Gabarito Prova2 2013-1

Cálculo III\n29 Prova\n02/09/2013\n\n1) Calcular \\[ \\iint \\text{not} (\\mathbf{F}) \\cdot \\mathbf{n} dS \\], onde\n\\[ \\mathbf{F}(x,y,z) = (e^{-z}) \\mathbf{i} + (x^2+y^2) \\mathbf{j} + 2yz \\mathbf{k} \\]\n\\[ S: \\{ x^2+y^2 \\leq 1, \\; z=0 \\} \\]\n\nSolução\n\nConsidera \\[ \\mathcal{S} = S \\cup S_1, \\text{ onde } \\]\n\\[ S_1 \\text{ é dada por: } \\]\n\\[ S_1: \\{ z=1 \\} \\]\n\\[ (x,y) \\in D: x^2+y^2 \\leq 1 \\]\n\n\\[ S_1 \\text{ orientada por } \\mathbf{n}_1 = (0,0,1) \\mathbf{e} \\]\n\n\\[ dS_1 = \\sqrt{1+(x_1^2)+(z_1^2)} dx_1 dy_1 = \\sqrt{1+0+0} dx_1 dy_1 = dx_1 dy_1 \\]\n\\[ (dS_1 = dx_1 dy_1) \\]\n\nSeja \\[ W \\text{ o sólido limitado por } \\mathcal{S} \\; (\\partial W = \\mathcal{S}) \\]\nPelo Teorema de Gauss, temos:\n\n\\[ \\iint (\\mathbf{F} \\cdot \\mathbf{n}) dS = \\iiint \\text{div} (\\text{not} \\mathbf{F}) dV \\]\n\\[ \\mathcal{S} = \\partial W \\]\n\\[ \\text{div} (\\text{not} \\mathbf{F}) = 0 \\]\n\n\\[ \\therefore \\iint \\text{not}(\\mathbf{F}) \\cdot \\mathbf{n} dS = 0 \\]\n\\[ \\mathcal{S} \\]\n\n-1 - \\[ \\iint \\text{not} (\\mathbf{F}) \\cdot \\mathbf{n} dS = - \\iint \\text{not} (\\mathbf{F}) \\cdot \\mathbf{n}_1 dS_1 \\]\n\n\\[ \\text{not} (\\mathbf{F}) = \\left| \\begin{array}{ccc}\n\\frac{\\partial z}{\\partial x} & \\frac{\\partial z}{\\partial y} & 1 \\\\\n e^{-z} & \\frac{\\partial}{\\partial y}(x^2+y^2) & 2y \\\\\n\\frac{\\partial}{\\partial z}(x^2+y^2) &\\frac{\\partial}{\\partial y}(z) & z^2 \\\\\n\\end{array} \\right| = (2z-x, 0, 2z+1) \\]\n\n(Em \\[ S, \\text{ temos } z=1 \\]) logo:\n\n\\[ \\iint \\text{not} (\\mathbf{F}) \\cdot \\mathbf{n} dS_1 = \\iint (2-z,x,0,2)(0,0,1) dx dy \\]\n\n\\[ = \\iint_{D} 2 dx dy = 2 \\text{área}(D) \\]\n\n\\[ = 2\\pi (\\sqrt{2})^2 = \\frac{6}{\\pi} \\]\n\n\\[ \\therefore \\iint \\text{not} (\\mathbf{F}) \\cdot \\mathbf{n} dS = \\frac{-4}{6\\pi} \\]\n\n-2 - \\[ \\mathbf{F}(x,y) = \\frac{-y}{x^2+y^2} \\mathbf{i} + \\frac{x}{x^2+y^2} \\mathbf{j} \\text{ definido em } D=\\mathbb{R}^2-\\{(0,0)\\} \\]\n\nCalcule:\n\na) \\[ \\int_{C_1} \\mathbf{F} \\cdot d\\mathbf{n} \\text{ onde } C_1: x^2+y^2=a^2, \\text{ orientada positivamente } \\]\n\nSolução\n\nNote que a região limitada por \\[ C_1 \\text{ não está contida em } D \\{(0,0) \\not\\in D\\} \\ldots \\]\n\nNão podemos aplicar o Teor. de Green.\n\nParametrizamos \\[ C_1, \\; x=a \\cos t, \\; y=a \\sin t, \\; 0 \\leq t \\leq 2\\pi \\]\n\nEntão \\[ dx = -a \\sin t dt, \\; dy = a \\cos t dt \\]\n\n\\[ \\int_{C_1} \\mathbf{F} \\cdot d\\mathbf{n} = \\int \\frac{-a\\sin t}{a^2} dx + \\frac{a\\cos t}{a^2} dy \\]\n\n\\[ = \\int_{0}^{2\\pi} [(\\frac{-a\\sin}{a^2}) (-a \\sin t) + \\frac{a\\cos t}{a^2}(a\\cos t)] dt \\]\n\n\\[ = \\int_{0}^{2\\pi} dt = 2\\pi \\]\n\n(b) \\[ \\int_{C_2} \\mathbf{F} \\cdot d\\mathbf{n} \\text{ onde } C_2 \\text{ é uma curva fechada, } C_{1} \\text{ para parte, que envolve a origem e está orientada positivamente.} \\]\n\n-3 - 4\nS: \\( \\Psi(u,v) = (u,v,2u+v) \\), \\( (u,v) \\in D: \\{ 0 \\leq u \\leq 1 \\; 0 \\leq v \\leq 4 \\} \\)\\n\\( \\delta(x,y,z) = x+y+z \\) (densidade)\\nDetermina a massa de S.\\n\nSolução:\\nM = \\( \\iiint_{S} \\delta(x_i,y_i,z) ds = \\iiint_{S} (x+y+z) ds \\)\\n\n= \\( \\iiint_{D} (u+v+(2u+v)) \\frac{\\partial \\Psi}{\\partial u} \\frac{\\partial \\Psi}{\\partial v} dudv \\)\\n\n= \\( \\iiint_{D} (3u+2v) \\| \\frac{\\partial \\Psi}{\\partial u} \\| \\frac{\\partial \\Psi}{\\partial v} dudv \\)\\n\n\\( \\frac{\\partial \\Psi}{\\partial u} = (1,0,2) , \\frac{\\partial \\Psi}{\\partial v} = (0,1,1) \\)\\n\\( \\frac{\\partial \\Psi}{\\partial u} \\times \\frac{\\partial \\Psi}{\\partial v} = (-2,-1,1) , \\| \\frac{\\partial \\Psi}{\\partial u} \\times \\frac{\\partial \\Psi}{\\partial v} \\| = \\sqrt{4+1+1} = \\sqrt{6} \\)\\n\nLogo M = \\( \\sqrt{6} \\iiint_{D} (3u+2v) dudv = \\sqrt{6} \\int_{0}^{1} \\int_{0}^{4} (3u+2v) dvdu \\)\\n\n= \\( \\sqrt{6} \\int_{0}^{1} (3u v + uv^{2})|_{v=4}^{v=0} du = \\sqrt{6} \\int_{0}^{1} (3u(4) + u(0)) du = \\sqrt{6} \\int_{0}^{1} (12u) du = \\sqrt{6} \\left(12\\frac{u^{2}}{2}\\right)|_{0}^{1} = \\sqrt{6} |12\\frac{1}{2}| = 6\\sqrt{6} \\)\\n= \\( \\frac{4\\sqrt{6}}{3}u^{3}|_{0}^{1} = \\frac{4\\sqrt{6}}{3} \\)