Test #3
April 10, 2015
Solutions
1. Reverse the order of integration and evaluate.
Z
1/2
0
Z
1/4
2
4y cos(x ) dx dy =
y2
Z
1/4
Z
1/4
Z
√
x
4y cos(x2 ) dx dy
0
0
2x cos(x2 ) dx
0
1/4
= sin(x2 )
=
0
= sin(1/16)
2. Changing to polar coordinates we find
Z
1
−1
Z
√
1−x2
2
√
− 1−x2
(1 +
x2
+
y 2 )2
2π
1
2
r dr dθ
2 2
0
0 (1 + r )
Z 2π
1 1
=
−
dθ
1 + r2 0
0
Z 2π
1
=
dθ
2
0
= π.
Z
dy dx =
Z
3. Evaluate the following integral.
Z
0
1
Z
√
0
x
Z
√
1−x2
1
4yz dz dy dx =
Z
1
=
Z
0
0
0
= 1/4
1
Z
√
x
2y(1 − x2 ) dy dx
0
x(1 − x2 ) dx
4. Verify Green’s theorem in flux form for the region in the first quadrant bounded by x = 0, y = x
and the unit circle and the vector field given by F = hx, yi.
Solution. We have to evaluate
Z π/2 Z
π/4
1
∇ · h−y, xi r dr dθ =
0
I
hx, yi · n ds.
The left side is π/4. The right side has three pieces.
• On the line y = x, x(t) = t = y(t), 0 ≤ t ≤ 1, r = ht, ti, r′ = h1, 1i, n = h1, −1i. Thus
Z
Z 1
F · n ds =
ht, ti · h1, −1i dt = 0.
0
• On the curve portion x(t) = cos t, y(t) = sin t, π/4 ≤ t ≤ π/2. So n = hcos t, sin ti and the
flux on this portion is
Z
Z π/2
F · n ds =
hcos t, sin ti · hcos t, sin ti dt = π/4.
π/4
• On the line x = 0, x(t) = 0, y(t) = 1 − t, 0 ≤ t ≤ 1, r = h0, 1 − ti, r′ = h0, −1i, n = h−1, 0i.
Thus
Z
Z 1
F · n ds =
h0, 1 − ti · h−1, 0i dt = 0.
0
The sum is π/4 as expected.
5. Verify Green’s theorem in circulation form for the region in the first quadrant bounded by x = 0,
y = x and the unit circle and the vector field given by F = h−y, xi.
Solution. We have to evaluate
Z π/2 Z
π/4
1
∇ × hx, yi r dr dθ =
0
I
hx, yi · T ds.
The left side is π/4. The right side has three pieces as in the previous problem.
• As above, on y = x, r′ = h1, 1i. Thus
Z
Z 1
′
F · r dt =
h−t, ti · h1, 1i dt = 0.
0
• On the curve portion x(t) = cos t, y(t) = sin t, π/4 ≤ t ≤ π/2.
Z
Z π/2
′
F · r dt =
h− sin t, cos ti · h− sin t, cos ti dt = π/4.
π/4
• On the line x = 0, x(t) = 0, y(t) = 1 − t, 0 ≤ t ≤ 1, r′ = h0, −1i. Thus
Z
Z 1
′
F · r dt =
ht − 1, 0i · h0, −1i dt = 0.
0
The sum is π/4 as expected.
2