MATH 404 ANALYSIS II - SPRING 2009 SOLUTIONS to HOMEWORK 6

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1 MTH 404 NLYI II - PRING 2009 OLUTION to HOMEWORK 6 Problem Let f(x, y) (y+) 2 and let {(x, y) R 2 x > 0 and x < y < 2x}, B {(x, y) R 2 x > 0 and x 2 < y < 2x 2 } how: f does not exist and B f exists Find the value of B f olution: The set It suffices to find a sequence (T N ) of compact Jordan mea- B T P Figure surable subsets T N of such that N T N and T N f For example, let T N be a triangle bounded by the straight lines y x + /(2N), y 2x /(2N) and the vertical line y N Then T N has vertices at the points (/N, 3/(2N)), (N, N + /(2N)) and (N, 2N /(2n)) The triangle T T N is contained in the rectangle I J where I [/N, N and J 3/(2N), 2N /(2N) Using the Fubini s theorem, [ [ 2x /(2N) f f T f T (y + ) 2 dy dx T N /N [ I J I x+/(2n) x + + /(2N) 2x + /(2n) dx ln(x + + /(2N)) N /N 2 ln(2x + /(2n)) N /N from which it follows that f as N T The set B Take for example sets P N defined as follows let P N the region bounded by the parabolas y x 2 +/(2N), y 2x 2 /(2N) and the vertical line y N This line intersects parabolas at ( N, N + /(2N)) and N, 2N /(2N)) and the parabolas intersect at the point (/ N, 3/(2N)) The region P N is a subset of the rectangle I J where I [/ N, N and J 3/(2n), 2N /(2N) Clearly, P N is compact Jordan measurable and B N P N Write P P N Then by

2 2 the Fubini s theorem, [ [ 2x 2 /(2n) f f P f P P I J I x 2 +/(2N) y + ) 2 dy dx N [ / N x /(2N) 2x 2 + /(2N) ( ) x N arctan + /(2N) + /(2N) / N ( ) 2x N arctan 2 /(2N) /(2N) / N Evaluating the function at the end points and taking the limit as N we get that P f converges to ( 2)π/2 o f is integrable over B and B f ( 2)π/2 Problem 2 Let U {(x, y) R 2 x 2 + y 2 < } and let f(x, y) (x 2 +y 2 ) 3/2 for (x, y) (0, 0) Determine if the function f is integrable over U \ {(0, 0)} and over R 2 \ U olution: Let a,b {(x, y) R 2 a < (x, y) < b} For (a, b) (0, 2π), let g : B : g() be defined by g(r, ϕ) (r cosϕ, r sin ϕ) Then B a,b \ L where L {(x, y) y 0 and x 0} Moreover, g is one-to-one on and Dg(r, ϕ) r > 0 so that g is a diffeomorphism between and B By the change of variables theorem (using the fact that f is continuous for (x, y) 0, that the boundary of a,b and L have measure 0 in R 2 ) and the Fubini s theorem, f f f f (f g) det Dg a,b a,b B g() drdϕ 2π r2 b a dr 2π r2 ( a b The set U Let C N {(x, y) R 2 /N (x, y) /N}, then C N is compact, Jordan measurable, and U N C N Then by the above calculations ( ) ( f 2π C N /N 2π N N ) /N N showing that f is not integrable on U The set R 2 \ U Consider C N {(x, y) + /N (x, y) < N} Then C N is compact, Jordan measurable, and R 2 \ U N C N, and ( f 2π C N + /N ) ( N 2π N + N ) N so that f is integrable on R 2 \ U and R 2 \U f Problem 3 Let π k : R n R be the projection onto the kth factor,ie, π k (x) x k If is a bounded Jordan-measurabel subset of R n with nonzero volume, define the centroid c() of to be the point in R n whose kth coordinate is equal to c() k : π k )

3 3 (a) is said to be symmetric with respect to the subspace x k 0 if g() where g : R n R n is defined by g(x,,x n ) (x,,x k, x k, x k+,, x n ) how that if is symmetric with respect to the subspace x k 0, then c() k 0 (b) Let U {(x, y, z) R 3 z > 0 and x 2 + y 2 + z 2 < r 2 } Use the spherical coordinates and the change of variables theorem to compute c(u) olution: (a) This follows from the change of variables theorem First note that π k g π k and that Dg(x) is is diagonal matrix with all entries equal to except the kth diagonal entry is equal to o, Dg and c() k π k Hence c() k 0 as claimed g() π k (π k g) det Dg π k c() k (b) The set U is symmetric with respect to to subspaces x 0 and y 0 Hence c(u) x 0 and c(u) y 0 To calculate c(u) z we use the change of variables theorem and the spherical coordinates Let (0, r) (0, π/2) (0, 2π) and z ϕ r (x, y, z) y x ψ Figure 2 pherical coordinates g : R 3 be defined by g(r, ϕ, ψ) (ρ sin ϕcosψ, ρ sinϕcosψ, ρ cosϕ) Then g() B : {(x, y, z) x 2 + y 2 + z 2 < r 2 and z > 0} \ L, where L {(x, y, z) R 3 y 0, x 0}, is open, g : B is one-to-one and Dg(ρ, ϕ, ψ) ρ 2 sinϕ > 0 on o, g is a diffeomorphism Using v(u) v(b 3 (r))/2 2πr 3 /3 (see Problem 5), the change of variables theorem, we find that c(u) z v(u) U π z 3 2πr 3 3 2π r4 4 2π r 3 (π z g) det Dg 3 2πr 3 π/2 0 cosϕsin ϕdϕ 3r 8 ρ 3 cosϕsin ϕ

