# SOLUTION OF QUIZ 6 April 19, 2013 ```SOLUTION OF QUIZ 6
MINGFENG ZHAO
April 19, 2013
1. [5 Points] Let E be an open Jordan region in Rn and f : E → R be continuous on E. Prove that
for every x ∈ E,
lim
r→0+
1
Vol(Br (x))
Z
f dV = f (x).
Br (x)
Proof. For any x ∈ E and fix, since E is open, then there exists some r0 &gt; 0 such that
Br0 (x) ⊂ E.
Since f is continuous on E, then for any &gt; 0, there exists some δ ∈ (0, r0 ) such that for all
y ∈ Bδ (x), we have
|f (x) − f (y)| &lt; .
So for all 0 &lt; r &lt; δ, we get
=
=
≤
≤
=
Z
1
f (y) dV − f (x)
Vol(Br (x)) Br (x)
Z
Z
1
1
f (y) dV −
f (x) dV Vol(Br (x)) Br (x)
Vol(Br (x)) Br (x)
Z
1
[f (y) − f (x)] dV Vol(Br (x)) Br (x)
Z
1
|f (y) − f (x)| dV
Vol(Br (x)) Br (x)
Z
1
dV
Vol(Br (x)) Br (x)
.
1
2
MINGFENG ZHAO
Hence we know that
1
Vol(Br (x))
lim
r→0+
Z
f dV = f (x).
Br (x)
2. [5 Points] Evaluate the iterated integral:
Z
0
Z
1
Proof. First look at
p
1
Z
1
√
Z
1
p
x3 + z dzdxdy.
x3
y
x3 + z dz, we have
x3
Z
1
p
x3 + z dz
1
3
1
3
2
(x
+
z)
1 + 12
x3
i
h
3
3
2
(x3 + 1) 2 − (x3 + x3 ) 2
3
√
3
2 3
4 2 9
2
(x + 1) −
x2 .
3
3
=
x3
=
=
Z
1
Let f (x, y) =
p
x3 + z dz, then we know that
x3
Z
0
1
Z
1
√
1
Z
Z
f (x, y) dxdy =
y
x2
f (x, y) dydx.
0
0
So we get
Z
0
1
Z
1
√
y
Z
1
x3
p
x3
#
√
3
2 3
4 2 9
2
2
=
(x + 1) −
x
dydx
3
3
0
0
#
√
Z 1&quot;
3
4 2 9 2
2 3
2
(x + 1) 2 &middot; x −
x2 &middot; x
dx
=
3
3
0
√ Z
Z
3
2 1 3
4 2 1 13
2
=
(x + 1) 2 &middot; x dx −
x 2 dx
3 0
3
0
√ Z
Z
3
2 1 1 3
4 2 1 13
&middot;
(x + 1) 2 d(x3 + 1) −
x 2 dx
=
3 3 0
3
0
h 5
i 4 √2
2
1
1
2
=
&middot;
&middot; 2 −1 −
&middot;
9 1 + 32
3
1 + 13
2
√
√
4
8 2
=
&middot; [4 2 − 1] −
45
45
√
8 2
4
=
−
45
45
Z
+ z dzdxdy
1
Z
x2
&quot;
SOLUTION OF QUIZ 6
3
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009