Spherical and Cylindrical Coordinates – On a cone – Exercise 4611

Exercise

Calculate the integral

Tx2+y2dxdydz\int\int\int_T \sqrt{x^2+y^2} dxdydz

Where T is bounded by the surfaces

z=1,x2+y2=z2z=1, x^2+y^2=z^2

Final Answer


Tx2+y2dxdydz=π6\int\int\int_T \sqrt{x^2+y^2} dxdydz=\frac{\pi}{6}

Solution

Coming soon…

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