Compute the flux of the vector field F rightarrow = 8x^2 y^2 zk rightarrow throu

Question

Compute the flux of the vector field F rightarrow = 8x^2 y^2 zk rightarrow through the surface S which is the cone squareroot x^2 + y^2 = z, with 0 lessthanorequalto z lessthanorequalto R, oriented downward. Parameterize the cone using cylindrical coordinates (write theta as theta). x(r, theta) = y(r, theta) = z(r, theta) = with lessthanorequalto r lessthanorequalto and lessthanorequalto theta lessthanorequalto With this parameterization, what is dA rightarrow? dA rightarrow = Find the flux of F rightarrow through S. flux =

Explanation / Answer

z = sqrt(x^2+y^2) and z [0 , R]

and in cylindercal coordinate system :

x=rcos(theta) , y = rsin(theta) and z = z and r= sqrt(x^2+y^2)

=> r = sqrt(x^2+y^2) = z = 0 and r = sqrt(x^2+y^2) = z = R

hence the bounds for r is r E [0 , R]

and x = Rcos(theta) , y = Rsin(theta) , z = z

and theta E [0 , 2pi]

b> dA = rdrdzd(theta)

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