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398 CHAPTER 12 Vector Integral Calculus

V
z
n

2
2
2
x + y + z = 4
1 ≤ z ≤ 2
z
Σ j

Σ

y
y
2
2
x + y = 3
x x

FIGURE 12.17 Cylinder with base j and FIGURE 12.18 Surface in Example 12.23.
height V n t.

Sum these quantities over the entire surface and take a limit as the surface elements are chosen
smaller, as we did for the mass of a shell. We get

flux of V across in the direction of n = V · ndσ.

The flux of the fluid (or any vector field) across a surface is therefore computed as the surface
integral of the normal component of the field to the surface.

EXAMPLE 12.23
2
2
2
We will calculate the flux of F = xi + yj + zk across the part of the sphere x + y + z = 4
between the planes z = 1 and z = 2.
2
2
The plane z = 1 intersects the sphere in the circle x + y = 3, z = 1. This circle projects
2
2
onto the circle x + y = 3inthe x, y - plane. The plane z = 2 hits the sphere at (0,0,2) only.

2
2
2
2
Think of as specified by z = S(x, y) = 4 − x + y for (x, y) in D, the disk 0 ≤ x + y ≤ 3
(Figure 12.18).
To compute the partial derivatives ∂z/∂x and ∂z/∂y we can implicitly differentiate the
equation of the sphere to get
∂z
2x + 2z = 0
∂x
so
∂z x
=− .
∂x z
Similarly,
∂z y
=− .
∂y z
The vector
x y
i + j + k
z z
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October 14, 2010 14:53 THM/NEIL Page-398 27410_12_ch12_p367-424