4 4 Problem 4 Let be an bounded open Jordan-measurable subset of R n and let p (p,, p n ) R n with p n > 0 Define the subset of R n by setting {x R n x ( t)a + tp where t (0, ) and a {0}} (a) Define a diffeomorphism g between and (0, ) (b) Calculate the volume in terms of the volume v() of (c) Express the centroid c() in terms of the centroid of c() and p and show that c() lies on the line segment connecting (c(), 0) and p olution: (a) We write a (b, 0 with b for a point a {0} Then define g : (0, ) by g(b, t) ( t)(b, 0) + tp ( t)a + tp Clearly, g is onto If g(b, t) g( b, s), then ( t)(b, 0) + tp ( s)( b, 0) + sp Then tp n sp n implying that t s and this in turn implies that b b o, g is one-to-one g is also of class C Moreover, t t t p a t t t p 2 a 2 Dg(b, t) t t t p n a n p n so that det Dg(b, t) ( t) n p n > 0 Consequently, the inverse g is also of class C umming up, g is a C -diffeomorphism (b) det Dg (0,) g( (0,)) (0,) ( t) n p n p n v() 0 ( t) n p n n v() (c) Using the change of the variables theorem and Fubini s theorem, we calculate for k n, c() k π k π k π k g det Dg g( (0,)) (0,) [( t)a k + tp k )( t) n p n and Hence n v() (0,) (0,) n n + c() k + n c() n n v() ( t) n a k + n v() ( n n + ) p k (0,) π n π n g( (0,)) (0,) c() t( t) n p n t( t) n p k n n + c() k + ( n n + (0,) ) p n n ( (c(), 0) + n ) p n + n + ( n ) p k n + π n g det Dg

5 5 Problem 5 Let B n (r) be a closed ball in R n of radius r and centered at 0 (a) how that v(b n (r)) α n r n where α n v(b n ()) (b) Calculate α and α 2 (c) Compute α n in terms of α n 2 (d) Find the formula for α n by considering two cases: n is odd and n is even olution: (a) Consider g : B n () R n defined by g(x) rx Then g(b ()) B n (r) so that g is a diffeomorphism between B n ( and B n (r) with Dg(x) r n By the change of variables theorem, v(b n (r)) det Dg r n r n B n () B n (r) g(b n ()) B n () B n () (b) ince B () (, ), v(b ()) 2 Let (0, ) (0, 2π) and g : R 2 be defined by g(r, ϕ) (r cosϕ, r sin ϕ) Then g is one-to-one, B : g() B 2 () \ I where I {(x, y) x 0 and y 0} o g() is open In addition, Dg(ϕ, r) r > 0 on o g is a diffeomorphim between and B and by the change of variables theorem and Fubini s theorem, [ det Dg r r dr dϕ π (0,) B (0,2π) (c) For z R n write z (x, y) where x R n 2 and y R 2 Now if z B n (), then z 2 x 2 + y 2 < Hence y < implying that y B 2 () and x 2 < y 2 which implies that x B n 2 ( y 2 ) Next note that B n () I where [, n 2 and I [, 2 For every y I 2, then χ B n ()(x, y) 0 if y B 2 () and if y B 2 (), then χ Bn ()(x, y) for x B n 2 ( y 2 ) and otherwise χ Bn ()(x, y) 0 o, χ B n ()(, y) is integrable over and { 0 y B 2 (), χ B n ()(x, y)dx v(b n 2 ( y 2 ) 2 )), y B 2 () ( y 2 ) n 2 2 αn 2 χ B2 () o by the Fubini s theorem [ χ B n () χ B n ()(x, y) dx dy I 2 α n 2 ( y 2 ) n 2 2 χb2 () dy α n 2 I 2 B 2 () ( y 2 ) n 2 The integral on the right-hand side can be calculated using the change of variables theorem using g : (0, ) (0, 2π) R 2, g(r, ϕ) (r cosϕ, r sin ϕ) We have ( y 2 ) n 2 2 ( r 2 ) n 2 drdϕ π ( r 2 ) n 2 2π 2 (2r)dr B 2 () 0 n o, α n 2π n α n 2 2

6 6 (d) From (c) and α 2 and α 2 π it follows that α 2k πk k! and α 2k+ 2 k+ 3 (2k + ) πk