Evaluate the integral

Mathematics
1. Show that the line integral is independent of path and evaluate the integral:
R
C
2x sin y + e
3z dx + x
2
cos y dy + 3xe3z + 5 dz, where C is any path from (1, 0, 0), to
(2, p/2, 1).
2. Calculate R
C F · dr where
F(x, y) = 
2xy
(x
2 + y
2
)
2
,
y
2 – x
2
(x
2 + y
2
)
2

and C is any positively oriented simple closed curve that encloses the origin.
3. (a) Let f = f(x, y, z), F = F(x, y, z). Verify the identity (assume the necessary
partial derivatives exist and are continuous):
div(f F) = f div F + F · ?f (1)
(b) Let r = hx, y, zi and r = ||r|| =
p
x
2 + y
2 + z
2
. Verify
i. ? × r = 0
ii. ? · (rr) = 4r.
4. The velocity vector field for the two-dimensional flow of an ideal fluid around a
cylinder is given by
F(x, y) = A

1 –
x
2 – y
2
(x
2 + y
2
)
2
, –
2xy
(x
2 + y
2
)
2

for some positive constant A.
(a) Show that F is irrotational i.e show that curl F = 0.
(b) Show that F is incompressible i.e show that div F = 0.
5. Zill section 9.12 p.536 #29.
6. Evaluate the line integral by two methods: by using Green’s theorem and by
evaluating directly. H
C
x
2
y
2 dx + xy dy, where C consists of the arc of the parabola
y = x
2
from (0, 0) to (1, 1) followed by the line segment from (1, 1) to (0, 1) followed
by the line segment from (0, 1) to (0, 0).
7. (a) Use the divergence-flux form of Green’s theorem (see Green’s theorem handout
on Sakai equation (2) in the pdf.) to prove Green’s first identity:
Z Z
D
f4g dA =
I
C
f
?g
?n ds –
Z Z
D
?f · ?g dA, (2)
1
MATH 352 – 012 Homework–3 Due: 2:00pm Thursday 4/7/16
where D and C satisfy the hypotheses of Green’s theorem and the appropriate
partials of f and g exist and are continuous. The derivative ?g
?n is called the
normal derivative of g and is given by the directional derivative ?g · n =
?g
?n,
and 4g = ? · ?g = ?2
g is the Laplacian of g.
(b) A function g is called a harmonic function on D if it satisfies Laplace’s equation
4g = 0 on D. Use Green’s first identity with the same hypotheses as in part (a)
to show that if g is harmonic on D then:
I
C
?g
?n ds = 0.

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