Calculating Triple Integrals – Bounded by surfaces – Exercise 4559

Exercise

Calculate the integral

\int\int\int_T y\cos(z+x) dxdydz

Where T is bounded by the surfaces

y=\sqrt{x},y=0,z=0,x+z=\frac{\pi}{2}

Final Answer


\int\int\int_T y\cos(z+x) dxdydz=\frac{{\pi}^2}{16}-\frac{1}{2}

Solution

Coming soon…

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