried to calculate it by integrating $z$ over $x,y$ and transforming the whole thing in polar coordinates with radius from $(-2,2)$ and angle $(0,2\pi)$ but I am getting zero but I have plotted the two graphs and they intersect so they should have an area (I believe so).

Thanks a lot !

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    What value of $z$ are you using? – 2017-01-25
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    I was integrating over z, I mean I was rewriting it something like this Integral(1/8*(r^3/)*cos2(x)) from (r,--2,2),(x,0,2pi). the way I wrote it (up) is exactly how the problem was given to me no extra information – 2017-01-25

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