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:

-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

Final answer:

To set up a doubleintegral for calculating the flux of the vector field through the given surface, parameterize the surface using the equation provided and the given condition. Calculate the cross product of the partial derivatives of x and y to find the normal vector. Finally, set up the double integral for the flux using the vector field and the normal vector.

Explanation:

To set up a double integral for calculating the flux of the vector field through the given surface, we first need to parameterize the surface. Given the equation of the surface z = 3(x^2 + y^2) and the condition x^2 + y^2 ≤ 25, we can parameterize the surface as follows:

x = rcosθ, y = rsinθ, z = 3r^2

We can now calculate the cross product of the partialderivatives of x and y to find the normal vector, which is: n = (3rcosθ, 3rsinθ, 1)

Finally, the double integral for calculating the flux through the surface is:

∬ F · n dA = ∬ (x, y, z) · (3rcosθ, 3rsinθ, 1) dA

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