First we consider how to parameterize a surface (similar to a parameterized curve for line integrals). Surfaces will need two parameters.

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1 Math 55 - Vector Calculus II Notes 14.6 urface Integrals Let s develop some surface integrals. First we consider how to parameterize a surface (similar to a parameterized curve for line integrals). urfaces will need two parameters. With u, v as the parameters, a parametrization of surface in 3 has the form r(u, v) x(u, v), y(u, v), z(u, v), where the range of r (on some rectangular domain in the uv-plane) is. Examples: (1) r(u, v) cos (u), sin (u), v on the rectangle given by u π and v H. This is a parametrization of the curved surface of a right circular cylinder of radius about the z-axis from z to z H (so the height is H). () r(u, v) v cos (u)/h, v sin (u)/h, v on the rectangle given by u π and v H. Here we have circular traces (at a fixed z v) that vary in radii from at z to at z H. Hence this is a parametrization of the curved surface of a right circular inverted cone of radius from z (at the vertex) to z H (so the height is H). (3) r(u, v) ρ sin (u) cos (v), ρ sin (u) sin (v), ρ cos (u) on the rectangle given by u π and v π. pherical coordinates gives us this one! Here u is playing the role of φ and v is playing the role of θ. This is a parametrization of the sphere of radius ρ centered at the origin. (4) uppose we want to parameterize some part of the graph of z f(x, y) above a rectangle in the xy-plane. Then we use the graph parametrization: r(u, v) u, v, f(u, v) over the rectangle in the uv-plane.

2 urface integrals of scalar functions: uppose we have a scalar function f(x, y, z) defined on a smooth surface in 3 parameterized by r(u, v) x(u, v), y(u, v), z(u, v), over a rectangle given by a u b and c v d, where the functions x(u, v), y(u, v) and z(u, v) are assumed to have continuous partials. Partition the rectangle into n subrectangles (ordered by k 1,,..., n) with sides of length u and v. The area of each rectangle is A u v. The kth rectangle, k, with area A corresponds under r to a curved patch k of the surface with area k. Let (u k, v k ) be the lower left corner of rectangle k. Let x k x(u k, v k ), y k y(u k, v k ) and z k z(u k, v k ). From this we derive a iemann sum, which adds up the function s values multiplied by the areas of the surface patches: n f(x k, y k, z k ) k. k1 In order to be able to turn this into an integral (by having n ) which we can evaluate, we need to express k in terms of A. Let t u r u and t v r v. When evaluated at (u k, v k ) these are tangent vectors to the surface at (x k, y k, z k ) in the patch of the surface (in directions given by varying one of u or v while holding the other one fixed). We want to consider the tangent vectors t u u and t v v so that as n these vectors get arbitrarily small. These vectors are approximations of r (in the respective directions) along two sides of the patch k, so the magnitude of their cross product is the area of a parallelogram that is a decent approximation to k : ee picture on next page for a visualization. k t u u t v v t u t v u v. Note: The vector t u t v is orthogonal (aka normal) to the surface at (x k, y k, z k ). The factor t u t v is similar to the speed factor in a path/line integral. o the iemann sum becomes: n f(x k, y k, z k ) t u t v A. k1 This gives us the integral of f over the surface, denoted f d, by taking the limit as n : f d f(x(u, v), y(u, v), z(u, v)) t u t v da. Note: If f 1 then the integral is the area of the surface.

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4 Let s use a surface integral to find the surface area of the sphere of radius : Let s use the parametrization r(u, v) sin (u) cos (v), sin (u) sin (v), cos (u), where u π and v π. Then t u cos (u) cos (v), cos (u) sin (v), sin (u) and t v sin (u) sin (v), sin (u) cos (v),. o t u t v sin (u) cos (v), sin (u) sin (v), sin (u) cos (u). Then t u t v (sin (u) cos (v)) + (sin (u) sin (v)) + (sin (u) cos (u)) sin 4 (u) cos (v) + sin 4 (u) sin (v) + sin (u) cos (u) sin 4 (u) + sin (u) cos (u) sin (u) using that sin (u) on [, π]. Note: This is the Jacobian of spherical coordinates where ρ! π o the sphere s area is 1 d sin (u) du dv π ( cos (u) π ) 4π. Let s evaluate the surface integral of f(x, y) x + y over the top hemisphere H {(x, y, z) x + y + z 1, z } of radius 1. We can use the parametrization above with 1 and the restriction u π to get straight to it: H x + y d π π sin 3 (u) du π π π 1 1 [(sin (u) cos (v)) + (sin (u) sin (v)) ](sin (u)) du dv (1 cos (u)) sin (u) du 1 x ( dx) 1 x dx π(1 1/3) 4π 3. ince the area of H is π it follows that the average value of x + y on H is /3.

5 Let s develop a formula that will work for the surface integral of f(x, y, z) on a surface g determined by the graph of z g(x, y): We begin with the graph parametrization. r u, v, g(u, v) on a region (at this point we drop the assumption that must be a rectangle, the construction still holds for more general regions). Then t u 1,, g u and t v, 1, g v so it follows that t u t v g u, g v, 1. Hence, t u t v (g u ) + (g v ) + 1. Thus f d g f(u, v, g(u, v)) (g u ) + (g v ) + 1 da. We could leave things in terms of x and y instead of u and v: f d g f(x, y, g(x, y)) (g x ) + (g y ) + 1 da. Let s find the surface integral of f(x, y, z) e z over the surface that is the part of the plane 3x + y + z 6 in the first octant. The function g(x, y) in this case is z g(x, y) 6 3x y. o this surface integral is f d e 6 3x y da, where is the region in the first quadrant bounded by 3x + y 6. o f d 1 e x e 6 e 3x e y 14 dy dx 1 e6 14 e 6 e 3x dx 1 ( e6 14 e e 6 1 ) dx 3 (e 3x 6 1)e 3x dx 14 6 (e6 7).

