Surface Integral

Surface Integral

• Surface integral is a definite integral taken over a surface. It can be thought of as the double integral analog of the line integral. Given a surface, one may integrate over its scalar fields and vector fields.

• Surface integrals have applications in physics, particularly with the classical theory of electromagnetism.

• The definition of surface integral relies on splitting the surface into small surface elements.

Surface Integral

• In line integral, we learned how to integrate along a curve. We will now learn how to perform integration over a surface in R3, such as a sphere or a paraboloid.

• Similar to how we used a parametrization of a curve to define the line integral along the curve, we will use a parametrization of a surface to define a surface integral. We will use two variables, u and v, to parametrize a surface G in R3 : x = x(u,v), y = y(u,v), z = z(u,v), for (u,v) in some region R in R2

Surface Integral

Surface Integral

• In this case, the position vector of a point on the surface G is given by the vector-valued function

Surface Integral

• Now take a point (u,v) in R as, say, the lower left corner of one of the rectangular grid sections in R.

• Suppose that this rectangle has a small width and height of Δu and Δv, respectively. The corner points of that rectangle are (u,v), (u + Δu,v), (u+Δu,v+Δv) and (u,v+Δv). So the area of that rectangle is A = ΔuΔv.

• Then that rectangle gets mapped by the parametrization onto some section of the surface G which, for Δu and Δv small enough, will have a surface area (call it dS) that is very close to the area of the parallelogram which has adjacent sides r(u+ Δu,v) − r(u,v) (corresponding to the line segment from (u,v) to (u + Δu,v) in R) and r(u,v+ Δv) − r(u,v)

Surface Integral

• We have

• so the surface area element dσ is approximately

Surface Integral

• DefinitionLet G be a surface in R3 parametrized by x = x(u,v), y = y(u,v), z = z(u,v), for (u,v) in some region R in R2. Let r(u,v) = x(u,v)i + y(u,v)j + z(u,v)k be the position vector for any point on G, and let f(x, y, z) be a real-valued function defined on some subset of R3 that contains G. The surface integral of f(x, y, z) over G is

In particular, the surface area S of G is

( ) ( ) ( ) ( )( ), , , , , , ,G R

f x y z dS f x u v y u v z u v dudvu v¶ ¶

= ´¶ ¶òò òòr r

1G

S dS=òò

Surface Integral

• In special case of a surface G described as a graph z = h(x, y) of a function h defined on a region R in the xy-plane, we may use x and y (rather than u and v) as the parameters.

• The surface integral of f(x, y, z) over G is

( ) ( )( )22

, , , , , 1G R

h hf x y z dS f x y h x y dxdy

x y

æ öæ ö¶ ¶ ÷ç÷ç ÷= + +ç÷ç ÷÷ çç ÷çè ø¶ ¶è øòò òò

Example 1

• A torus T is a surface obtained by revolving a circle of radius a in the yz-plane around the z-axis, where the circle’s center is at a distance b from the z-axis (0 < a < b), as in figure. Find the surface area of T.

Solution

• the torus can be parametrized as:

Solution

Example 2

• Evaluate where G is the part of the plane 2x – y + z = 3 above the triangle R with vertices (0, 0), (1, 0), and (1, 1)

( )G

xy z dS+òò

Flux of a Vector Field Through a Surface

• Let G be such a smooth, two sided surface, and assume that it is submerged in a fluid with a continuous velocity field F(x,y,z). If ΔS is the area of a small piece of G, then F is almost constant there, and the volume ΔV of fluid crossing this piece in the direction of the unit normal n is

ΔV F . n ΔS• We conclude that

Flux of F across G = G

dSòòF.n

Example 3

• Find the upward flux of F = – y i + x j + 9 k across the part of the spherical surface G determined by ( ) 2 2 2 2, 9 , 0 4z f x y x y x y= = - - £ + £

Flux of a Vector Field Through a Surface

• TheoremLet G be a smooth, two sided surface given by z = f(x,y), where (x, y) is in R, and let n denote the upward unit normal on G. If f has continuous first order partial derivatives and F = M i + N j + P k is a continuous vector field, then the flux of F across G is given by

[ ]x y

G R

dS Mf Nf P dxdy= - - +òò òòflux F = F.n

Flux of a Vector Field Through a Surface

• Proof:If we write H(x, y, z) = z – f(x, y), we obtain

It follows from definition of surface integral that

2 2 1

x y

x y

ffHH ff

- - +Ñ= =

Ñ + +

i j kn

( )

( )

2 2

2 21

1

x yx y

G R x y

x y

R

ffdS M N P ff dxdy

ff

Mf Nf P dxdy

- - += + + + +

+ +

= - - +

òò òò

òò

i j kF.n i j k .

Exercise

1. Evaluate the surface integral , where F(x, y, z) = yz i + xz j + xy k and G is the part of the plane x + y + z = 1 with x ≥ 0, y ≥ 0, and z ≥ 0, with the outward unit normal n pointing in the positive z direction (see figure)

G

dSòòF.n

Exercise

2. Calculate the flux of F = y i – x j + 2 k across G where G is the surface determined by

3.

21 , 0 5z y x= - £ £