Consider the vector field F=. Find the flux of F across the part of the paraboloid x^2+y^2+z=2 that lies above the plane z=1, orientated upwards.
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Note that the paraboloid intersects the plane when x^2 + y^2 + 1 = 2.
==> x^2 + y^2 = 1 (this is for the region of integration).
Next, rewrite the paraboloid as z = 2 - x^2 - y^2
So, ∫∫s F · dS
= ∫∫ · <-z_x, -z_y, 1> dA
= ∫∫ <1^(1^2) * arcsin(y/1), 2x/1, 1> · <2x, 2y, 1> dA
= ∫∫ (2x arcsin y + 4xy + 1) dA.
Now, convert to polar coordinates.
∫(r = 0 to 1) ∫(θ = 0 to 2π) (2r cos θ arcsin(r sin θ) + 4r^2 cos θ sin θ + 1) * r dθ dr
= ∫(r = 0 to 1) ∫(θ = 0 to 2π) (2r^2 cos θ arcsin(r sin θ) + 4r^3 cos θ sin θ + r) dθ dr
= ∫(r = 0 to 1) (0 + 2πr) dr
= π.
I hope this helps!
==> x^2 + y^2 = 1 (this is for the region of integration).
Next, rewrite the paraboloid as z = 2 - x^2 - y^2
So, ∫∫s F · dS
= ∫∫
= ∫∫ <1^(1^2) * arcsin(y/1), 2x/1, 1> · <2x, 2y, 1> dA
= ∫∫ (2x arcsin y + 4xy + 1) dA.
Now, convert to polar coordinates.
∫(r = 0 to 1) ∫(θ = 0 to 2π) (2r cos θ arcsin(r sin θ) + 4r^2 cos θ sin θ + 1) * r dθ dr
= ∫(r = 0 to 1) ∫(θ = 0 to 2π) (2r^2 cos θ arcsin(r sin θ) + 4r^3 cos θ sin θ + r) dθ dr
= ∫(r = 0 to 1) (0 + 2πr) dr
= π.
I hope this helps!