p
p
√
√
1.
x = 2 u − 2/3 v, y = 2 u + 2/3 v to evaluate the integral
RR Use2 the transformation
(x − xy + y 2 ) dA, where R is the region bounded by the ellipse x2 − xy + y 2 = 2.
R
√
∂x ∂y √
2
2 ∂u
∂u
p
p
Jacobian of the transformation is ∂x ∂y = = √23 + √23 = √43 .
2/3 − 2/3
∂v
∂v
The function x2 − xy + y 2 in new variables becomes
p
p
p
p
p
√
√
√
( 2 u − 2/3 v)2 − ( 2 u − 2/3 v)( 2/3 v + 2/3 u) + ( 2 u + 2/3 v)2 =
√
√
2
2
2
(2u2 − 4/ 3 uv + v 2 ) − (2u2 − v 2 ) + (2u2 + 4/ 3 uv + v 2 ) = 2u2 + 2v 2 .
3
3
3
Therefore, the curve x2 − xy + y 2 = 2 becomes 2u2 + 2v 2 = 2, which is equivalent to u2 + v 2 = 1.
The integral is equal then to
ZZ
4
(2u2 + 2v 2 ) · √ du dv.
3
u2 +v 2 ≤1
Passing to polar coordinates, we get that it is equal to
Z 2π Z 1
8
1
4π
8
√
r2 · r dr dθ = √ · 2π · = √ .
4
3 0
3
3
0
2.
R Find a function f such that ∇f =t F for F = y i + (x + z) j + y k. Use it to find the integral
F · dr along the curve C : r(t) = te i + (1 + t) j + t k, 0 ≤ t ≤ 1.
C
∂f
∂f
We have to find a function f (x, y, z) such that ∂f
∂x = y, ∂y = x + z, ∂z = y. It follows from
the first equality, that f is of the form xy + g(y, z). Substituting this into the second equality, we
get that x + ∂g(y,z)
= x + z. It follows that g(y, z) is of the form yz + h(z), hence f is of the form
∂y
xy + yz + h(z). Substituting this into the third equation, we get that y + h0 (z) is equal to y, which
implies that h(z) is a constant. It follows that we can take f (x, y, z) = xy + yz = y(x + z).
0
Beginning of the
R curve C is the point (0 · e , 1 + 0, 0) = (0, 1, 0), its end is (e, 2, 1). It follows
that the integral C F · dr is equal to f (e, 2, 1) − f (0, 1, 0) = 2(e + 1) − 1(0 + 0) = 2(e + 1).
3. Find the area of the region bounded by the curve with parametric equation hx(t),
R y(t)i =
3
3
hcos
t
i
+
sin
t
ji,
0
≤
t
≤
2π,
using
Green’s
Theorem,
by
evaluating
one
of
the
integrals
−y dx,
C
R
R
1
x
dy,
or
(−y
dx
+
x
dy).
2 C
C
Let us evaluate the third integral (one can also use the first two).
1
2
Z
2π
(− sin3 t · (cos3 t)0 + cos3 t(sin3 t)0 ) dt =
0
1
2
Z
2π
(3 sin3 t cos2 t sin t + 3 cos3 t sin2 t cos t) dt =
0
3
2
3
2
Z
Z
2π
(sin4 t cos2 t + cos4 t sin2 t) dt =
0
2π
sin2 t cos2 t(sin2 t + cos2 t) dt =
0
3
8
Z
0
3
2
Z
2π
sin2 t cos2 t dt =
0
2π
sin2 2t dt =
3
8
Z
0
2π
1 − cos 4t
3π
dt =
.
2
8
4. Find the area of the surface given by the parametric equation r(u, v) = uv i + (u + v) j + (u − v) k
for u2 + v 2 ≤ 1.
Let us find the area element dS = |r0u × r0v | du dv. We have
i j k r0u × r0v = v 1 1 = h−2, u + v, v − ui.
u 1 −1 p
√
Its length is 4 + (u + v)2 + (v − u)2 = 4 + 2u2 + 2v 2 . Hence, the area is equal to
ZZ
p
4 + 2(u2 + v 2 ) du dv.
u2 +v 2 ≤1
Let us pass to polar coordinates:
Z 2π Z 1 p
0
4 + 2r2 · r dr dθ = 2π
Z
0
1
p
4 + 2r2 · r dr.
0
Make a substitution t = 4 + 2r2 . Then dt = 4r dr, so that the area is equal to
6
Z
π
π 2 3/2 2π 6 √
= (63/2 − 43/2 ).
t dt =
t
4 4
2 3
3
t=4
RR
5. Evaluate S F · dS for F = −y j + z k, where S is the part of the paraboloid z = x2 + y 2 below
the plane z = 1 with upward orientation.
The surface is the graph of the function g(x, y) = x2 + y 2 . It follows that dS = h−2x, −2y, 1i.
The integral is equal to
ZZ
ZZ
2
2
h0, −y, x + y i · h−2x, −2y, 1i dx dy =
2y 2 + x2 + y 2 dx dy.
x2 +y 2 ≤1
x2 +y 2 ≤1
Let us pass to polar coordinates:
Z 2π Z 1
Z
(2r2 sin2 θ + r2 )r dr dθ =
0
0
2π
0
Z
1
r3 dr
0
Z
Z
1
(2 sin2 θ + 1)r3 dr dθ =
0
2π
(2 sin2 θ + 1) dθ =
0
1
4
Z
2π
(1 − cos 2θ + 1) dθ =
0
1
4
Z
2π
(2 − cos 2θ) dθ =
0
1
· 2 · 2π = π.
4
6. Use Stokes’ Theorem to evaluate C F · dr, where F = z 2 i + y 2 j + xy k, where C is the triangle
with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 2) oriented counterclockwise as viewed from above.
R
Curl of the vector field is
i
∂/∂x
z2
j
∂/∂y
y2
k
∂/∂z
xy
= hx, 2z − y, 0i.
Let us use the plane of the triangle as our surface. Equation of the plane is x + y + z/2 = 1. We
can write it as z = 2 − 2x − 2y, and parametrize it by (x, y). Then dS = h2, 2, 1i dx dy. It follows
that the integral is equal to
ZZ
ZZ
ZZ
hx, 2(2−2x−2y)−y, 0i·h2, 2, 1i dx dy =
(2x+8−8x−8y−2y)dx dy =
(8−6x−10y)dx dy,
T
T
T
where T is the projection of the triangle onto the xy-plane, i.e., the triangle in the xy plane with
vertices (1, 0), (0, 1), and (0, 0). It follows that the integral is equal to
Z
1
1−y
Z
Z
8 − 6x − 10y dx dy =
0
0
1
(8(1 − y) − 3(1 − y)2 − 10y(1 − y)) dy =
0
Z
0
1
(8 − 8y − 3 + 6y − 3y 2 − 10y + 10y 2 ) dy =
Z
0
1
(7y 2 − 12y + 5) dy =
7 12
4
−
+5= .
3
2
3