Math 212 Multivariable Calculus - Midterm
II
April 7th, 2003
SOLUTIONS
(1) Evaluate the following double integral.
Z 1 Z √1−y2 p
3
2
1−x
dxdy.
√
0
−
1−y 2
Solution: The intergal as it stands is very hard
to calculate. Let us change the order of integration,
noting that this region is a Type 3 region. Changing
from a Type 2 region as given in the integral to a
Type 1 region we have
−1 ≤ x ≤ 1
and
0≤y≤
p
1 − x2
So
√
R 1 R √1−x2 √
R 1 R 1−y2 √
3
3
2
2
√
( 1 − x ) dx dy = −1 0
( 1 − x ) dy dx
0
− 1−y 2
R √ 2 R1 √
1−x
2
= −1 ( 1 − x ) 0
dy dx
R1
= −1 (1 − x2 )2 dx
= (x − x3 /3 + x5 /5)|1−1
= 16/15
(2) Recall (from elementary school arithmetics) that the
πr2 h
volume of a cone of radius r and height h is
.
3
Using triple integral, show that the volume of the
1
2
8π
. (Hint : Triple
3
integral of 1 over W is the volume of W .)
cone of radius 2 and height 2 is
Solution: Express the cone in cylindrical coordinates: it is all points (r, θ, z) where
0 ≤ θ ≤ 2π
0≤z≤2
0≤r ≤2−z
Then the required volume is:
R 2π R 2 R 2−z R 2π R 2
2
dr dz dθ = 0
(2 − z) /2dz dθ
0
0
0
R02π
= (4/3) 0 dθ
= 8π
3
(3) Let P be the parallelogram in the xy-plane spanned
by the two vectors (1, 2) and (1, 1) (based at the origin). Evaluate the following double integral.
Z Z
2x2 − 3xy + y 2 dA.
P
Solution: Let us change coordinates. We will find
a one-to-one onto linear map from [0, 1] × [0, 1] to the
parallelogram P .
Suppose this map is
a b
u
T (u, v) =
= (au + bv, cu + dv)
c d
v
We will choose T (1, 0) = (1, 1) and T (0, 1) = (1, 2).
With this choice, solving for a, b, c and d we get a =
1, b = 1, c = 1 and d = 2 i.e. the map is
1 1
u
= (u + v, u + 2v)
T (u, v) =
1 2
v
3
So the coordinate functions are
x(u, v) = u + v
and
y(u, v) = u + 2v
Calculating,
∂(x, y) 1 1 =
∂(u, v) 1 2 Then:
RR
2
2
P (2x − 3xy + y )dA
R 1 R 1
2
2
= 0 0 (2(u + v) − 3(u + v)(u + 2v) + (u + 2v) )du dv
R 1 R 1
= 0 0 (−uv)du dv
= −1/4
(4) Let W be the region in the second quadrant (x ≤
0, y ≥ 0) bounded by z = 1 and z = xR2 R+ Ry 2 . Use
cylindrical coordinate system to evaluate
W x dV.
Solution: In cylindrical coordinates the region over
which we are integrating is given by:
0≤r≤1
π/2 ≤ θ ≤ π
r2 ≤ z ≤ 1
Thus
RRR
R 1 R π R 1
0
π/2
r2 (r cos θ)rdz dθ dr
W xdV =
R
R1 2
π
2
= 0 r (1 − r ) π/2 cos θdθ dr
R
= − 01 r2 (1 − r2 )dr
= −2/15
4
(5) Let c be the helix c(t) = (cos t, sin t, t), 0 ≤ t ≤ π/2.
Let f (x, y, z) = xy. Evaluate the following integral.
Z
f ds.
c
Solution: Calculate
c0 (t) = (− sin t, cos t, 1)
and so
p
√
2
2
||c (t)|| = (− sin t) + (cos t) + 1 = 2
0
Thus
√
R
R π/2
2)dt
f
ds
=
(cos
t)(sin
t)(
c
0
(use substitution u = sin t)
√ R1
= √2 0 udu
=
2/2
(6) (a) Let c be the path defined by c(t) = (cos3 t, sin3 t, sin 2t ),
0 ≤ t ≤ π. Evaluate the following line integral.
Z
F · ds
c
where F(x, y, z) = (2xyz, x2 z, x2 y).
Solution: If we do this integral directly it becomes very hard. But we notice that
F(x, y, z) = ∇f (x, y, z)
where
f (x, y, z) = x2 yz
So we calculate
R
R
F
·
ds
=
c (∇f ) · ds
c
= f (c(π)) − f (c(0))
= 0−0
= 0
5
(b) Let c be the cuspidal curve c(t) = (t2 , t3 ), −1 ≤
t ≤ 1. Evaluate the following integral.
Z
xy dx + x2 dy.
c
R
Solution: Calculate
R1 2 3
2
xydx
+
x
dy
=
(t )(t )(2t) + (t4 )(3t2 )dt
c
R−1
1
= −1 5t6 dt
= 10/7