f(p i )Area(T i ) F ( r(u, w) ) (r u r w ) da

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1 MAH 55 Flux integrals Fall Review 1.1. Surface integrals. Let be a surface in R. Let f : R be a function defined on. efine f ds = f(p i Area( i lim mesh(p as a limit of Riemann sums over sampled-partitions. A sampled-partition of, P, is a division of the surface into pieces, i, followed by a choice of sample point, p i, in each i. By a division of a surface into pieces we mean i i = and i j lies in the boundary of each piece. he associated Riemann sum is i f(p iarea( i. he mesh of the sampled-partition is ɛ provided every i is contained in the ball of radius ɛ centered at p i. If f is continuous, f ds is well-defined, and hence a number. 1.. Surface integrals as double integrals. Parametrize S as r(u, w: R where is some bounded region in the plane. We can evaluate f at points in S as f ( r(u, w. he remaining issue is to figure out Area( i. At a point r(u, w we can look at the tangent plane to S at the point. It is r(u, w + rr u (u, w + tr w (u, w In the tangent plane we have the parallelogram with vertices r(u, w, r(u, w + r u (u, w du, r(u, w +r w (u, w dw and r(u, w +r u (u, w du+r w (u, w dw. he area of this parallelogram is r u (u, w r w (u, w du dw = r u (u, w r w (u, w da and we use this as the approximation to Area( i. ( f ds = P f ( r(u, w r u r w da. Flux integrals One important source of functions to integrate over surfaces comes using vector fields and the dot product. Let be an oriented surface. Given any vector field F defined on, define F ds = F N ds where N is the unit normal field to, which is why has to be oriented. Since F and N are both vector fields defined on, F N is a function defined on and hence the right hand surface integral is defined. Just as in the line integral case, the flux integral looks like it is going to be the most difficult of the surface integrals, but in fact it is the easiest. First recall, so F ds = N = r u r w r u r w F ( r(u, w r u r w r u r w r u r w da = 1 F ( r(u, w (r u r w da

2 . Orientation An orientation for a surface is a continuous choice of unit normal vector for the surface. Every surface which can be parametrized has two choices of a unit normal vector at every point, hence if there is an orientation, there are at least two. he boundary of a solid E in three space has orientations. By E we mean the boundary of E with the orientation for which the normal vector points out of E. Remark Also just as in the line integral case, the surface integral does not depend on the orientation, but F N changes sign if the orientation does. Hence the value of the flux integral depends on the orientation and changes sign if the orientation changes. Below we will have occasion to do flux integrals over surfaces that some in several disjoint pieces. hen we will have to orient each piece and if we switch the orientation on just one piece then only the sign of the answer for that piece changes. In this case, if there are k pieces, there are k possible orientations. Consider the case in which E is the region between two shpheres. his is an example in which E is in two pieces, the inner sphere and the outer sphere. When we write E we mean the boundary plus an orientation on it. he orientation on the outer sphere points outward but the orientation on the inner sphere points inward. hey both point out of E. As surfaces, each sphere has two orientations for a total of altogether on the union of the two. Once you have parametrized the surface, you have a natural choice for normal vector, the cross product of the two partial derivatives. If the surface is orientable, this works. If the surface is not orientable, nothing works. We saw the Möbius strip as an example of a non-orientable surface. In general, any surface without boundary is orientable. Additionally, any subset of an oriented surface is oriented. Hence level sets are orientable and surfaces of revolution are orientable. Examples include graphs, cylinders, cones,....

3 Example: Find. Examples f ds where is the surface consisting of the part of the plane x+y +z = 6 lying in the first octant and f(x, y, z = x + y + z. Step 1: First parametrize the surface: r(x, y = x, y, 6 x y ; (x, y in where is the triangle in the first quadrant of the xy plane below the line x + y = 6. Step : Calculate the normal field. r x = 1,, r y = 1,, i j k r x r y = det 1 =,, 1 1 Step : Reduce to a double integral. his is not a flux integral so there is no need to worry about the orientation. However we need to compute the length of the normal field. r x r y = = f ds = (x + y + z da = (x + y + 6 x y da = (6 x y da Step : Evaluate the double integral. x raw a picture: x + y = 6. (6 x y dy dx = ((1 x x + ( x (8x 5x + x 7 You may certainly evaluate this double integral as dx dy. 6y xy y x dx = 8x x + dx = 9 = ( 15 + = 11 ( 8 1x + x dx = 9

4 Example: Find F ds where is the surface in the last example with downward normal and where F = x, y, z. he region and the calculation of r x r y =,, 1 are the same as for the last example. Step : Reduce to a double integral. his is a flux integral so we need to check the orientation. he normal vector at all points is,, 1 which points up since the z coordinate is positive. We wanted downward so F ds = x, y, z,, 1 da = (x + y + 6 x y da = 6 da Step : so Evaluate the double integral. 6 da = 6 F ds = 18 = 18 Example: Let be the part of z = x y above the xy plane and let F = y, x, z. Let the normal field be upward. Step 1: Parametrize as r(x, y = x, y, x y for : x + y. Step : Calculate the normal field. r x (x, y = 1,, x r y (x, y =, 1, y i j k r x r y = det 1 x = x, y, 1 1 y Step : Reduce to a double integral. Check the orientation since this is a flux integral. When (x, y = (,, r x r y =,, 1 which points up so the normal field is the one we wanted. F (r x r y = y, x, x y x, y, 1 = xy + xy + x y = xy + x y F ds = (xy + x y da x +y Step : Evaluate the double integral. In this case, polar coordinates looks good. π π (r cos(θ sin(θ + r r drdθ = (r cos(θ sin(θ + r r drdθ = π ( r cos(θ sin(θ + r r dθ = π ( 16 cos(θ sin(θ + dθ = 8 sin (θ π =

