Chapter 9: Vector Calculus
Section 9.6: Surface Integrals
Working in Cartesian coordinates, obtain the flux of the field F=x i+y j+z k through the surface of the unit sphere centered at the origin. Use a Cartesian representation of the surface.
A Cartesian representation of the unit sphere is fx,y,z≡x2+y2+z2=1.
A unit outward normal on the sphere is ∇f/∇fx=a|f(x)f=1. The normalized gradient, evaluated on the sphere, is then
22x2+y2+z2xyz ⇒ xyz≡N
According to Table 9.6.1, the element of surface area can be taken as ∇f|fz| dA; so, on the sphere itself, 2x2+y2+z22 z becomes 1/z. On both hemispheres, z=1−x2−y2. Now, the projection of each hemisphere onto the plane z=0 is the unit disk, so there are two integrals over this disk to consider, each the integral of
F·N dσ = xyz·xyz 1z dA = x2+y2+z2z dA = 11−x2−y2 dA
The two integrals are therefore the same, so in polar coordinates, the flux integrals become
2∫02 π∫011−r2−1/2 r dr dθ=4 π
Maple Solution - Interactive
The display in Table 9.6.12(a) is obtained by accessing the MultiInt command through the Context Panel system after loading the Student MultivariateCalculus package, and after making the choices indicated in the dialogs in Figures 9.6.12(a, b).
Tools≻Load Package: Student Multivariate Calculus
Context Panel: Student Multivariate Calculus≻Integrate≻Iterated
Figure 9.6.12(a) First dialog
Figure 9.6.12(b) Second dialog
Table 9.6.12(a) Stepwise evaluation of the flux through a sphere
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