Question

Compute the flux of F⃗ =xi⃗ +yj⃗ +zk⃗ through just the curved surface of the cylinder x2+y2=9 bounded below by the plane x+y+z=2, above by the plane x+y+z=4, and oriented away from the z-axis

Answer 1

Answer:

36π

Explanation:

Close the surface (call it S) by including the cylindrical caps using the given planes.

Now, we can apply the Divergence Theorem:

∫∫s F · dS

= ∫∫∫ div F dV

= ∫∫∫ (1+1+1) dV

= ∫(r = 0 to 3) ∫(θ = 0 to 2π) ∫(z = 1 - r cos θ - r sin θ to 5 - r cos θ - r sin θ) 3 dz dθ dr, via cylindrical coordinates

= ∫(r = 0 to 3) ∫(θ = 0 to 2π) 12 dθ dr

= 72π.

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If the surface was not originally closed, we need to subtract the flux contributions from the caps.

(i) Bottom cap via z = 1 - x - y (with downward pointing normal)

∫∫s₁ F · dS

= ∫∫ <x, y, z> · -<-z_x, -z_y, 1> dA

= ∫∫ <x, y, 1 - x - y> · <-1, -1, -1> dA

= ∫∫ -1 dA

= -9π, being the area inside x^2 + y^2 = 9.

(ii) Upper cap via z = 5 - x - y (with upward pointing normal)

∫∫s₁ F · dS

= ∫∫ <x, y, z> · <-z_x, -z_y, 1> dA

= ∫∫ <x, y, 5 - x - y> · <1, 1, 1> dA

= ∫∫ 5 dA

= 5 * 9π

= 45π.

Hence, the flux for the capless surface equals

72π - (-9π) - 45π = 36π.

I hope this helps!