5 Remark: In general, parametrize the surface and do the integral using that parametrization. After you do the calculation, check to see if you did it with the correct orientation or not. If you did it with the required orientation you are done. If you did it with the other orientation, then the correct answer is the negative of the one you worked out. 5 Example: Let be the cone z = x + y below z = together with the disk x + y, z =. Orient the surface so that the normal is outward. Integrate the field F = x, y, over this surface. Step 1: Parametrize as 1 were 1 is parametrized in spherical coordinates by φ = π and ρ r 1 (ρ, θ = ρ cos(θ, ρ sin(θ, ρ = cos(θ, sin(θ, 1 ; 1 : θ π, ρ. he surface is parametrized by r (x, y = x, y, ; : x + y. Write the surface as a union of surfaces which you can parametrize and then F ds = F ds + F ds. o each surface integral separately. 1 Step 1 : Compute the normal field for 1. cos(θ sin(θ (r 1 ρ =,, ρ sin(θ ρ cos(θ (r 1 θ =,, i j k cos(θ sin(θ (r 1 ρ (r 1 θ = det ρ sin(θ ρ cos(θ = ρ cos(θ, ρ sin(θ, ρ cos (θ + sin (θ = ρ cos(θ, sin(θ, 1 Step 1 : Reduce to a double integral. F ( (r 1 ρ (r 1 θ = ρ cos(θ, ρ sin(θ, ( ρ cos(θ, sin(θ, 1 = ρ Check the orientation. We were asked to use the outward normal. If we use the point r 1 (, we get the cone point which is most likely singular, so let s use r 1 (1, =,,. Hence (r 1 ρ (r 1 θ = 1 1,, 1. his points upward whereas outward on this cone points down. Hence we have the opposite orientation to the one we want. ( ( F ds = ρ da = ρ da 1 1 1

6 6 Step 1 : o the double integral. F ds = ρ da = 1 θ π ρ π ρ dρ dθ = π ρ dθ = π 8 dθ = 16π Repeat for Step : Compute the normal field for r. (r x = 1,, = i (r y =, 1, = j (r x (r y = k =,, 1 Step : Reduce to a double integral. F ( (r x (r y = x, y,,, 1 = F ds = da = and there was no need to check the orientation. Note however that it points up and up is the correct direction. Also there is no need for a separate step to evaluate the double integral. Final Step: F ds = 16 π. Example: Let the unit cube in the first octant. Orient the surface so that the normal is outward from the solid cube. Integrate the field F = x, y, z over this surface. Step 1: here are six surfaces to parametrize. Let be the unit square in the first quadrant of uw space. he six faces of the cube can be parametrized by (1 C 1 : r [] (u, w = u, w, ; ( C : r [1] (u, w = u, w, 1 ; ( C : s [] (u, w = u,, w ; ( C : s [1] (u, w = u, 1, w ; (5 C 5 : t [] (u, w =, u, w ; (6 C 6 : t [1] (u, w = 1, u, w for (u, w. Step : (1 r [] u (u, w = 1,, ; r [] w (u, w =, 1, ; r [] u r [] w =,, 1 ( r [1] u (u, w = 1,, ; r [1] w (u, w =, 1, ; r [1] u r [1] w =,, 1 ( s [] u (u, w = 1,, ; s [] w (u, w =,, 1 ; s [] u s [] w =, 1, ( s [1] u (u, w = 1,, ; s [1] w (u, w =,, 1 ; s [1] u s [1] w =, 1, (5 t [] u (u, w =, 1, ; t [] w (u, w =,, 1 ; t [] u t [] w = 1,, (6 t [1] u (u, w =, 1, ; t [1] w (u, w =,, 1 ; t [1] u t [1] w = 1,, Steps & : Reduce to a double integral. (1 F ( r [] u (r [] w = u, w,,, 1 = ; F ds = da = C 1

7 ( F ( r [1] u (r [1] w = u, w, 1,, 1 = 1; F ds = 1 da = Area of( = 1 C ( F ( s [] u (s [] w = u,, w, 1, = ; F ds = da = C ( F ( s [1] u (s [1] w = u, 1, w, 1, = 1; F ds = ( 1 da = 1 C (5 F ( t [] u (t [] w =, u, w 1,, = ; F ds = da = C 5 (6 F ( t [1] u (t [1] w = 1, u, w 1,, = 1; F ds = 1 da = 1 C 6 Step + : etermine the normal vectors and add up the answers. (1 he normal vector points up, we want down. ( he normal vector points up, we want up. ( he normal vector points in negative y direction, we want negative. ( he normal vector points in negative y direction, we want positive. (5 he normal vector points in positive x direction, we want negative. (6 he normal vector points in positive x direction, we want positive. Hence F ds = ( ( 1 ( + 1 = 7