Integral of the surface for the vector field $F (y, x, z) = (y, x, z)$ the surface $z = 9-x ^ 2-y ^ 2$ and the plane $z = 0$.

Asked Oct 12 '13 at 03:15

Active Oct 12 '13 at 03:15

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Calculate the integral of the surface facing positive, for the vector field $F (y, x, z) = (y, x, z)$ located on the surface $z = 9-x ^ 2-y ^ 2$ and the plane $z = 0$.

Well I try to solve this exercise with this integral

$\int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int_{0}^{9-x^2-y^2}1 dz dy dx=\frac{81\pi}{2}$

Am I ok??

asked Oct 12 '13 at 03:15

Salvattore

Yes, everything is correct. Congratulations! You could be more specific though: you are nto calculating the integral of the surface, but rather the flux of the vector field $F(x,y,z)$ over the surfaces. – Mark Fantini Dec 21 '13 at 06:31

Integral of the surface for the vector field $F (y, x, z) = (y, x, z)$ the surface $z = 9-x ^ 2-y ^ 2$ and the plane $z = 0$.

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