MATH 3170 W15 Test 2 Solutions
Z
1.
a) Show that the integral
1
(y − x2 ) dx + (x + y 2 ) dy is the same for any smooth
C
curve C from (−1, 1) to (0, 3).
Solution The vector field is F(x, y) = (y − x2 )i + (x + y 2 )j where P (x, y) = y − x2
and Q(x, y) = x + y 2 . Since ∂Q
= ∂P
= 1, we see that F is conservative. This
∂x
∂y
R
means that F is also path independent. Thus C (y − x2 ) dx + (x + y 2 ) dy is the
same for any smooth curve C from (−1, 1) to (0, 3).
b) Evaluate the integral.
Solution We can choose any curve we want from (−1, 1) to (0, 3). Let C be the
line segments L1 from (−1, 1) to (0, 1) and L2 from (0, 1) to (0, 3). Then
Z
Z
2
2
(y − x ) dx + (x + y ) dy =
(y − x2 ) dx + (x + y 2 ) dy
C
ZL1
(y − x2 ) dx + (x + y 2 ) dy
+
L
Z 02
Z 3
2
=
(1 − x ) dx +
y 2 dy
−1
1
2
1
28
=
+9− = .
3
3
3
I
2. Use Green’s Theorem to evaluate
F · dr where F(x, y) = y 2 cos xi + (x2 + 2y sin x)j
C
and C is the triangle with vertices (0, 0), (2, 0) and (2, 6).
Solution We have F(x, y) = P (x, y)i + Q(x, y)j where P (x, y) = y 2 cos x and Q(x, y) =
x2 + 2y sin x. By Green’s theorem, we have
I
Z Z
∂Q ∂P
F · dr =
(
−
) dA
∂y
C
D ∂x
Z Z
Z 2 Z 3x
=
2x dA =
2x dy dx
D
0
0
Z 2
=
6x2 dx = 16.
0
3. Let F(x, y, z) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k be a vector field where P, Q, R
have continuous partial derivatives. Show that div(curl F) = 0.
Solution See notes.
MATH 3170 W15 Test 2 Solutions
2
4. Using a line integral, find the area of the region in the plane bounded by the line y = x
and the curve y = x2 .
Solution Let C be the closed curve defined by L1 and L2 , where L1 is the portion of
the curve y = x2 from (0, 0) to (1, 1), and L2 is the line segment
H from (1, 1) to (0, 0).
1
The area of the region D enclosed by the curves is given by 2 C −y dx + x dy. We
parameterize the curves L1 and L2 as below.
L1 : r(t) =< t, t2 >, 0 ≤ t ≤ 1,
L2 : r(t) =< 1 − t, 1 − t >, 0 ≤ t ≤ 1.
We have
Z
Z
I
1
1
−y dx + x dy +
−y dx + x dy
−y dx + x dy =
2 C
2
L1
L2
Z 1
Z 1
1
2
=
−(1 − t) · (−1) + (1 − t) · (−1) dt
(−t ) · 1 + t · (2t) dt +
2
0
0
Z
1 1 2
1
=
t dt = .
2 0
6
5. Find the surface area of the paraboloid given by z = 2(x2 + y 2 ), 0 ≤ z ≤ 8.
RR
Solution The surface area is given by
dS. Using polar coordinates, we see that
S
r = (r cos θ, r sin θ, 2r2 ), 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π is a parametrization of S. We have
rθ =< −r sin θ, r cos θ, 0),
rr =< cos θ, sin θ, 4r >
and
rθ × rr =< 4r2 cos θ, 4r2 sin θ, −r >
Thus we have
Z Z
Z Z
krθ × rr k dA
dS =
S
Z
D
2π Z 2
=
0
√
r 16r2 + 1 dr dθ
0
3
1
= 2π ((65) 2 − 1)
48
MATH 3170 W15 Test 2 Solutions
3
6. Let F(x, y, z) = 4xi − 2y 2 j + z 2 k be a vector field. Let S be the surface which is the
portion of the cylinder
x2 + y 2 = 4 between z = 0 and z = 3. Evaluate the vector
Z Z
F · dS.
surface integral
S
Solution We can parameterize S as follows:
r(u, v) =< 2 cos(u), 2 sin(u), v >, 0 ≤ u ≤ 2π, 0 ≤ v ≤ 3.
From this, we see that ru =< −2 sin u, 2 cos u, 0 > and rv =< 0, 0, 1 > . Taking the
cross product, we obtain ru × rv = 2 cos ui + 2 sin uj. Now we see that
Z Z
Z Z
F · dS =
F(r(u, v)) · (ru × rv ) dA
S
Z ZD
=
< 8 cos u, −8 sin2 u, v 2 > · < 2 cos u, 2 sin u, 0 > dA
Z 3 ZD π
(16 cos2 u − 16 sin3 u) du dv
=
0
Z 2π0
16 cos2 u du
= 3
Z0 2π
1 + cos(2u)
= 48
du = 48π.
2
0
Bonus Is there a vector field F(x, y, z) such that curl F = 2i + j + 3k.?
Solution Try F = zi + 3xj + 2yk.