Worksheet #6
Math 223-01 (Salomone)
August 6, 2007
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| {z }
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Problem 1. (8 points) For each of the following integrals, circle the technique(s) you could use to evaluate it.
There may be more than one answer.
(a) (2 points) x
S
F·ndA where F(x,y,z)=hx,2y,3ziand Sis the surface of a 2 ×2×2 cube centered at the
origin, oriented outwards.
Path Shape Green’s Stokes’ Divergence None of
Independence Independence Theorem Theorem Theorem the above
Here, Sis a closed surface, so the divergence theorem is in play. But ∇ · F=6,0, so this field is not shape independent and hence
F,∇ × Gso Stokes’ theorem is not possible.
(b) (2 points) x
S
F·ndA where F(x,y,z)=∇×hx2−y,y2−z,z2−xiand Sis the upper hemisphere of radius 5
centered at the origin, oriented upwards.
Path Shape Green’s Stokes’ Divergence None of
Independence Independence Theorem Theorem Theorem the above
Here, Sis not a closed surface, so divergence theorem won’t work. But since F=∇ × Gis a curl vector field, it must be shape
independent! And since the vector potential Gis known, you can use Stokes’ theorem to turn this surface integral into a line integral.
(c) (2 points) ZC
F·dr, where F(x,y,z)=hyz,xz,xyiand Cis the triangular path from (1,0,0) to (0,2,0) to (0,0,3)
and back to (1,0,0).
Path Shape Green’s Stokes’ Divergence None of
Independence Independence Theorem Theorem Theorem the above
First of all, notice this is a line integral! The curve Cis closed, so we may use Stokes’ theorem. Moreover, ∇ × F=h0,0,0iso this field
is path-independent. This combination implies the integral is zero, in fact.
(d) (2 points) x
S
F·ndA where F(x,y,z)=hy+z,x+z,x+2yiand Sis the unit disk x2+z2≤1 in the xz-plane,
oriented in the positive y-direction.
Path Shape Green’s Stokes’ Divergence None of
Independence Independence Theorem Theorem Theorem the above
Sis not a closed surface, here; it does not divide space into an inside and outside. Thus the divergence theorem is not an option. But
∇ · F=0, so this vector field is shape independent, i.e. there is a vector field Gsuch that F=∇ × G. Since we don’t know what Gis,
however, using Stokes’ theorem to convert to a line integral is not an option.
