It's about finding the area of the paraboloid z=x²+y² beneath z=2. The area integral is(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int\int_{\{(u,v): u^2+v^2<2\}}\sqrt{1+4(u^2+v^2)}dudv[/tex]

A polar change of variable seems to fits nicely:

[tex]=\int_0^{\sqrt{2}}\int_0^{2\pi}\sqrt{1+4r^2}rd\theta dr[/tex]

Then the change of variable [itex]\xi=1+4r^2[/itex] brings the area to

[tex]2\pi\int_1^9\sqrt{\xi}\frac{d\xi}{8}=\frac{2\cdot 2 \pi}{3 \cdot 8}[\xi^{3/2}]_1^9=\frac{13\pi}{3}[/tex]

Looks good?

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# A double integral

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