November 10, 2014
1
(a)
In polar coordinates, x = r cos θ and y = r sin θ. Thus we have
r2 cos2 θ
a2
+
r2 sin2 θ
b2
=1
(b)
r
Area =
R 2π R
0
0
1
cos2 θ + sin2 θ
2
a
b2
rdrdθ
(c)
Z
2π
r
Z
0
1
cos2 θ + sin2 θ
a2
b2
rdrdθ
0
π
2
Z
=4 ·
1
2(
0
π
2
Z
=4 ·
cos2
a2
θ
2( a12 +
π
2
=4 ·
0
Z
=2ab ·
+
π
2
0
=2ab
tan2 θ
)
b2
1
2( a12
Z
sin2 θ
)
b2
sec2 θ
0
Z
+
tan2 θ
)
b2
dθ
d tan θ
1
(1 +
a2 tan2 θ
)
b2
π/2
d(arctan(
0
dθ
d(
a tan θ
)
b
a tan θ
))
b
=2ab · (π/2 − 0)
=πab
We can evaluate the integral from 0 to π/2 and then multiply the result by 4 because
the ellipse is symmetric with respect to both x-axis and y-axis.
1
2
We only have to compute lim f (E)(1 − E 2 ). Set x = 1 − E. Then
E→1−
lim f (E)(1 − E 2 )
E→1−
= lim f (1 − x)x(2 − x)
x→0+
tanh−1 (1 − x)
x(2 − x)
1−x
x→0+
2 − x x(2 − x)
1
)
= lim ln(
x
1−x
x→0+ 2
2−x
= lim x ln(
)
x
x→0+
=0
= lim
Thus lim S(E) = 2π. This makes sense because when E → 1− , the ellipsoid will go to
E→1−
the two-sided unit disk whose surface area is 2π
3
(a)
Area =
R arctan ha R
0
a
cos θ
0
rdrdθ =
1
2
R arctan ha
0
a2
dθ
cos2 θ
= 21 a2 ·
h
a
= 21 ah
(b)
p
2 x2 + y 2 + y = 1
The graph is on the next page.
(c)
Area =
R 1/2 q 4 4
1
2
9 − 3 x − 3 )dx =
−1/2 (
2π
√
9 3
−
1
6
(d)
Area =
RπR
0
1
2+sin θ
0
rdrdθ =
1
2
Rπ
1
0 (2+sin θ)2 dθ.
R π/2
R π/2
1
Set φ = θ−π/2. Then the integral is equal to 12 −π/2 (2+cos
dφ = −π/2
φ)2
R 1 1+v2
2π
√ − 1
Set v = tan φ/2. The integral is equal to −1 (3+v
2 )2 dv =
6
9 3
2
1
d(φ/2).
(1+2 cos2 φ/2)2
2
2 0.5
2 ((x +y )
)+y−1 = 0
1
0.8
0.6
0.4
y
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−1
−0.5
0
x
0.5
4
(a)
+∞
Z
I=
−∞
Z +∞
=2
2
e−x dx
2
e−x dx
0
<2
+∞
X
2−i
i=0
=4
Thus it is a finite positive number.
(b)
Z
+∞
V =
2
2πxe−x dx
0
+∞
Z
= −π
0
=π
(c)
3
2
d(e−x )
1
Suppose the point (x, y, z) is on the surface and its corresponding point on the xz−plane
is (x0 , 0, z0 ). Then we have z = z0 and x2 + y 2 = x20 . Since (x0 , 0, z0 ) is on the curve
2
2
2
2
z = e−x , we see that z0 = e−x0 . Thus z = e−(x +y ) .
(d)
R +∞
2
2
2
A(x) = −∞ e−(x +y ) dy = e−x I
(e)
This is the method of slices.
(f) R
R +∞
√
2
2
+∞
V = −∞ e−x Idx = I · I = I 2 . Thus −∞ e−x dx = π.
5
(a)
Choose an integer n that is bigger than r. Then
xn
1
≤ n! lim x = 0
x→+∞ ex
x→+∞ e
0 ≤ lim xr e−x ≤ lim
x→+∞
(b)
From (a) we see that
lim xr+2 e−x = 0. Thus there exists a integer N such that
x→+∞
xr+2 e−x ≤ 1 when x ≥ N . Then
Z +∞
Z N
Z
r −x
r −x
x e dx =
x e dx +
0
0
+∞
r −x
x e
r
Z
+∞
dx ≤ N · N +
N
N
Thus the improper integral converges.
(c)
Z
+∞
A(n + 1) =
xn+1 e−x dx
0
Z +∞
=−
xn+1 d(e−x )
0
Z +∞
=
e−x d(xn+1 )
0
Z +∞
= (n + 1)
xn e−x dx
0
= (n + 1)A(n)
4
1
1
dx ≤ N r+1 +
2
x
N
(d)
+∞
Z
A(0) =
e−x dx = 1
0
Thus A(n) = n!.
(e)
√
Set x = u2 . Then dx = 2udu = 2 xdu.
1
A(− ) = 2
2
Z
+∞
2
e−u du =
√
π
0
6
6
The Taylor Series for F 0 (x) = 3x2 e−x at 0 is
+∞
P
i=0
for F (x) is
+∞
P
i=0
3·(−1)i 6i+3
i!(6i+3) x
5
3·(−1)i 6i+2
x
.
i!
Thus the Taylor Series