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(MATH100)[2000](s)final~2385^_10001.pdf
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2000 (spring) Final solutions of math 100
Solutions:
(1) We only consider the .rst octant instead of the whole ellipsoid. Let V (x, y, z)= xyz be the volume of rectangular on the .rst octant. Since the maximum volume must occur in the boundary, we can only consider in the region
222
x+ y+ z=1.x 0,y 0,z 0.
32 42 52
Then, we can get a two variables function
22
f(x, y)=5xy 1 . x. y, x 0,y 0
32 42
22
3xy
5y(1 .. )
.f 32 42
.x 22
xy
1 ..
32 42 22
x3y
5x(1 .. )
.f 32 42
.y 22
xy
1 ..
32 42
.f .f 34
By solving = 0 and = 0, we have x = ,y = or x =0,y = 0.
.x .y 22 34
You need to check x = ,y = which is the maximum of f(x, y) by
22
second partial derivatives. So the maximum volume of rectangular box is 34 5

V =8 = 60 2 in the ellipsoid.

22 2
(2) We cannot integrate it directly. We should change dxdy to dydx.
. 3 . 1 . 1 . 3x
sin(x 2)dxdy = sin(x 2)dydx 0 y/3 00 . 1
=3x sin(x 2)dx 0 . 1
3
= sin(x 2)d(x 2)
2 0 3
= . cos(x 2)|1
2 0 3
= (1 . cos(1))
2
1
(3)
x
.
2
1.x
. 1
V = dzdydx
00 x.1
. 1
x
=1 . x2 . (x . 1)dydx
00
. 1
= x 1 . x2 . x(x . 1)dx
0
. 1 . 1
.1
1 . x2d(1 . x 2) . (x 2 . x)dx
=
0 2
0
1
.1 x3 x2
2)3/2 .
= (1 . x +
3 32
0
111
= . +
332 1
=
2
(4) Let u = x . y, v = x + y.
|J| =
.u .u
.x .y
.v .v .x .y
= 1 .1
11 = |2| =2
(x + y) sin(x . y)dA = v sin u|J|dA
RR

. 1
2
=2v sin udvdu
0 .1

2 sin u
2
= v
1
du
.1
0
=0
(5) . from 0 to 1.

r
. from 0 to arctan = arctan 3 = .
z 3
. from 0 to 2.
. 2 . /3 . 1 V = 2 sin ddd 00 0 . 2 . /3
sin
= dd 00 3
. 2
/3
. cos
d
=
0 3
0
. 2
11
= (1 . )d
3 0 2

=
3 (6) We use Greens Theorem and R represents the region inside the ellipse
2
y
x2 + =1.
2 2
x
[(y 3 + y) . e cos y]dx +[ x 3 + x(1 + 4y 2)+ e x sin y]dy
C 3

. 2 .
x
=[ x 3 + x(1 + 4y 2)+ e x sin y] . [(y 3 + y) . e cos y]dA
.x 3.y
R
= (2x 2 +1+4y 2 + e x sin y) . (3y 2 +1+ e x sin y)dA
R
= (2x 2 + y 2)dA
R

Let x= x, y= y/ 2 and dx. = dx, 2dy. = dy. Then, the region becomes (x.)2 +(y.)2 = 1.

= [2(x .)2 + 2(y .)2]2dx.dy.
R . 2 . 1

= 22r 3drd 00

=2
(7) By Divergence Theorem,
.. ...
F ndS = F dV .div .
G

.. . 2.x 2.y 2

= (3x 2 +3y 2 + 1)dV R 2 x2+y2.1
..
= (3x 2 +3y 2 + 1)[(2 . x 2 . y 2) . (2 x2 + y2 . 1)]dA
R . 2 . 1 = (3r 2 + 1)(3 . r 2 . 2r)rdrd 00 . 1
=2 (.3r 5 . 6r 4 +8r 3 . 2r 2 +3r)dr 0
16 23
=2(.. +2 . +)
25 32
34

=
15