Math 251
Exam III (training)
Instructor: Volodymyr Nekrashevych,
Fall Term 2012
1. Use the
RR transformation x = 2u + 3v, y = 3u − 2v to evaluate the
integral R (x+y) dA, where R is the square with vertices (0, 0), (2, 3),
(5, 1), and (3, −2).
We have
∂(x, y) 2 3 =
= −4 − 9 = −13.
∂(u, v) 3 −2 If (x, y) = (0, 0), then (u, v) = (0, 0). The point (x, y) = (2, 3)
oviously corresponds to (u, v) = (1, 0). The point (x, y) = (5, 1) corresponds to (u, v) = (1, 1), and (x, y) = (3, −2) corresponds to (0, 1).
It follows that the square R corresponds to the square with vertices
(0, 0), (1, 0), (0, 1), and (1, 1) in the uv-plane.
In general, when the points in the uv-plane are not obvious, you just
have to solve the corresponding systems for u and v.
The function x + y becomes 2u + 3v + 3u − 2v = 5u + v. It follows
that the integral is equal to
Z 1Z 1
Z 1Z 1
Z 1Z 1
(5u + v) · 13 du dv = 65
u du dv + 13
v du dv =
0
0
0
0
0
0
65 13
+
= 39.
2
2
R
2. Evaluate the integral C xy 4 ds, where C is the right half of the
circle x2 + y 2 = 16.
The curve is parametrized by hx, yi = h4 cos θ, 4 sin θi for θ ∈ [−π/2, π/2].
We have then ds = |h−4 sin θ, 4 cos θi| = 4, hence
Z
Z π/2
4
xy ds =
4 cos θ · 44 sin4 θ · 4 dθ =
−π/2
C
6
Z
π/2
sin4 θ d sin θ =
4
−π/2
5 π/2
213
sin
θ
6 2
=
4
·
=
.
46 ·
5 θ=−π/2
5
5
3. Determine whether or not F is a conservative vector field. If it is,
find a function f such that F = ∇f .
F = (x2 + y)i + (y 2 + x)j.
2
2
We have ∂(y∂x+x) = 1, and ∂(x∂y+y) = 1, which implies that the field is
conservative.
∂f
2
2
We have to find a function f such that ∂f
∂x = x + y and ∂y = y + x.
From the first equality, we conclude that f (x, y) = x3 /3 + xy + C(y)
for some function C. Let us substitute this into the second equation:
∂x3 /3 + xy + C(y)
= y 2 + x,
∂y
3
= x + C 0 (y), hence C 0 (y) = y 2 , so that we can take
but ∂x /3+xy+C(y)
∂y
C(y) = y 3 /3. Consequently, we can take f (x, y) = x3 /3 + xy + y 3 /3.
R
4. Use Green’s Theorem to evaluate the line integral C x2 y dx−3y 2 dy,
where C is the circle x2 + y 2 = 1.
RR ∂(−3y2 )
−
By Green’s Theorem, the line integral is equal to
D
∂x
RR 2
∂x2 y
2
2
D x dA, where D is the disc x + y ≤ 1. Passing
∂y dA = −
to polar coordinates, we get
ZZ
Z 2π Z 1
2
−
x dA = −
r2 cos2 θ · r dr dθ =
D
0
0
Z 2π
Z 1
2
−
cos θ dθ ·
r3 dr =
0
Z 2π 0
Z 1
1 + cos 2θ
π
−
dθ ·
r3 dr = − .
2
4
0
0
RR
5. Evaluate the integral S Fp· dS, where F = −xi − yj + z 2 k, and
S is the part of the cone z = x2 + y 2 between the planes z = 1 and
z = 2 with upward orientation.
Let us parametrize the cone using the polar coordinates in the xyplane. The parametrization is then given by
hr cos θ, r sin θ, ri,
p
since z = x2 + y 2 = r.
i
j
sin θ
dS = cos θ
−r sin θ r cos θ
Then
k 1 dr dθ =
0
h−r cos θ, −r sin θ, r(cos2 θ+sin2 θi dr dθ = h−r cos θ, −r sin θ, ri dr dθ.
Note that we write the derivatives over r above the derivatives over θ,
since this will give us the upper orientation (evidenced by the positive
value of the third coordinate). If you get negative value of the third
coordinate, then it means that your vector is pointing down, so you
have to change the signs.
The integral is
ZZ
~=
F · dS
Z 2πSZ 2
(−r cos θ)(−r cos θ) + (−r sin θ)(−r sin θ) + r2 · r dr dθ =
0
1
Z 2π Z 2
2
(r2 + r3 ) dr dθ = 2π · (r3 /3 + r4 /4)r=1 =
0
1
2π(8/3 + 4 − 1/3 − 1/4) =
73π
.
6
Alternatively, one can parametrize the cone by x, y, use
* p
+ *
+
p
2
2
2
2
∂ x +y
∂ x +y
x
y
dS = −
,−
, 1 = −p
, −p
,1
∂x
∂y
x2 + y 2
x2 + y 2
and pass to polar coordinates in the integral after that.
RR
6. Use the Divergence Theorem to calculate the surface integral S F ·
dS, where F = x3 i + y 3 j + z 3 k, and S is the sphere x2 + y 2 + z 2 = 1.
We have div(F) = 3x2 + 3y 2 + 3z 2 . Passing to spherical coordinates,
we get that divergence is equal to 3ρ2 , and hence we have to compute
the integral
Z 2π Z π Z 1
3ρ2 · ρ2 sin φ dρ dφ dθ =
0
0
0
Z π
Z 1
1 12π
.
3 · 2π ·
sin φ dφ ·
ρ4 dρ = 6π · 2 · =
5
5
0
0