28.2. SURFACES OF THE FORM z = f (x,y) 561

Example 28.1.7 Let U = [0,2π]× [0,2π] and for (t,s) ∈U, let

f (t,s) = (2cos t + cos t coss,−2sin t − sin t coss,sins)T .

Find∫f(U) hdV where h(x,y,z) = x2.

Everything is the same as the preceding example except this time it is an integral of afunction. The area element is (coss+2) dsdt and so the integral called for is

∫f(U)

hdA =∫ 2π

0

∫ 2π

0

 x on the surface︷ ︸︸ ︷2cos t + cos t coss

2

(coss+2) dsdt = 22π2

28.2 Surfaces of the Form z = f (x,y)The special case where a surface is in the form z = f (x,y) ,(x,y) ∈ U , yields a simpleformula which is used most often in this situation. You write the surface parametrically inthe form f (x,y) = (x,y, f (x,y))T such that (x,y) ∈U . Then

f x =(

1 0 fx)T

, f y =(

0 1 fy)T

and∣∣f x ×f y

∣∣=√1+ f 2y + f 2

x so the area element is√1+ f 2

y + f 2x dxdy.

When the surface of interest comes in this simple form, people generally use this areaelement directly rather than worrying about a parametrization and taking cross products.

In the case where the surface is of the form x = f (y,z) for (y,z) ∈ U , the area el-

ement is obtained similarly and is√

1+ f 2y + f 2

z dydz. I think you can guess what the

area element is if y = f (x,z). It also generalizes immediately to higher dimensions wherexki = f

(x1, ...,xki−1 ,xki+1 , ...,xn

).

There is also a simple geometric description of these area elements. Consider the sur-face z = f (x,y). This is a level surface of the function of three variables z− f (x,y). Infact the surface is simply z− f (x,y) = 0. Now consider the gradient of this function ofthree variables. The gradient is perpendicular to the surface and the third component ispositive in this case. This gradient is (− fx,− fy,1) and so the unit upward normal is just

1√1+ f 2

x + f 2y(− fx,− fy,1). Now consider the following picture.

knθ

θ

dA

dxdy

In this picture, you are looking at a chunk of area on the surface seen on edge and so itseems reasonable to expect to have dxdy = dAcosθ . But it is easy to find cosθ from thepicture and the properties of the dot product.

cosθ =n ·k|n| |k|

=1√

1+ f 2x + f 2

y

.