Question

Set up a double integral for calculating the flux of the vector field ????⃗ (????⃗ )=????⃗ , where ????⃗ =⟨x,y,z⟩, through the part of the upward oriented surface z=3(x2+y2) that lies above the disk x2+y2≤25.

Answer 1

Answer:

-937.5π

Step-by-step explanation:

F (r) = r = (x, y, z) the surface equation z = 3(x^2 + y^2) z_x = 6x, z_y = 6y the normal vector n = (- z_x, - z_y, 1) = (- 6x, - 6y, 1)

Thus, flux ∫∫s F · dS is given as;

∫∫ <x, y, z> · <-z_x, -z_y, 1> dA

=∫∫ <x, y, 3x² + 3y²> · <-6x, -6y, 1>dA , since z = 3x² + 3y²

Thus, flux is;

= ∫∫ -3(x² + y²) dA.

Since the region of integration is bounded by x² + y² = 25, let's convert to polar coordinates as follows:

∫(θ = 0 to 2π) ∫(r = 0 to 5) -3r² (r·dr·dθ)

= 2π ∫(r = 0 to 5) -3r³ dr

= -(6/4)πr^4 {for r = 0 to 5}

= -(6/4)5⁴π - (6/4)0⁴π

= -937.5π