MATH503 - HOMEWORK #2. v k. + v i. (fv i ) = f v i. + v k. ) + u j (ɛ ijk. ) u j (ɛ jik

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1 1 KO UNIVERITY Mon April 18, ollege of Arts and ciences Handout # Department of Physics Instructor: Alkan Kabakço lu MATH - HOMEWORK # 1 ( f) = Using Einstein notation and expressing x, y, x as e 1, e, e, we can write the above expression as ( v) = [ ( f)] i = ɛ ijk ( f) e k j f f f = ɛ ijk = ɛ ijk = ɛ ikj = [ e j e k e k e j e k e ( f)] i = j ( v) = v k v k ɛ ijk v k = ɛ ijk = ɛ jik = e j e j e j e ( v) = i (f v) = f ( v) + v ( f) (f v) = (fv i ) = f v i + v i f = f ( v) + v ( f) ( u v) = v ( u) u ( v) ( u v) = ɛ ijk u j v k = ɛ ijk (u j v k + v k u j ) = v k (ɛ ijk u j ) + u j (ɛ ijk v k ) = v k (ɛ kij u j ) u j (ɛ jik v k ) = v ( u) u ( v) 911 alculate c F ( r) dr for F = [cosh x, sinh y, e z ], : r = [t, t, t ] from [,, ] to [, 4, 8] Let us change the integration variable to t The limits are then from t = to t = Note that dr = dr dt dt = [1, t, t ]dt Then, F dr = F dr dt dt = 911 Evaluate the line integral (cosh t + t sinh t + t e t ) dt = sinh + (cosh 4 1) + (e 8 1) 99 f( r)ds with the arc length as parameter, where f( r) = x + y + z First make a change of parameters to the arc length s : s(t) = : r(t) = cos(t)ˆx + sin(t)ŷ + tẑ, t 4π t r (t) dt = t sin (t) + cos (t) + 4dt = t

2 Then x = cos(t) = cos( s ), y = sin(t) = sin( s ), z = s The line integral is f( r)ds = 4π [cos ( s ) + sin ( s ) + 4s ]ds = 4π (1 + 64π ) heck for path independence in the following dierential forms and integrate from (,, ) to (a, b, c) if path independent 91 sinh(xz)(zdx xdz) is path independent if it is exact, ie, if there exists a F (x, z) such that df = F F x dx + z dz = sinh(xz)(zdx xdz) Then we must have which does not hold, since F xz = [z sinh(xz)] = [ x sinh(xz)], z x [z sinh(xz)] = sinh(xz) + xz cosh(xz), z x [ x sinh(xz)] = sinh(xz) xz cosh(xz) [z sinh(xz)] z 914 (yzdx + xzdy + xydz) is clearly exact since it is the dierential of F (x, y, z) = xyz + const Then it can be integrated easily: (a,b,c) (,,) (yzdx + xzdy + xydz) = (a,b,c) (,,) df = F (a, b, c) F (,, ) = abc F ( r) d r for F = [sin(y), cos(x)], the boundary of 946 Evaluate using the Green's theorem the line integral c the region R : the triangle with vertices A : (, ), B : (π, ), : (π, 1) Let us rst calculate the normal vector: r B r A = πˆx, r r A = πˆx + ŷ N = ( r B r A ) ( r r A ) = πẑ The unit normal vector is ẑ, and the area of the triangle is A = 1 N = π In order to use the Green's theorem we rst calculate F : F = F y x F x = [sin(x) + cos(y)] ẑ y The line integral is c F ( r) d r = = π π dx x/π dy [sin(x) + cos(y)] [ x π sin(x) + sin( x π )]dx = πcos( 1 π ) π 1

