Surface Integrals – On a cone – Exercise 4068

Exercise

Calculate the integral

$\int\int_S (x^2+y^2+z^2) ds$

Where S is part of the cone

$z=\sqrt{x^2+y^2}$

That is on the circle

$x^2+y^2\leq 2x$

Final Answer

$\int\int_S (x^2+y^2+z^2) ds=3\sqrt{2}\pi$

Solution

Coming soon…

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