Final Review Sheet Hints—Math 222
Prof. Frank, Spring 2005
(1) Let θ∈[π, 2π] and let φ∈[0, π]. Then Φ(θ, φ) = (4 + 3 cos θsin φ, 5 + 3 sin θsin φ, −5 +
3 cos φ). The easy way to get the unit normal is to write it as 1
3(x−4, y −5, z + 5), but you
could also compute −1
3Tθ×Tφ=1
3Tφ×Tθ.
(2) You can get the derivative everywhere except where u=v. There are two cases. If u > v,
then Tu×Tv= (1,−1,1). If u < v, then Tu×Tv= (1,1,−1). You can graph the figure
by thinking about what happens on the line u=vand then thinking about what happens
to each boundary segment of D. When you’re done, go to the surfaces applet and plot it!
(You’ll need to use that |u−v|=p(u−v)2.)
(3) Since ||Tu×Tv|| =√3, you don’t need to break it into two regions. The area is √3.
(4) In both cases we need a parametrization Φ : D→R3for which Φ(D) = S. Once we
have that we can compute the normal vector Tu×Tv. When we compute ZZS
f dS as
in §7.5, we are dealing with a real-valued function f:R3→Rand the defintion is
ZZS
f dS =ZZD
f(Φ(u, v))||Tu×Tv|| dudv. Orientation is not a concern for this type
of integral since it is only the length, and not the direction, of the normal vector which
plays a role. When we compute ZZS
F·dSas in §7.6, we are integrating a vector field
F:R3→R3and the definition is ZZS
F·dS=ZZD
F(Φ(u, v)) ·(Tu×Tv)dudv. Orientation
is an issue here; switching the orientation will introduce a negative sign to the integral. If
we consider f:R3→Rto be a mass density function, then computing ZZS
f dS will give
us the total mass of the surface S(recall the “big wobbly ball” example). If instead we
consider F:R3→R3to be the velocity vector field of a fluid, then the integral ZZS
F·dS
gives us the flux across the surface S.
(5) The trick here is to notice that n=Tu×Tv
||Tu×Tv|| and reduce to an integral from §7.5.
(6) Similarly, notice that T(t) = c0(t)
||c0(t)|| and reduce to a path integral from §7.1.
(7) In the first case, let P(x, y ) = 0 and Q(x, y) = x, so that Z∂D
x dy =Z∂D
P dx +Q dy. By
Green’s theorem that equals ZZD
∂Q
∂x −∂P
∂y dA =ZZD
1dA, which is the area of D. Try
the second part yourself.
(8) If you write F(x, y ) = (P(x, y), Q(x, y)), then the scalar curl of Fis ∂Q
∂x −∂P
∂y , which is a
scalar quantity which depends on (x, y). The curl of Fis a vector field in R3which gives
an indication of the strength and direction of the rotation of the vector field. So we must
think of Fas embedded in R3by writing F(x, y, z)=(P(x, y), Q(x, y),0). Go ahead and
compute ∇ × F, you’ll see that you get ∂Q
∂x −∂P
∂y k.
(9) Since Sis closed, it has no boundary, so the integral over the boundary of Sis zero. Now
apply Stokes’ theorem.