Surface Integrals

10.6 Surface Integrals

To define a surface integral, we take a surface S, given by a parametric representation as just discussed,

(1) r (u, v) ⫽ [x (u, v), y (u, v), z (u, v)] ⫽ x (u, v)i ⫹ y (u, v)j ⫹ z (u, v)k

where (u, v) varies over a region R in the uv-plane. We assume S to be piecewise smooth (Sec. 10.5), so that S has a normal vector

N⫽r u ⴛr v and unit normal vector

n⫽

ƒNƒ at every point (except perhaps for some edges or cusps, as for a cube or cone). For a given

vector function F we can now define the surface integral over S by

冮 S 冮 F • n dA ⫽ 冮 冮 (u, v)) • N R

F (r (u, v) du dv.

Here N⫽ƒNƒn by (2), and ƒNƒ⫽ƒr u ⴛr v ƒ is the area of the parallelogram with sides r u and , by the definition of cross product. Hence r v

n dA ⫽ n ƒ N ƒ du dv ⫽ N du dv.

And we see that dA ⫽ ƒ N ƒ du dv is the element of area of S.

CHAP. 10 Vector Integral Calculus. Integral Theorems Also F•n is the normal component of F. This integral arises naturally in flow problems,

F ⫽ rv . Recall, from Sec. 9.8, that the flux across S is the mass of fluid crossing S per unit time. Furthermore, is the density of the fluid r and v the velocity vector of the flow, as illustrated by Example 1 below. We may thus call the surface integral (3) the flux integral.

where it gives the flux across S when

We can write (3) in components, using F⫽ [F 1 , F 2 , F 3 ], N ⫽ [N 1 , N 2 , N 3 ], and n⫽ [cos a, cos b, cos g] . Here, a , b, g are the angles between n and the coordinate axes; indeed, for the angle between n and i, formula (4) in Sec. 9.2 gives cos a ⫽ n • i> ƒ n ƒ ƒ i ƒ ⫽ n • i, and so on. We thus obtain from (3)

In (4) we can write cos a dA ⫽ dy dz, cos b dA ⫽ dz dx, cos g dA ⫽ dx dy. Then (4) becomes the following integral for the flux:

F • n dA ⫽

(F 1 dy dz ⫹ F 2 dz dx ⫹ F 3 dx dy).

We can use this formula to evaluate surface integrals by converting them to double integrals over regions in the coordinate planes of the xyz-coordinate system. But we must carefully take into account the orientation of S (the choice of n). We explain this for the integrals

of the -terms, F 3

(5 r )

F 3 cos g dA ⫽

F 3 dx dy.

If the surface S is given by z⫽h (x, y) with (x, y) varying in a region in the xy-plane, R and if S is oriented so that cos g ⬎ 0 , then (5 r ) gives

(5 s )

F 3 cos g dA ⫽ ⫹ F 3 (x, y, h (x, y)) dx dy.

But if cos g ⬍ 0, the integral on the right of (5 s ) gets a minus sign in front. This follows if we note that the element of area dx dy in the xy-plane is the projection ƒ cos g ƒ dA of the element of area dA of S; and we have cos g ⫽ ⫹ ƒ cos g ƒ when cos g ⬎ 0, but

cos g ⫽ ⫺ ƒ cos g ƒ when cos g ⬍ 0. Similarly for the other two terms in (5). At the same time, this justifies the notations in (5).

Other forms of surface integrals will be discussed later in this section.

EXAMPLE 1 Flux Through a Surface

Compute the flux of water through the parabolic cylinder S: 2 , 0 ⬉ x ⬉ 2, 0 ⬉ z ⬉ 3 (Fig. 245) if the velocity vector is

speed being measured in meters sec. (Generally, > , but water has the density

[3z 2 ,

6, 6xz], 1 ton>m . )

1 g>cm 3 3

SEC. 10.6 Surface Integrals

Fig. 245. Surface S in Example 1

Solution. Writing x⫽u and z⫽v , we have y⫽x 2 ⫽ u 2 . Hence a representation of S is

(0 ⬉ u ⬉ 2, 0 ⬉ v ⬉ 3). By differentiation and by the definition of the cross product, N⫽r u ⴛr v ⫽ [1, 2u, 0] ⴛ [0, 0, 1] ⫽ [2u, ⫺1, 0]. On S, writing simply F (S) for F [r (u, v)], we have F (S) ⫽ [3v 2 , 6, 6uv]. Hence F (S) • N ⫽ 6uv 2 ⫺ 6. By

r ⫽ [u, u 2 , v]

integration we thus get from (3) the flux

冮 S 冮 F • n dA ⫽ 冮 冮 (6uv

6) du dv ⫽ 冮 (3u 6u) ` dv

0 0 0 u⫽0

冮 3 12) dv ⫽ (4v 12v) ` ⫽ 108 ⫺ 36 ⫽ 72 [m >sec]

(12v 2 ⫺

0 v⫽0

or 72,000 liters>sec. Note that the y-component of F is positive (equal to 6), so that in Fig. 245 the flow goes from left to right.

Let us confirm this result by (5). Since N⫽ƒNƒn⫽ƒNƒ [cos a, cos b, cos g] ⫽ [2u, ⫺1, 0] ⫽ [2x, ⫺1, 0] we see that cos a ⬎ 0, cos b ⬍ 0, and cos g ⫽ 0 . Hence the second term of (5) on the right gets a minus sign,

and the last term is absent. This gives, in agreement with the previous result,

6 dz dx ⫽ 4 (3z 2 ) dz ⫺ 6 # 3 dx ⫽ 4 # 3 3 ⫺ 6 冮 # 3 冮 # 2 ⫽ 72. 䊏 S 冮

F • n dA ⫽

3z 2 dy dz ⫺

EXAMPLE 2 Surface Integral

Evaluate (3) when F⫽ [x 2 , 0, 3y 2 ] and S is the portion of the plane x⫹y⫹z⫽ 1 in the first octant (Fig. 246).

Solution. Writing x⫽u and y⫽v , we have z⫽ 1 ⫺ x ⫺ y ⫽ 1 ⫺ u ⫺ v. Hence we can represent the

plane x⫹y⫹z⫽ 1 in the form r (u, v ) ⫽ [u, v , 1 ⫺ u ⫺ v]. We obtain the first-octant portion S of this plane by restricting x⫽u and y⫽v to the projection R of S in the xy-plane. R is the triangle bounded by the two coordinate axes and the straight line x⫹y⫽ 1, obtained from x⫹y⫹z⫽ 1 by setting z⫽ 0 . Thus 0 ⬉ x ⬉ 1 ⫺ y, 0 ⬉ y ⬉ 1 .

Fig. 246. Portion of a plane in Example 2

CHAP. 10 Vector Integral Calculus. Integral Theorems By inspection or by differentiation,

N⫽r u ⴛr v ⫽ [1, 0, ⫺1] ⴛ [0, 1, ⫺1] ⫽ [1, 1, 1]. Hence By F (S) • N ⫽ [u 2 , 0, 3v 2 ] • [1, 1, 1] ⫽ u 2 ⫹ 3v 2 . (3),

1 1ⴚv

(u 2 冮 ⫹ 冮 冮 冮 3v 2 ) du dv ⫽ 冮 冮 (u 2 ⫹ 3v 2 ) du dv

F • n dA ⫽

(1 ⫺ v) 3 ⫹ 3v 2 冮 1 c (1 ⫺ v) d dv ⫽ .