Question

Find \( \int \int_SF.Nds \) , that is find the flux of F across S. If S is closed , use the positive (outward) orientation.

\( F(x,y,z)=(2x,2y,z);S \) is the part of the paraboloid \( z=4-x^2-y^2 \) above the xy-plane ; N point upward.

Answer 1

Solution

we have \( F(x,y,z)=(2x,2y,z);S \) is the part of the paraboloid 

\( z=4-x^2-y^2 \) above the xy-plane ; N point upward.

we have \( F(x,y,z)=(2x,2y,z) \) where M=2x,N=2y,P=z

Let's z=g(x,y)=\( 4-x^2-y^2=>g_x(x,y)=-2x,g_y(x,y)=-2y \)

=>\( \int \int_sF.Nds=\int \int_R\big(-Mg_x-Ng_y+P\big)dA \)

                          \( = \int \int_R\big(4x^2+4y^2+4-x^2-y^2 \big)dA \)

                          \( =\int \int_R\big(3(x^2+y^2)+4\big)dA \)

We use polar coordinate where \( 0 \leq \theta \leq2\pi,0 \leq r \leq2 \)

                          \( =\int_0^{2\pi} \int_0^2\big(3r+4\big)rdrd\theta=40\pi \)

Answer

Therefore, \( \int \int_SF.Nds =40\pi \)