6 Let s extend our derivation of the surface integral of a scalar function to that of a vector field. The surfaces we consider are orientable. A surface in 3 is orientable if for any closed curve C on the surface if a person walks the curve, the person ends up on the same side as he started. For instance, a Möbuis strip is not orientable. ee the picture below. At ant point P of any orientable surface, there are two possible unit normal vectors; this creates a way to find one set of unit vectors on one-side of the surface versus another set of unit vectors on the other side. Picking one of these is considered to give an orientation to the surface.

7 The most common surface integral of a vector field is a flux integral. The basic intuition is that flux integral we are about to derive measures how much of the vector field passes through the surface. Consider a vector field F f, g, h in 3 that is continuous on a region of 3 and represents a flow (e.g. some fluid). Given a smooth surface we are interested in how much of F goes through (that is, the flux). We define the flux integral (relative to some orientation of ) as F d F n d, where d n d and n is a unit normal to (at each point on where it is evaluated). uppose we have a parametrization r(u, v) x(u, v), y(u, v), z(u, v) for where (u, v) vary over a region of the uv-plane. Then n ± t u t v t u t v, so F n d F n t u t v da F (±t u t v ) da This works as long as t u t v almost everywhere, which means it can be zero at some points, as long as that collection of points forms a subset of whose area is. The choice of ± is based on the desired orientation of the surface. This is also reduced to fq x gq y + h da in the special case of the surface being z q(x, y) since r(x, y) x, y, q(x, y) implies t x t y q x, q y, 1 (here using the upward orientation).

8 Using the upward orientation, let s find the flux of the radial vector field F x, y, z through the surface that is the part of the paraboloid z 1 x y above the xy-plane. We could use the special case with the graph parametrization, but instead let s use the following parametrization: ince x + y 1 z we set 1 z v to get r(u, v) v cos (u), v sin (u), 1 v. Here the region in the uv-plane is determined by u π and v 1. Then t u v sin (u), v cos (u), and t v cos (u), sin (u), v so it follows that t u t v v cos (u), v sin (u), v. Looking at the third component we see a downward orientation. No problem, just negate. F n d v cos (u), v sin (u), 1 v v cos (u), v sin (u), v da π 1 v 3 + v v 3 da v 3 + v dv du π(1/4 + 1/) 3π/.

9 The heat flow vector field for a heat conducting object is F k T where T (x, y, z) is the temperature in the object and k > is a constant that depends on the material. Let s compute the outward flux of the heat flow across the surface of a spherical object (of radius ) where k 1 and T (x, y, z) e x y z is the temperature inside. We use the parametrization r(u, v) sin (u) cos (v), sin (u) sin (v), cos (u), where is the rectangle u π and v π. We already have t u t v sin (u) cos (v), sin (u) sin (v), sin (u) cos (u), which points outward. o since T (x, y, z) T (x, y, z) x, y, z and T (r(u, v)) e T n d e sin (u) cos (v), sin (u) sin (v), cos (u) sin (u) cos (v), sin (u) sin (v), sin (u) cos (u) da 3 e π 3 e (π) 4π 3 e sin (u) du sin 3 (u)(cos (v) + sin (v)) + sin (u) cos (u) du dv sin 3 (u) + sin (u) cos (u) du 4π 3 e ( cos (u) π ) 8π3 e.

10 Consider the radial vector field F r. Let be the sphere of radius centered at the r p origin. Let s compute the outward flux of F across the sphere in two different ways: First, let s do it with the parametrization s(u, v) sin (u) cos (v), sin (u) sin (v), cos (u) where is the rectangle u π and v π. ecall that for this parametrization, t u t v sin (u) cos (v), sin (u) sin (v), sin (u) cos (u). F n d 1 1 p sin (u) cos (v), sin (u) sin (v), cos (u) sin (u) cos (v), sin (u) sin (v), sin (u) cos (u) da p 3 π p 3 π p 3 4π p 3 π sin 3 (u)(cos (v) + sin (v)) + sin (u) cos (u) du dv sin 3 (u) + sin (u) cos (u) du sin (u) du Now let s do it by symmetry and the graph parametrization of z q(x, y) for the half of x +y +z that satisfies z. We just double the flux across the top-half of the sphere: We can find the partials z x and z y implicitly: z x (x + y + z ) x /(x + y + z ) z x/z and z y (x + y + z ) y /(x + y + z ) z y/z. o we can use z x y and n z x, z y, 1. F n d 1 x +y p x, y, x y p x x y, x + y x +y x y + x y da p x y da p 4π p 4π p 3 x +y π ( r 1 r dr dθ r ) y x y, 1 da

11 Finally sometimes we use the definition F d F n d directly: Let s calculate the (upward) flux integral of the vector fields F 1 x, y, 3z and F y, z, x through the disk D {(x, y, z) : z 1, x + y 1}. We can use n,, 1 obviously. o for F 1, we get D F 1 n d D 3z d 3 1 d 3π. D And for F with the parametrization r(x, y) x, y, 1 over the unit disk D we get (noting that d da): D F n d x da D π 1 r cos (θ) r dr dθ 1 4 π (1/)(1 + cos (θ)) dθ π 4.