0 2 c f dr 0 1 2 1 0 dr d 0 0 3 a 2r

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ∂ ∂x ∂y ∂z xz 2xy 3xy 5 = [3x, x − 3y, 2y ] F • dr = − 3 2 r (x, y ) = [x, y, 3 − 3x − y ] with 0 ≤ x ≤ 1, 0 ≤ y ≤ 3 − 3x 3−3x 1 C F • dr = 0 0 [3x, x − 3y, 2y ] • ([1, 0, −3] × [0, 1, −1]) dy dx = 7 2 (b) curl F = [x, 2z − y, 0] r (x, y ) = [x, y, 2 − 2x − 2y ] with 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x 1−x 1 F • dr = C 0 0 [x, 2z − y, 0] • ([1, 0, −2] × [0, 1, −2]) dy dx = 4 3 (c) curl F = [5, 2, 4] r (r, θ) = [r cos θ, r sin θ, r cos θ + 4] with 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π 2π C F•dr = 0 2 [5, 2, 4]•([cos θ, sin θ, cos θ] × [−r sin θ, r cos θ, −r sin θ]) dr dθ = −4π 0 (d) r (r, θ) = r cos θ, r sin θ, 1 − r2 with 0 ≤ r ≤ 1, 0 ≤ θ ≤ π 2 curl F = [2y, −2x, 0] = [2r sin θ, −2r cos θ, 0] π 2 C F • dr = 0 1 π 2 1 = 0 dr dθ = 0 0 3. (a) [2r sin θ, −2r cos θ, 0] • ([cos θ, sin θ, −2r] × [−r sin θ, r cos θ, 0]) dr dθ 0 0 i. S is part of a sphere and it is an OPENED surface. Since we can only apply the Divergence theorem when the surface is CLOSED, then we have to add another surface at the bottom and make the surface to be closed. So, we introduce a new surface S1 for the bottom part of the CLOSED surface. E is the region inside the surface, which is upper part of the sphere and E= (ρ, φ, θ| 0 ≤ ρ ≤ 1, 0 ≤ φ ≤...
View Full Document

Ask a homework question - tutors are online