18.024 HW 9
April 15, 2010
11.18 6 We assume that the density is 1 everywhere. Compute the integrals for total mass and x̄,
ȳ as follows:
Z a Z log x
dydx = a log a − a + 1
m(S) =
1
Z
0
a Z log x
x̄m(S) =
1
Z
ȳm(S) =
1
0
a Z log x
0
1
xdydx = (2a2 log a − a2 + 1)
4
1
ydydx = −1 + a − a log a + a(log a)2
2
This then implies that
x̄ =
ȳ =
2a2 log a − a2 + 1
4(a log a − a + 1)
−1 + a − a log a + 12 a(log a)2
a(log a)2
=
−1
a log a − a + 1
2(a log a − a + 1)
11.18 7 Again assume that the density is 1 everywhere and compute the integrals
Z 1 Z 1+x−2√x
1
m(S) =
dydx =
6
0
0
Z 1 Z 1+x−2√x
1
x̄m(S) =
xdydx =
30
0
0
Z 1 Z 1+x−2√x
1
ȳm(S) =
ydydx =
30
0
0
So this gives
1
5
1
ȳ =
5
x̄ =
H
11.22 Use Greens theorem to evaluate C y 2 dx + xdy when
b. C is the square with vertices (±1, ±1) then by Greens theorem this integral equals
Z 1Z 1
(1 − 2y)dxdy = 4
−1
−1
1
c. C has the vector equation a(t) = 2 cos3 ti + 2 sin3 tj, 0 ≤ t ≤ 2π.
This is the graph of the “hypocycloid with 4 cusps”, also known as an “astroid” (this is interesting
in itself, check out the wikipedia page for some cool animations). In any case, we need to find a
Cartesian equation for the parametric equation a(t). We do this using the relation cos2 t+sin2 t = 1
to write
x2/3 + y 2/3 = 22/3 cos2 t + 22/3 sin2 t = 22/3
then solving for y gives y = (22/3 − x2/3 )3/2 . Since the integrand does not depend on x and the
region is symmetric about the x and y axis, we integrate only along the right half and then multiply
this quantity by 2. This gives
2 Z (22/3 −x2/3 )3/2
Z
(1 − 2y)dydx = 3π/2
2
−(22/3 −x2/3 )3/2
0
11.22 4 Compute:
Z Z
Z Z ∂v
∂v
∂u ∂u
v(x, y) +
u(x, y)dxdy
−
−
f · gdxdy =
∂x ∂y
∂x ∂y
R
Z Z ∂u
∂v
∂v
∂u
=
v(x, y) +
u(x, y) −
u(x, y) +
v(x, y) dxdy
∂x
∂y
∂y
R ∂x
Now let P = Q = u(x, y)v(x, y) then this equals
Z Z ∂Q ∂P
−
dxdy
∂x
∂y
R
Using Greens theorem on this and the parametrization of the circle (cos t, sin t) 0 ≤ t ≤ 2π then we
have
Z
2π
(u(cos t, sin t)v(cos t, sin t), u(cos t, sin t)v(cos t, sin t)) · (− sin t, cos t)dt
0
Since this is a contour integral over the boundary, we use the fact that on this u(x, y) = 1 and
v(x, y) = y to write
Z 2π
=
(sin t, sin t) · (− sin t, cos t)dt
0
Z
=
2π
(− sin2 t + sin t cos t)dt = −π
0
11.22H 5 If f and g are
H continuously differentiable on an open connected set S in the plane show
that f ∇g · dα = − g∇f dα for every piecewise smooth Jordan curve C in S.
Write out both integrals explicitly:
I
I
∂g
∂g
f ∇g · dα = f dx + f dy
∂x
∂y
I
I
∂f
∂f
− g∇f · dα = g dx + g dy
∂x
∂y
2
Check that the hypotheses of Green’s theorem are satisfied and apply:
I
Z Z Z Z ∂f ∂g
∂2g
∂f ∂g
∂2g
∂f ∂g ∂f ∂g
f ∇g · dα =
+f
−
−f
dydx =
−
dydx
∂x ∂y
∂x∂y
∂y ∂x
∂y∂x
∂x ∂y
∂y ∂x
where in the final step we use the equality of mixed partials. Similarly for the second integral:
I
Z Z Z Z ∂g ∂f
∂g ∂f
∂2f
∂g ∂f
∂2f
∂g ∂f
− g∇f ·dα = −
dxdy =
dxdy
+g
−
−g
−
∂x ∂y
∂x∂y ∂y ∂x
∂y∂x
∂y ∂x ∂x ∂y
and these two quantities are equal.
R
R
18.22 7 If f = Qi − P j show that C P dx + Qdy = C f · nds. Use equation given above for
line integrals with respect to arc length:
Z
Z
Z
X 0 (t)i + Y 0 (t)j
1
P dx + Qdy = (P i + Qj) ·
ds =
(P X 0 (t) + QY 0 (t))ds
0 (t)k
0 (t)k
kα
kα
C
C
C
on the other hand
Z
f · nds =
C
1
0
kα (t)k
Z
(Q, −P ) · (Y 0 (t), −X 0 (t))ds =
C
1
0
kα (t)k
Z
(QY 0 (t) + P X 0 (t))ds
C
and these are equal.
11.22 8 Use the definition of the normal derivative to write:
I
I
I ∂g
∂g
∂g
f
ds =
f ∇g · nds =
f i + f j · nds
∂x
∂y
C ∂n
C
by the previous problem this equals
Z
−f
C
now apply Green’s theorem
Z Z ∂g
∂g
dx + f dy
∂y
∂x
∂f ∂g
∂2g
∂f ∂g
∂2g
+f 2 +
+f 2
∂x ∂x
∂x
∂y ∂y
∂y
Z Z
=
(f ∇2 g + ∇f · ∇g)dxdy
I f
C
∂f
∂g
−g
∂n
∂n
dxdy
ds
for this we can just use the linearity of the line integral to split this integral up and then apply the
previous problem twice.
3