21.2. SURFACES OF THE FORM z = f (x,y) 393

and so |f t ×f s|= (coss+2) so the area element is (coss+2) dsdt and the area is∫ 2π

0

∫ 2π

0(coss+2) dsdt = 8π

2

Example 21.1.7 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

21.2 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 =

 10fx

 , f y =

 01fy

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 element 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).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