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Math 100 -Introduction to Multivariable Calculus
FINAL EXAMINATION
Spring Semester, 2001
Time Allowed: 2 Hours. Total Marks: 100
Student Name:
Student Number:
1. (20 marks) Show that for any continuous function f(x, y),
. a .. x .. a .. a .
f(x, y)dydx = f(x, y)dxdy.
00 0 y
2.
(15 marks) Find the angle , which is formed by the x-axis and the ray as in the following .gure, such that the two shaded regions have the same area. Here the two circles are C1 : x2 + y2 = 4 and C2 : x2 +(y . 1)2 = 1.
3.
(15 marks) Evaluate the line integral
.
.
.
.
C2 .
.
.
.
.
=
.
C1
. (x + y)dx . (x . y)dy
,
22 + y
where C is the pentagon inscribed in the circle x2 + y2 = R2 (R> 0) along the counterclockwise direction as shown in the .gure.
C x
.
.
.
.
.
2
x2 + y= R2
1
4. (20 marks) Use Stokes theorem to show that for the di.erentiable
vector .eld
F.(x, y, z)= f(x).i + g(y).j + h(z).k,
we have
F d.r = F d.r,
C1 C2
where C1 and C2 are two piecewise smooth paths with the same end points.
5. (15 marks) Use the direct approach to compute the surface integral
F .ndS
where
F (x, y, z)=(x + y 2 + e z).i +(y + sin(z . x)).j + z.k,
and is the upper-hemisphere x2 + y2 + z2 = 1, z 0, with the normal vector along positive z-axis.
6. (15 marks) Add a suitable surface to use the divergence theorem to evaluate the surface integral
F .ndS
where
2 z)..
F (x, y, z)=(x + y + ei +(y + sin(z . x)).j + zk,
and is the upper-hemisphere x2 + y2 + z2 = 1, z 0, with the normal vector along positive z-axis.
1. From the .rst iterated integral, we know that the region of the integral is a triangular region. Thus, if we .rst integrate it in y, we have the right-hand side.
y = x
.
.
.
..
a a
2. By the condition, we have
. . 2 sin . /2 . 2
r drd = r drd.
00 2 sin
i.e.,
. cos sin = . + . cos sin
2
Hence, = /4.
3. In any region not containing (0, 0), the vector .eld x + yx . y
F = i . j
x2 + y2 x2 + y2
is conservative since
2
. . x + y . x2 . 2xy . y. . x . y .
= =(.1)
.y x2 + y2 (x2 + y2)2 .x x2 + y2
Thus, by the Greens theorem,
F d.r = F d.r,
CCa
where Ca is the circle x2 +y2 = a2 along the counterclockwise direction. The latter integral can be evaluated as
. .. (x + y)dx . (x . y)dy
F d.r =
Ca Ca x2 + y2
. 2 (a cos t + a sin t)(.a sin t) . (a cos t . a sin t)(a cos t)
= dt
a2 = .2.
0
4
4. It is easy to see that for the given vector .eld,
. .i .j.k .
...
Curl F.= det . =0.
..x .y .z .
.
f(x) g(y) h(z)
Since C1 and C2 . for a closed path (C2 . is the path C2 but with the opposite direction), thus, by Stokes theorem, we have
F d.r =0,
C1+C.
2
Hence
F d.r = . F d.r = F d.r.
C.
C12 C2
5. The surface is
.r = x.i + y.j + .1 . x2 . y2.k
We calculate
..r..r x y