Surface Integral

Surface Integral The surface integral of f over the surface S is RR Pm Pn f (x, y, z) dS = limm,n→∞ i=1 j=1 f (Pij∗ ) △ Sij S Formula: RR f (x, y, z) dS = S RR − − − f (→ r (u, v)) | → ru X → r v | dA D Formula RR f (x, y, z) dS = S Exercise 1. Evaluate RR f (x, y, g(x, y)) D RR q ∂z 2 ∂z 2 ( ∂x ) + ( ∂y ) + 1 dA x2 z dS if S is the portion of the cone z 2 = x2 + y 2 that lies between S the planes z = 1 and z = 4. (Swok Sec 18.5 Ex 1) Exercise 2. Evaluate RR (xz/y) dS if S is the portion of the cylinder x = y 2 that lies in the first S octant between the planes z = 0, z = 5, y = 1, and y = 4. (Swok Sec 18.5 Ex 2) Exercise 3. Evaluate RR (z + y) dS if S is the part of the graph of z = √ 1 − x2 in the first octant S between the xz-plane and the plane y = 3. (Swok Sec 18.5 Ex 3) Class RR Exercise 1. Evaluate the surface integral. (#6-20 even) xyz dS, S is the cone with parametric equations x = u cos v, y = u sin v, z = u, (a) S 0 ≤ RR u ≤ 1, 0 ≤ v ≤ π/2 (b) (x2 + y 2 ) dS, S is the surface with vector equation: S → − r (u, v) = < 2uv, u2 − v 2 , u2 + v 2 >, u2 + v 2 ≤ 1 (c) (d) (e) (f) RR S RR S RR S RR xz dS, S is the part of the plane 2x + 2y + z = 4 that lies in the first octant y dS, S is the surface z = 23 (x3/2 + y 3/2 ), 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 z dS, S is the surface x = y + 2z 2 , 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 y 2 dS, S is the part of the sphere x2 + y 2 + z 2 = 4 that lies inside the cylinder S 2 x2 + RRy = 1 and above the xy-plane (g) xz dS, S is the boundary of the region enclosed by the cylinder y 2 + z 2 = 9 and the planes S x = RR 0 and x + y = 5 (h) (x2 + y 2 + z 2 ) dS, S is the part of the surface of the cylinder x2 + y 2 = 9 between the S planes z = 0 and z = 2, together with its top and bottom disks. − Definition: If it is possible to choose a unit normal vector → n at every such point (x, y, z) so that → − − n varies continuously over S, then S is called an oriented surface and the given choice of → n provides S with an orientation. Definition: For a closed surface, that is, a surface that is the boundary of a solid region E, the convention is that the positive orientation is the one for which the normal vectors point outward from E. → − Definition: If F is a continuous vector field defined on an oriented surface S with unit nor→ − → − mal vector n , then the surface integral of F over S is RR → RR → − → − − → F · dS = F ·− n dS. S S → − This integral is called the flux of F across S. Formula: RR → RR → − → − − − − F · dS = F · (→ ru X → r v ) dA. S D Formula: RR → RR − → − ∂g ∂g F · dS = (−P ∂x − Q ∂y + R) dA. S D → − Exercise 4. Consider the radial vector field F = < x, y, z >. Find the upward flux of the field across the hemisphere x2 + y 2 + z 2 = 1, for z ≥ 0. (Briggs Sec 14.8 Ex 8a) → − Exercise 5. Consider the radial vector field F = < x, y, z >. Find the upward flux of the field across the paraboloid z = 1 − x2 − y 2 , for z ≥ 0. (Briggs Sec 14.8 Ex 8b) Exercise 6. Let S be the part of the graph of z = 9 − x2 − y 2 with z ≥ 0. → − → − → − → − → − If F (x, y, z) = 3x i + 3y j + z k , find the flux of F through S. (Swok Sec 18.5 Ex 5) Class Exercise 2. Evaluate the surface integral RR → − → − → − F · d S for the given vector field F and S → − the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. (#24-32 even) p → − → − → − → − (a) F (x, y, z) = −x i + −y j + z 3 k , S is the part of the cone z = x2 + y 2 between the planes z = 1 and z = 3 with downward orientation → − → − → − → − (b) F (x, y, z) = xz i + x j + y k , S is the hemisphere x2 + y 2 + z 2 = 25, y ≥ 0, oriented in the direction of the positive y-axis → − → − → − → − (c) F (x, y, z) = xy i + 4x2 j + yz k , S is the surface z = xey , 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, with upward orientation → − → − → − → − (d) F (x, y, z) = x i + y j + 5 k , S is the boundary of the region enclosed by the cylinder x2 + z 2 = 1 and the planes y = 0 and x + y = 2 → − → − → − → − (e) F (x, y, z) = y i + (z − y) j + x k , S is the surface of the tetrahedron with vertices (0,0,0), (1,0,0), (0,1,0), (0,0,1)