Evaluate the surface integral F · dS for the given vector field F and the oriented surface S.?
Evaluate the surface integral F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = x^2 i + y^2 j + z^2 k S is the boundary of the solid half-cylinder 0 ≤ z ≤ sqrt(25 − y^2), 0 ≤ x ≤ 2
More Questions From This User See All
Answers & Comments
Verified answer
Since S is closed (being the boundary of the entire SOLID region), we can apply the Divergence Theorem.
∫∫s F · dS
= ∫∫∫ div F dV
= ∫∫∫ (2x + 2y + 2z) dV
Now, convert to cylindrical coordinates (with x playing the usual role of z):
∫(r = 0 to 5) ∫(θ = 0 to π/2) ∫∫(x = 0 to 2) (2x + 2r cos θ + 2r sin θ) * (r dx dθ dr)
= ∫(r = 0 to 5) ∫(θ = 0 to π/2) (4r + 4r^2 cos θ + 4r^2 sin θ) dθ dr
= ∫(r = 0 to 5) (2πr + 8r^2) dr
= (πr^2 + (8/3)r^3) {for r = 0 to 5}
= 25π + 1000/3.
I hope this helps!