Exercícios - Integrais Duplas e Triplas - 2024 1

Em coordenadas polares, x = r\cos\varphi \ e \ y = r\sin\varphi Interior: \ x^2 + y^2 = \sqrt{x^2 + y^2} \rightarrow r \rightarrow x^2 - r = r + r\sin\varphi → \ x^2 = r(1 + \sin\varphi) = r \rightarrow r = 1 + \sin\varphi Exterior: \ x^2 + y^2 \rightarrow J \rightarrow r = 1 então, 1 \le r \le 1 + \sin\varphi Intersecção, r = 1 = 1 + \sin\varphi \rightarrow \sin\varphi = 0 \rightarrow \varphi = 0 \ ou \varphi = \pi Portanto, \iint\limits_K \frac{d\h}{\sqrt{x^2 + y^2}} = \int_0^\pi \int_1^{1+\sin\varphi} \frac{r}{\sqrt{1-1^2}}drd\varphi = \int_0^\pi \int_1 r \ dr \ d\varphi = \int_0^\pi \int_1^{1+\sin\varphi} d\varphi = \int_0^{\sin\varphi} d\varphi = 2 \boxed{\iint\limits_K \frac{dA}{\sqrt{x^2+y^2}} = 2} \int_{-1}^1 \int_0^y \int_0^y (z+y+1) dx dy = - \int_{-1}^1 \int_0^y 2y x + x \Big|_0^y dy = \int_{-1}^1 \int_0^y (z+y^2+y) dz dy = \int_{-1}^{1+} 2z y^3+4y \Big|_0^ydy = \int_{-1}^1 (z y^3+y^2 y^3) dy = \frac{y^4}{2} + \frac{y^3}{3} \Big|_{-1}^1 = \frac{2}{3} \boxed{\int_{-1}^1 \int_0^y \int_0^y (z+y+1) dx dz dy = \frac{2}{3}} Em coordenadas cilíndricas, x = r\cos\varphi, \ y = r\sin\varphi \ \ e \ z = z x^2+y^2 -1 = 1 \rightarrow u = 1 \rightarrow u = 1, \int_{-1}\le u \le 1 x^2+y^2 -1 = 0 \rightarrow u^2 -1 \rightarrow u = 2 g = \sqrt{x^2+y^2} \rightarrow g = \sqrt{x^2} \rightarrow u \le g \le 6 - u^2 e também, 0 \le \varphi < 2\pi \int_0^{\sqrt{\pi}/6} \int_0^{\sqrt{\pi}/6} \cos\varphi dy dx = \frac{1}{4} Volume V_3 = \int_0^2 \int_0^2 \int_0^{l-x-y} dy dz dx = \int_0^2 \int_0^2 z \Big| \int_0^ldy dx V_s = \int_0^2 \int_0² \int_0^{l-x-y} dy dz dx = \int_0^2 2y - y^4 - \frac{y^2}{2} \Big|_0^{x-y} dx \boxed{V_s = \frac{-1}{3}} Calcule \int \int \int \frac{1}{x^2 + y^2 + z^2} dV onde E é a região no primeiro octante entre as esferas x^2 + y^2 + z^2 = 9 e x^2 + y^2 + z^2 = 1. Escolha uma opção: \circ \frac{2\pi}{3} \circ 2\pi \circ \frac{\pi}{3} \circ \pi \circ \frac{\pi}{2} Seja I = \int_0^{\pi} \int_0^x \int_0^{\frac{1}{\sqrt{x}}} \rho^3\sin\varphi d\rho d\varphi dx + \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \int_0^1 \rho^2 \sin\varphi d\varphi d\rho d\varphi. Calcule o valor de I. Sugestão: rescreva I em termo de uma única integral tripla em coordenadas cartesianas. Escolha uma opção: \circ I = \frac{1}{3\pi} \circ I = \frac{1}{2\pi} \circ I = \frac{\pi}{3} \circ I = 2\pi \circ I = \frac{1}{\pi} Calcule \int \int_R x^2y^2dA onde R é a região do primeiro quadrante delimitada pela curva y = \frac{1}{x} e pelas retas y = 1 e y = 2. Escolha uma opção: \circ 0 \circ 3 \circ 1 \circ 3 \circ 3 A integral iterada \int_0^{\sqrt[3]{\ln 3}} \int_{2x}^{\sqrt[3]{\ln 3}} e^{y^2} dydz vale Escolha uma opção: \circ 20 \circ 8 \circ 1 \circ \ln 3 \circ 81 Seja V_S o volume do sólido S delimitado pelo cilindro y = x^2, pelo cilindro x = y^2, pelo plano z = 0 e pelo paraboloide z = 5(x^2 + y^2). Então V_S vale Escolha uma opção: \circ \frac{6}{35} \circ \frac{1}{5} \circ 1 \circ \frac{1}{6} \circ \frac{9}{7} Calcule \int_{0}^{1} \int_{\sqrt{4-x^2}}^{\sqrt{4-x^2}} \frac{1}{\sqrt{x^2+y^2}} dydx + \int_{0}^{2} \int_{1}^{\sqrt{4-x^2}} \frac{1}{\sqrt{x^2+y^2}} dydx Sugestao: use coordenadas polares Escolha uma opcao: \( \frac{3}{2} \) ◯ 3 / 2 ◯ 2\pi ◯ \frac{\pi}{2} ◯ 3 ◯ \pi Calcule \iiint_E \frac{1}{\sqrt{x^2+y^2+z^2}} dV onde E e a regiao de \mathbb{R}^3 entre os dois cilindros \( x^2+y^2 = 1 \) e \( x^2+y^2 = 4 \) acima do semicone \( z = \sqrt{x^2+y^2} \) e abaixo do paraboloide \( z = 6-x^2-y^2 \). Escolha uma opcao: ◯ \frac{3\pi}{2}\ ◯ \frac{7\pi}{2}\ ◯ 4\pi ◯ \pi ◯ \frac{13\pi}{3}\ Calcule \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{0}^{\sqrt{2-x^2-y^2}} \frac{z}{\sqrt{x^2+y^2}} zdz dydx Escolha uma opcao: ◯ \frac{1}{2} ◯ \frac{\pi}{4} ◯ \frac{\pi}{3} ◯ \pi ◯ \frac{\pi}{2} A integral \int_{0}^{1} \int_{y}^{1} \frac{1}{1+x^2} dx dy vale Escolha uma opcao: ◯ \frac{\ln(2)}{2} ◯ \frac{\pi}{4} ◯ \frac{1}{2} ◯ \frac{\sqrt{2}}{2} ◯ \frac{2}{3} Calcule \iiint_E (\sqrt{x^2 + y^2 + \sqrt{z^2}}) dV onde E={(x,y,z)\epsilon R^3: Z > (\frac{\sqrt{x^2+y^2}}{3}-x^2+y^2+z^2<1)}. Escolha uma opcao: ◯ \frac{3\pi}{4} ◯ \frac{5\pi}{8} ◯ \frac{8\pi}{5} \pi \frac{2}{3} ◯ \frac{3\pi}{8} Calcule \int_{-1}^{1} \int_{0}^{y} \int_{\sqrt{2-y}}^{\sqrt{1-x-y}} (-xz) dz dx dy Escolha uma opcao: ◯ \frac{3}{8} \frac{5\pi}{8} ◯ \frac{8\pi}{5} ◯ \pi \frac{2}{3} ◯ \frac{3\pi}{8} Calculo 2 1) J = \iint (x+2y)dA \text{ com } y=1 \text{ e } y=x^2 \text{Primeiro quadrante} \int_{0}^{1} \int_{x^2}^{1}(x+2y)dydx=\int_{0}^{1}[xy+y^2]_x^2^1dx = \int_{0}^{1}(x - x^3/3 + 1 -x^2)dx = [x^2/2 - x^4/12 + x - x^3/3]_0^1 2) \int_0^{\sqrt{\pi/6}} \int_x^{\sqrt{\pi/6}} \cos(y^2) dy dx = \int_0^{\sqrt{\pi/6}} [\cos(y^2)]_x^{\sqrt{\pi/6}} dx Torna-se Torna-se = [\cos(y^2)]_0^{\sqrt{\pi/6}} ∬∬ dV / √(x²+y²) E = ∫₀²π ∫₀² ∫ᵤ(⁶⁻ᵤ²) ᵤ / √(ᵘ²) dz dr dφ = = ∫₀²π ∫₀ᵈ ∫ᵘ(⁶⁻ᵘ²) dz dr - 2π ∫₁² ∫₁(⁶⁻ᵘ²) dz dr = = 2π ∫₀² (⁶⁻ᵘ²) - ᵤ³/3 - ᵤ²/2 ∣¹ = 13π/3 → ∬∬ dV / √(x²+y²) = 13π/3 7) Em coordenadas esféricas, x = rsenφcosθ, y = rsenφsenθ e z = rcosθ Interseção, x⁶ + y⁶ + z² = 1 → z⁶ + z² = 4 → 4z² - 4 → = 3z⁶ → z⁶ = 1 → z = ±1 3z⁶ = x⁶ + y⁶ = 1 copy r²sen²φcos²θ + r²sen²φsen²θ - 3r²cos²φ =r²senφ(cos²φ + senθ(φ) - 3r²cos²φ senθφ = 3 → Tgφ = √3 → φ = π/3 ou sêg, φ¹ = π/2 - π/3 = π/6 Portanto, ∬∬∬ 3zdV = ∫₀²π ∫₀π/6 ∫₀² r²cosφ . r²senφ dr dθdφ = ∫₀²π ∫₀π/6 cosφsenφ ∫₀³ r⁴ dr = 8π/15 (8 - 3√3) ∬∬∬ 3zdV = 8π/15 (8 - 3√3) 8) Em coordenadas esférias, x = rsenφcosθ, y = rsenφsenθ e z = rcosφ ∬∬∫ 1/ √(1-x²) ∫ 1/ √(1-x²-y²) ∫ 1/ √(1-x²-y²) r³ dφ dydx = ∫₀π/2 ∫₀π/2 ∫₀³ 1/r³cos³φ . r²senφ dr dθdφ = ∫₀π/2 ∫₀π/2 ∫₀¹ cos³φsenφ dφ = π/2 (1/6) ∫₀¹ cos³φ senφ dφ u = cosφ → du = -senφ dφ = -π/12 ∫₀¹ u³ du = -π/12 [u⁴/4]₀¹ = π/48 - ∫₀¹ ∫₀√(1-x²) ∫₀√(1-x²-y²) 3zdzdydx = π/48 9) Interseção, 2x⁴ = √(1-x) → 4x⁴ = 1x → x³ = 1 J = ∫∫ x y dA = ∫₀¹ ∫₀√(1-x) x/2 √(1-y²) dydx = ∫₀¹ x/2 √(1-x²-y²) dx J = ∫₀¹ (2x² - x²) dx = 2/3 x³ - 2/5 x⁵ ∣₀¹ J = 1/3 1) Volume V_s = \int_{-1}^{1} \int_{\sqrt{1-x^2}}^{\sqrt{1-x^2}} dz dy dx = \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} dz dy dx V_s = \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} 2\sqrt{1-x^2} dy dx = \int_{-1}^{1} 2y \sqrt{1-x^2} \Bigg| _{-\sqrt{1-x^2}} ^{\sqrt{1-x^2}} dx V_s = 8 \int_{0}^{1} (1-x^2) dx = 8 \left( x - \frac{x^3}{3} \right) \Bigg|_{0}^{1} = \frac{16}{3} V_s = \frac{16}{3} 2) x = r \cos \phi \ e \ y = r \sin \phi (x^2 + y^2)^{3/2} = (u^2)^{3/2} = u^3 e Tambem x + y = 1 \longrightarrow r \cos \phi + r \sin \phi = 1 fazendo r^2 \cos^2 \phi + 2r \cos \phi \sin \phi + r^2 \sin^2 \phi = 1 r^2 (\cos^2 \phi + \sin^2 \phi) + 2r \cos \phi \sin \phi = 1 r^2 + 2r \cos \phi \sin \phi = 1 e Tambem x + y = 2 \longrightarrow r \cos \phi + r \sin \phi = 2 fazendo r^2 \cos^2 \phi + 2r \cos \phi \sin \phi + r^2 \sin^2 \phi = 4 r^2 + 2r \cos \phi \sin \phi = 1 fazendo r^2 + 2r \cos \phi \sin \phi - 1 = 0 r = \frac{-2 \cos \phi \sin \phi \pm \sqrt{4r^2 \cos^2 \phi \sin^2 \phi + 4r^2}}{2} r = \frac{-2 \cos \phi \sin \phi \pm 2 \sqrt{\cos \phi \sin \phi + 1}}{2} Não consegui encontrar os limites de integração 3) Em coordenadas cilíndricas x = r\cos \phi , \ y = r\sin \phi \ e \ z = z x^2 + y^2 = 1 \longrightarrow r^2 = 1 \longrightarrow r = 1 x^2 + y^2 = 1 \longrightarrow r^2 = 1 \longrightarrow r = 2 z = 3 \sqrt{x^2 + y^2} - 1 , z = 3r - 1 z = x^2 + y^2 - 1 , z = r^2 Então, \iiint_{E} \frac{-\scriptstyle1_{\cdot}}{x^2 + y^2} dV = \int_{0}^{2\pi} \int_{1}^{3r} \int_{r^2}^{3r-1} \frac{-\scriptstyle1_{r}}{x^2 + y^2} dz dr d\phi = \int_{0}^{2\pi} \int_{1}^{3r} \int_{\sqrt{r}}^{3r} \frac{-\scriptstyle1_{r}}{r} dz dr d\phi = 2\pi \int_{1}^{3r} \frac{1}{r} \frac{3r^2 - r^2}{3r} dr = \frac{39\pi}{4} \iiint_{E} \frac{-\scriptstyle1_{\cdot}}{x^2 + y^2} dV = \frac{39\pi}{4} 4) Em coordenadas esféricas x = r \sin \phi \cos \theta , \ y = r \sin \phi \sin \theta \ e \ z = r \cos \phi x^2 + y^2 + z^2 = 1 \longrightarrow r^2 = 1 \longrightarrow r = 1 x^2 + y^2 + z^2 = 4 \longrightarrow r^2 = 4 \longrightarrow r = 2 g > 0 \longrightarrow 0 < \phi < \frac{\pi}{2} S = \frac{1}{3}\sqrt{3x^2+y^2} = \frac{1}{3}\sqrt{3(x^2+y^2)} = \frac{1}{3}\sqrt{3u^2us\phi} S = \frac{\sqrt{3}}{3} \Rightarrow us\phi = \frac{3}{\sqrt{3}} \Rightarrow uc = \sqrt{3} \Rightarrow \phi = \frac{\pi}{3} \Rightarrow \phi' = \frac{\pi}{2} - \frac{\pi}{3} \int \int \int_E \frac{dV}{x^2+ y^2+3z^2} = \int_0^{2\pi} \int_0^{\frac{\pi}{6}} \int_{\frac{1}{2}}^{\frac{u^2}{3us dc} c d\phi d\theta du} \\ \boxed{\int \int \int_E \frac{dV}{x^2+y^2+3z^2} = \frac{\pi}{2}} J4 \, \underbrace{Intersection} \text{,} \quad y^2 = 1 - x^2 = 3 - x^2 = 1 \Rightarrow x = \pm 1 \boxed{\int x^2 dydx = \frac{4}{15}} \Rightarrow \boxed{J = \frac{4}{15}} J5 \, \text{Observe that} \int_0^1 \int_0^x (c^{\frac{x}{y^4}+1}) dyd\theta = \int_0^1 \int_0^y (c^{\frac{y}{4}+1}) dxd\theta = 0 \int \int_0^y \int_0^x 2(cx) dxdydx = c \quad J6 \, \text{Volume} V_s = \int_0^1 \int_0^x (x^2+y^4)dxdx = \frac{1}{3} \Rightarrow \boxed{\int_0^1 V_s \equiv \frac{1}{3}} \\ J7 \, \text{We know that,} \int_0^\infty \int_0^x \int_0^y (2(x-u)) dudydx = \frac{2}{3} J8 \text{Cylindrical coordinates} x=us\phi, \quad y=us uc \phi \quad \text{and} \quad z = g u^2 + u^2 + z = 5 \Rightarrow z = \sqrt{5 - u^2} J = 1+\sqrt{u^2+v^2} \Rightarrow z = 1+u \text{Intersection} 1+u = \sqrt{5-u^2} \Rightarrow (1+u)^2 = 5 - u^2 1+2u+u^2 = 5-u^2 \Rightarrow 2u^2+2u=4 = 0 \quad u = 1 \text{Hence,}\int \int \int_E 3zdv = \int_0^{2\pi} \int_0^{d} \int^\sqrt{5-u^2}_{1+u} 3z3dzd\phi u = \frac{5\pi}{2} \boxed{\int \int \int_E 3zdV = \frac{5\pi}{3}} J9 \text{In cylindrical coordinates,} x = u\cos\phi, y = u\sin\phi \quad \text{and} \quad g = g \quad \sqrt[3]{x}^0 \int_{-\frac{1}{3}}^0 \int_0^y \int_{-\sqrt{1-x^2-y^2}}^{\sqrt{\frac{x^2+y^2}{3}}} 3^3 dzdydx = \quad \frac{\pi}{\sqrt{3}} \int \int_0^\frac{\pi}{0} \int\sqrt{4-u^2}^3 z g dzd\phi = \frac{5\pi}{2} \begin{aligned} \int\sqrt{3}^0 \int_{-\sqrt{3-x^2-y^2}}^ \sqrt{\frac{x^2+y^2}{3}} 3^3 dz dy dx = \frac{5\pi}{3} \end{aligned} 27 Temos que, ∫_0^(π/2) ∫_00^(1/cosϕ) ∫_0^1 r^2 senϕ dr dϕ dθ = π/3 ∫_π^(2π) ∫_1^π/2 ∫_1^senϕ r^2 senϕ dr dϕ dθ = 2π/3 I = π/3 + 2π/3 = 3π/3 → I = π 28 Temos que ∬_R x^2 y^3 dA = ∫_(-1)^2 ∫_x^4 x^3 y^3 dx dy = 1/3 → ⌠⌡_R x^2 y^3 dA = 1/3 29 Temos que J = ∬_R y dA - ∫_0^(ln2) ∫_0^(e^x) y dy dx = ⌠⌡_R (3/4) 30 Temos que, ∫_0^(ln4) ∫_√x^2 (3/(1 + y^3) dx dy = ln 9 ∫_0^π/2 ∫_0^y^(2) 3/(1 + y^3) dx dy 22 Interseção 2 - x^2 = x^2 ⟶ 2 = 2x^2 ⟶ x^2 = 1 ⟶ x = ±1 ⦽ V_s = ∫_-1^1 ∫_x^2^(1-x) (1 - x - y) dy dx = 8 ⟶ V_s = 8 26 Em coordenadas esféricas, x = r senϕ cosθ, y = r senϕ senθ e z = r cosϕ ∭_E (x^2 + y^2 + z^2)^(3/2) dV = ∭_E r cosϕ r^2 senϕ dr dϕ dθ / (r^2)^(3/2) = ∫_0^(π/2) ∫_0^(π/2) ∫_0^1 r cosϕ senϕ dr dϕ dθ ∛_E δ / (x^2 + y^2 + z^2)^(3/2) dV = π/2 29) Temos que, ∫_0^√ln(3) ∫_x^√ln(3) e^x² dydx = ∫_0^√ln(3) ∫_0^√ln(3) e^x² dxdy ∫_0^√ln(3) ∫_x^√ln(3) e^x² dydx = 20 30) Interseções x² = √x → x⁴ = 1 → x = ±1 V_s = ∫_0^1 ∫_x^√x ∫_0^√5 (x²+y²) dzdydx = C/7 → V_s = C/7 31) Em coordenadas polares: x = rcosφ = 1 e x = rcosφ = 0 então, 1.cosφ = 1.cosψ = 1 {cosφ = 1 {cosψ = 1 → cosφ = 1/1 → φ = arccos(1/1) Não consegui achar os limites de integração. 32) Em coordenadas cilíndricas, ∫∫∫_E dv/√x²+y² = ∫_0^(2π) ∫_0^2 ∫_r^(6-r²) r z dzdrdφ = ∫_0^(2π) ∫_0^2 (∫_r^(6-r²) z dz) drdφ = 13π/3 ∫∫∫ dv/√x²+y² = 13π/3 33) Em coordenadas cilíndricas ∫_-1^1 ∫_√(1-x²)^(√(2-x²-y²) ∫_0^√(x²+y²) z dzdydx = ∫_0^(2π) ∫_0^1 ∫_r^√(2-r²) 3rdzdrdφ ∫_-1^1 ∫_√(1-x²)^(√(2-x²-y²) z dzdydx = I/2 34) Temos que, ∫_0^J ∫_J^1 dxdy - ∫_0^J ∫_0^x dy dx = π/8 35) Em coordenadas esféricas, θ = √(x²+y²/z²) = √(u²+v²/z²) = r sinφ = r cosθ/√3 → sinφ = 1/√3 → tanφ = √3 → φ = π/3 → φ' = π/2 - π/3 - π/c ∫∫∫_E ← Não consegui ver o denominador do integrando 36) Em coordenadas esféricas, ∫_-1^0 ∫_√(1-x²)^(√(y²) ∫_0^√(x²+y²) (-x-z) dzdxdz = ∫_0^(π/2) ∫_0^(π/2) ∫_0^1 (x sinφ sinθ cosθ/√xsinφ) dφdθdr = \int_0^{\pi} \int_0^{\pi/2} \int_0^1 (-u^7 \sin^6 \phi \cos^6 \phi \sin \theta ) du \,d\phi \,d\theta = -\frac{1}{15\sqrt{2}}