3 941 Find the area of the region under one arch of the cycloid r = a(t sin(t))ˆx + a(1 cos(t))ŷ, t π Using the formula derived from the Green's theorem, A = 1 (xdy ydx) = 1 (x(t) dy(t) y(t) dx(t) )dt dt dt = a = πa π [(t sin(t))sin(t) (1 cos(t))(1 cos(t))]dt The minus sign on the third line is due to the counter-clockwise direction of the contour integral 94 Find the parameter curves and a normal vector of the surface r(u, v) = [u cos v, u sin v, u ] onstant v curves are simply rotations of the parabola (z = x, y = ) around the z-axis onstant u curves are circles given by x + y = z = const A normal vector of the surface at r(u, v) is N(u, v) = r u r v = [cos v, sin v, u] [ u sin v, u cos v, ] [cos v, sin v, 1 u ] 98 Find the unit normal vector of the surface g(x, y, z) = x y + z 1 = g = xˆx yŷ + zẑ g = x + y + z = r g g = x r ˆx y r ŷ + z r ẑ 96 Evaluate the ux integral F nda for F = [e y, e z, e x ], : r = [ cos u, sin u, v], u [, π ], v [, ] The surface of integration is the rst quadrant of the lateral surface of the cylinder with a radius r = around the z-axis, extending from z = to z = The normal vector of the surface facing 'outward' is N = r u r v = [ sin u, cos v, ] [,, 1] = [ cos u, sin u, ] N N = [cos u, sin u, ], da = N dudv = dudv Then, the ux integral is F nda = = dv π/ dv [ e6 1 6 du [e 6 sin u cos u + e v sin u] + e v ] = (e 6 1) + (1 e 4 ) c Given the surface : r(u, v) with the corresponding quadratic dierential form ds = Edu + F dudv + Gdv with coecients E = r u r u, F = r u r v, G = r v r v,

4 4 the length of the normal vector N can be written as N = r u r v = ( r u r v ) ( r u r v ) Using the identity a ( b c) = b ( c a), with a = ( r u r v ), b = r u, c = r v, we can rewrite above as But a ( b c) = b( a c) c( a b) : N = r u [ r v ( r u r v )] N = r u r u r v r v ( r v ru ) = r u r v r u r v = EG F The area of the corresponding surface is then given by A() = da = N dudv = where R indicates the region of integration for the variables u, v R EG F dudv, 989 how that a region T with boundary surface has the volume V = 1 r cos φda where r is the distance of a variable point P : (x, y, z) on from the origin O and φ is the angle between the directed line OP and the outer normal of at P V olume = dv = V V r dv = 1 r nda = 1 r cos φda 981 Volume of a ball or radius a: Pick the center of the ball to be the origin Then the surface is parametrized as : r = [a sin θ cos φ, a sin θ sin φ, a cos θ] Accordingly, The volume integral is V olume = 1 N = r θ r φ = [a cos θ cos φ, a cos θ sin φ, a sin θ] [ a sin θ sin φ, a sin θ cos φ, ] = a [sin θ cos φ, sin θ sin φ, sin θ cos θ] N cos θ N = [cos φ, sin φ, sin θ ], N = a sin θ = 4πa r nda = 1 r n( N dθdφ) = 1 π dφ π dθ a(sin θ + cos θ sin θ )a sin θ The same result can be obtained simply by using the result of the previous problem after noting that cos φ = 1 for all the points on the surface: V olume = 1 ada = a (4πa ) = 4πa 994 Evaluate the surface integral and the corresponding line integral using the tokes' Thm for the eld F = [,, x cos z] and the surface : x + y = 1, z [, π/4] and y

5 Direct evaluation: F = [, cos z, ] r = [x, 1 x, z] x N = r z r x = [,, 1] [1,, ] = [ x, 1, ] 1 x 1 x I = ( F ) nda = ( F ) Ndxdz = 1 1 dx dz( cos z) = 1 tokes' Thm can be used to obtain the same result In order to use it eectively, note rst that the eld F is along the z direction, hence perpendicular to the semicircular portions of the boundary curve Therefore, only the two straight (along the z-axis) portions of contribute to the integral: I = F dr = F (x = 1)dz F (x = 1)dz = ( cos z)dz cos zdz = 1 where the overall sign of the integral is determined by the orientation of the contour determined by the right-hand rule consistent with the choice of the normal direction (order of r z and r x in the cross product) in the previous direct evaluation