Double integral

Double integral

Postby magben » Wed Oct 13, 2021 11:29 am

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


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

[tex]\int \int\limits_{S} 2( x^{2}+ y^{2 }-2) \sqrt{(2x)^{2}+(2y)^{2}+1} dA[/tex]

[tex]\int \int\limits_{S} 2( x^{2}+ y^{2 }-2) \sqrt{4x^{2}+4y^{2}+1} dA[/tex]

Polar Coordinates:

[tex]x=rcos\theta[/tex]
[tex]y=rsin\theta[/tex]

[tex]dA=rd_rd\theta[/tex]


[tex]\int\limits_{0}^{2\pi} \int\limits_{0}^{\sqrt{2}}2( r^{2}-2) \sqrt{4r^{2}-1}rdrd\theta[/tex]

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