# Double integral

### Double integral

Calculate $$\int \int\limits_{S} 2zdS$$, where $$S$$ is the part of the paraboloid $$z^{2}+ y^{2}-2$$ below the $$xy$$ plane.

Guys, I don't know how to resolve this issue. I managed to do the following.

$$\int \int\limits_{S} 2( x^{2}+ y^{2 }-2) \sqrt{(2x)^{2}+(2y)^{2}+1} dA$$

$$\int \int\limits_{S} 2( x^{2}+ y^{2 }-2) \sqrt{4x^{2}+4y^{2}+1} dA$$

Polar Coordinates:

$$x=rcos\theta$$
$$y=rsin\theta$$

$$dA=rd_rd\theta$$

$$\int\limits_{0}^{2\pi} \int\limits_{0}^{\sqrt{2}}2( r^{2}-2) \sqrt{4r^{2}-1}rdrd\theta$$

Can anyone help me resolve this issue or present me with a different resolution?
magben

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