evaluate the surface integral ∫sf⋅ ds where f=⟨−4x,−3z,3y⟩ and s is the part of the sphere x2 y2 z2=16 in the first octant, with orientation toward the origin.∫∫SF⋅ dS=∫∫SF⋅ dS=

The value of the surface integral ∫sf⋅ ds over the given surface S is 2√2.

To evaluate the surface integral ∫sf⋅ ds, we first need to parameterize the surface S which is the part of the sphere [tex]x^{2}[/tex]+[tex]y^{2}[/tex]+[tex]z^{2}[/tex]=16 in the first octant.

One possible parameterization of S is:

x = r sinθ cosφ

y = r sinθ sinφ

z = r cosθ

where 0 ≤ θ ≤ π/2 and 0 ≤ φ ≤ π/2.

Next, we need to find the unit normal vector to the surface S. Since the surface is oriented toward the origin, the unit normal vector points in the opposite direction of the gradient vector of the function [tex]x^{2}[/tex]+[tex]y^{2}[/tex]+[tex]z^{2}[/tex]=16 at each point on the surface S.

∇( [tex]x^{2}[/tex]+[tex]y^{2}[/tex]+[tex]z^{2}[/tex]) = ⟨2x,2y,2z⟩

So, the unit normal vector to the surface S is

n = -⟨x,y,z⟩/4 = -⟨r sinθ cosφ, r sinθ sinφ, r cosθ⟩/4

Now, we can evaluate the surface integral using the parameterization and unit normal vector:

∫sf⋅ ds = ∫∫S f⋅n dS

= ∫0-π/2 ∫0-π/2 (-4r sinθ cosφ, -3r cosθ, 3r sinθ sinφ)⋅(-⟨r sinθ cosφ, r sinθ sinφ, r cosθ⟩/4) [tex]r^{2}[/tex] sinθ dθ dφ

= ∫0-π/2 ∫0-π/2 ([tex]r^{3}[/tex] [tex]sin^{2}[/tex]θ/4)(12 [tex]sin^{2}[/tex]θ) dθ dφ

= 3/4 ∫0-π/2 ∫0-π/2 [tex]r^{3}[/tex][tex]sin^{4}[/tex]θ dθ dφ

= 3/4 ∫0-π/2 [[tex]r^{3/2}[/tex](2/3)] dφ

= 3/4 (2/3) [tex]2^{3/2}[/tex]

= 2√2

Correct Question :

Evaluate the surface integral ∫sf⋅ ds where f=⟨−4x,−3z,3y⟩ and s is the part of the sphere [tex]x^{2}[/tex]+[tex]y^{2}[/tex]+[tex]z^{2}[/tex]=16  in the first octant, with orientation toward the origin.∫∫SF⋅ dS=?

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