Atividade - Calculo 3

2) Considere a função f(x,y) = 4 - x - 2y. Determine o volume do sólido limitado pelo gráfico de f e pela região R:\na)\n -1 ≤ x ≤ 0\n R: 0 ≤ y ≤ -x² + 1\n0 = ∫⁰∫(-x² + 1) (4 - x - 2y)dy dx = ∫⁰⁻¹ (4y - xy - y²) |_{y=-x² + 1}^{y=0} dx = ∫⁰⁻¹ (4(-x² + 1) - x(-x² + 1) - 0) dx = ∫⁰⁻¹ (4 - 4x² - x + x³) dx = |^{-1}_{0} (-x⁴ + x³ - 2x² + 3x) | = 0 - = -1 + 1 + 2 - 1 - 3 = -143/60 = 143 mu.v. 1b)\n y = ax + b\n (1 = a(0) + b -> b = 1)\n (0 = a(1) + b)\n b_0 = a + 1\n a = -1\nR: 0 ≤ x ≤ 1\n 0 ≤ y ≤ -x + 1\n ∫ⁱ∫(0) (4 - x - 2y) dy dx = ∫ⁱ_0 (4y - xy - y²) |_{y=-x + 1}^{y=0} dx = ∫ⁱ_0 (4(-x + 1) - x(-x + 1) - 0) dx = ∫ⁱ_0 (4 - 4x - x + x²) dx = |^{1}_{0} (-x³ + 2x² - 3x) = 0 - = -3 mu.v. ∬ x dx dy, R é delimitada por x² + y² - 4x = 0\n { x² + y² = r² }\n x = r cos θ\n y = r sen θ\n 0 ≤ r ≤ 4 cos θ\n R: -π/2 ≤ θ ≤ π/2\n ∬ x dx dy = ∫^{π/2}_{-π/2} ∫^{4cosθ}_{0} r cos θ dr dθ\n = ∫^{π/2}_{-π/2} (64 cos θ)/(3) dθ\n = ∫^{π/2}_{-π/2} (cos θ) dθ\n = [1/3] = (64/3) ∫^{π/2}_{-π/2} cos θ dθ\n = cos θ |^{π/2}_{-π/2} = 64/3 = sen θ cos³ θ + 3 \n4 [ 1/2 ∫(dθ) + 1 ∫(cos² θ dθ) ]\n\n= sen θ cos³ θ + 3 \n4 [ 1/2 θ + 1/2 ∫(cos u 1 du) ]\n \n= sen θ cos³ θ + 3 θ sen 20\n4 8 16\n\n= 64/3 √(π/2) cos⁴ θ dθ\n\n= 64/3 [ sen(π/2) cos³(√2) + 3 sen(√2) + 3 sen(2√(π/2)) - sen(-π/2) cos³(√(π/2)) ] \n\n= 64 [ (3π/2 + 3π/8) ] = 64·6π = 8π ∫ ∫ ∫ 12xy³z² dv, R: -1 ≤ x ≤ 2\n0 ≤ y ≤ 3\n0 ≤ z ≤ 2\n\n= ∫ ∫ 12xy³z² dz dy dx = ∫ ∫ 12xy² d²x=∫0 z³ dz\n-1 0\n0 -1 (3/4) \n\n= ∫ -1 2 [ 12xy² (3/4) - ∫0(3/4) ] = - ∫ (12xy² * 2²/4) dy dx = \n\n= ∫ ∫ (48xy² dy dx) = ∫0 (48x)(3/3) dy dx = ∫(432x) dx = ∫(432) dx\n= 432 [x²/2]_{-1}^{2}\n=432 [4 - 1] 2\n= 432·3/2 = 648 ∫ ∫ ∫ z dz dx dy\n∫ ∫ z dx dz dy = ∫ ∫ z dy dx = ∫0 y² z dy\n0 -y²\n\n= ∫ ∫ y² z dy = ∫0 y dy dx = ∫0 dx [ (√y³)_0^{-y²}] = (1/3)[ (x³/3) ]_{0}^{1} = (1/3)[x/0] 1 = 1/3 ∭ x dv, R:{(x,y,z) ∈ ℝ³ | 10 ≤ x ≤ 3; 0 ≤ y ≤ x; x-y ≤ z ≤ x+y} ∫₀³ x³ dx = \\left[\\frac{x^4}{4}\\right]₀³ = \\frac{3^4 - 0^4}{4} = \\frac{81}{4} = 20.25