(1) Evaluate the following integral by first changing the order of integration.
Z 4Z 2
cos(x3 ) dxdy
√
0
y
Solution: By changing the order of integration, we have
Z 2
Z 2 Z x2
Z 4Z 2
1
3
3
x2 cos(x3 ) dydx = sin(8),
cos(x ) dydx =
cos(x ) dxdy =
√
3
0
0
0
y
0
where we computed the final integral using the substitution u = x3 .
(2) Let R be the region bounded by
x + y = 0 x + y = 2 x − y = 0 x − y = 2.
By first applying a change of variables, evaluate
Z Z
2
2
(x + y)ex −y dxdy.
R
Solution: By inspection of the integral, we make the substitution
(0.1)
u=x+y
and v = x − y.
Using this substitution, and applying the change of variables formula, the
integral simplifies to
Z Z
ueuv |DT | dvdu,
R∗
∗
where T (R ) = R. However, T is a transformation that sends R∗ in the
uv-plane to the to R in the xy-plane, so to find T , we need to express x
and y in terms of u and v. By adding both equations in Eq.(0.1), we get
u v
u + v = 2x, or x = + .
2 2
By subtracting the second equation from the first in Eq.(0.1), we get
u v
u − v = 2y or y = − .
2 2
Therefore our transformation is T (u, v) = ( u2 + v2 , u2 − v2 ), and |DT |, the
absolute value of the determinant of the Jacobian, is
1 1 1
|DT | = det 12 2 1 = .
−2
2
2
To find R∗ , we use the fact that T −1 (x, y) = (x + y, x − y), which is our
original substitution. We simply map the vertices of R using T −1 to the
vertices of R∗ . The vertices of R are (0, 0), (1, 1), (2, 0), (1, −1) and we
find that
T −1 (0, 0) = (0, 0), T −1 (1, 1) = (2, 0), T −1 (2, 0) = (2, 2), T −1 (1, −1) = (0, 2).
2
So R∗ is the square with vertices (0, 0), (2, 0), (2, 2) and (0, 2). Plugging
all this information into the change of variables formula, we get
Z
Z Z
Z 2Z 2
1 2 2u
1
1 4
1
1
uv
uv
dvdu =
(e −1) du =
e −2 −
.
ue |DT | dvdu =
ue
2
2 0
2
2
2
R∗
0
0
(3) Given the vector field
F (x, y, z) = (−2 sin(2x)e5yz , 5z cos(2x)e5yz , 5y cos(2x)e5yz )
(a) Find a scalar
R function f such that ∇f = F .
(b) Evaluate C F · ds , from (0, 0, 0) to ( π2 , 1, 1) along the path c(t) =
( π2 t, t3 sin( π2 t), t4 cos(2πt)).
Solution:
(a) If there exists f such that ∇f = F , then
(0.2)
(0.3)
(0.4)
∂f
∂x
∂f
5z cos(2x)e5yz =
∂y
∂f
5y cos(2x)e5yz =
∂z
By integrating Eq(0.2) with respect to x, Eq.(0.3) with respect to y
and Eq(0.4) with respect to z, we have that
−2 sin(2x)e5yz =
f (x, y, z) = cos(2x)e5yz + f1 (y, z)
f (x, y, z) = cos(2x)e5yz + f2 (x, z)
f (x, y, z) = cos(2x)e5yz + f3 (x, y)
We pick f1 , f2 , f3 to make the above expressions equal, so clearly we
choose f1 = f2 = f3 = 0 and
f (x, y, z) = cos(2x)e5yz .
(b) Given that F = ∇f , we have that
Z
π
F · ds = f ( , 1, 1) − f (0, 0, 0) = −e5 − 1
2
C
(4) Let S be the portion of the unit sphere with z ≥ 12 .
(a) Parametrize S.
(b) Compute the area of S.
Solution:
(a) We can use our standard parametrization of the sphere with appropriate limits on φ and θ. We have that
Φ(φ, θ) = (sin(φ) cos(θ), sin(φ) sin(θ), cos(φ)).
3
where 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π3 . To determine the upper bound
on φ, we choose φ so that cos(φ) = 12 , which is when z = 12 on the
sphere.
(b) Using the formula for the area, we have that
Z
Z 2π Z π
3
kTφ × Tθ k dφdθ.
Area(S) =
1 dS =
0
0
S
We can easily compute kTφ × Tθ k:
q
kTφ × Tθ k = sin4 (φ) cos2 (θ) + sin4 (φ) sin2 (θ) + cos2 (φ) sin2 (φ) = sin(φ).
So
Z 2π Z
Area(S) =
0
π
3
Z
2π
Z
kTφ × Tθ k dφdθ =
0
0
0
π
3
√ !
3
sin(φ) dφdθ = 2π 1 −
.
2
(5) Find the area enclosed by the path
π π
c : [− , ] → R2 ,
2 2
c(t) = (2 cos(t), 5 sin(2t)).
Solution: By Green’s Theorem and the definition of a line integral,
Z
Z π
2
1
1
− ydx+ xdy =
Area Enclosed =
(5 sin(2t) sin(t) + 10 cos(t) cos(2t)) dt.
2
2
C
− π2
By applying the double angle formulas sin(2t) = 2 sin(t) cos(t) and cos(2t) =
cos2 (t) − sin2 (t), we have
Z π
Z π
2
2
(5 sin(2t) sin(t) + 10 cos(t) cos(2t)) dt =
10 cos3 (t) dt.
− π2
− π2
By writing cos2 (t) = 1 − sin2 (t), we have
π2
Z π
Z π
3
2
2
sin
(t)
= 40 .
10 cos3 (t) dt = 10
cos(t)−sin2 (t) cos(t) dt = 10 sin(t) −
π
3
3
− π2
− π2
−2
R
(6) Given F (x, y, z) = (y − 2z, z − x, 2x − y), compute C F · ds, where C
is the square with vertices (0, 0, 0), (1, 0, 0), (1, 1, 0), (0, 1, 0) traversed in
that order.
Solution: We could compute this directly by parametrizing each edge
of the square. Instead, we will apply Stokes’ Theorem. By Stokes’ Theorem, we have that
Z
Z Z
F · ds =
((∇ × F ) · n)dS,
C
S
4
where S is the interior square enclosed by C oriented with upward pointing, unit normal n. The normal is upward pointing given the orientation
of C. By direct computation, we find that
∇ × F = (−2, −4, −2),
and n = (0, 0, 1), given that S lies in the xy-plane. So
Z Z
Z Z
((∇ × F ) · n)dS =
(−2, −4, −2) · (0, 0, 1) dS = −2
1 dS = −2,
S
S
S
RR
where
1 dS = 1 given that S is the unit square. R R
S
(7) Given F (x, y, z) = (x + yz, −y + xz, 3z + xy), evaluate
F · dS, where
2
2
2
S is the sphere x + y + z = 4.
Z Z
Solution: We could compute this directly, but instead we use the
divergence theorem. Let V be the ball of radius 2 enclosed by S. Then
we
have
Z Z
Z Z Z
Z Z Z
Z Z Z
4π
F ·dS =
∇·F dV =
3 dV = 3
1 dV = 3 (2)3 = 32π,
3
S
V
RRR
4π
where we used the fact that ∇ · F = 3 and
1 dV = 3 (2)3 is the
volume of the ball of radius 2.