advertisement

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 > 0 such that Br0 (x) ⊂ E. Since f is continuous on E, then for any > 0, there exists some δ ∈ (0, r0 ) such that for all y ∈ Bδ (x), we have |f (x) − f (y)| < . So for all 0 < r < 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" 3 4 2 9 2 2 3 2 (x + 1) 2 · x − x2 · x dx = 3 3 0 √ Z Z 3 2 1 3 4 2 1 13 2 = (x + 1) 2 · x dx − x 2 dx 3 0 3 0 √ Z Z 3 2 1 1 3 4 2 1 13 · (x + 1) 2 d(x3 + 1) − x 2 dx = 3 3 0 3 0 h 5 i 4 √2 2 1 1 2 = · · 2 −1 − · 9 1 + 32 3 1 + 13 2 √ √ 4 8 2 = · [4 2 − 1] − 45 45 √ 8 2 4 = − 45 45 Z + z dzdxdy 1 Z x2 " SOLUTION OF QUIZ 6 3 Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT 06269-3009 E-mail address: mingfeng.zhao@uconn.edu