Questões - Integrais e Volumes de Sólidos 2022 1

1. Calcule as integrais duplas para as regiões indicadas. a) \int \int_D y \sin(xy) \, dx \, dy, \text{ onde } R = [1, 3] \times \left[0, \frac{\pi}{2}\right]. b) \int \int_D x \log(1 + x y) \, dx \, dy, \text{ onde } R = [0, 1] \times [0, 1]. c) \int \int_D (x + 2y) \, dx \, dy, \text{ onde } D \text{ é limitada por } y = x^{-2}, y = 1 \text{ e } y = 4. 2. Esboce a região de integração e mude a ordem de integração. a) \int_0^3 \int_{x^{-2}}^1 f(x, y) \, dy \, dx. b) \int_{-1}^1 \int_{y^{-1/2}}^{x^{-2}} f(x, y) \, dy \, dx. 3. Determine o volume do sólido limitado pelas superfícies z = 1 - y^2, x + z = 2 \text{ e } x = 2 \text{ para } z \geq 0. 4. Calcule o volume do sólido limitado pelos cilindros x^2 + y^2 = a^2 e x^2 + z^2 = a^2. 5. Calcule o volume do sólido acima do plano xy limitado pelo paraboloide z = x^2 + y^2 e pelo cilindro x^2 + y^2 = 2y. 1) a) \int_0^{\frac{\pi}{2}} \int_1^3 y \sin(xy) \, dx \, dy = I I = \int_0^{\frac{\pi}{2}} \int_y^{3y} \sin(u) \, \frac{du}{y} \, dy = \int_0^{\frac{\pi}{2}} [-\cos(u)]_y^{3y} \, dy = \int_0^{\frac{\pi}{2}} [\cos(y) - \cos(3y)] \, dy I = [\sin(y)]_0^{\frac{\pi}{2}} - \frac{1}{3} [\sin(3y)]_0^{\frac{\pi}{2}} I = \sin^2\left(\frac{\pi}{2}\right) - \sin^2(0) - \frac{1}{3} \sin\left(\frac{3\pi}{2}\right) + \frac{1}{3} \sin(0) I = \frac{1}{1} + \frac{1}{3} I = \frac{3 + 1}{3} , \; I = \frac{4}{3} b) \int_0^1 \int_0^1 x \ln(1 + xy)\, dx \, dy=I I=\int_0^1 \int_0^1 x \ln(1+xy) \, dy \, dx u = 1 + xy, :\, du = x \, dy Para y=0, u=1 y=1, u=1+x I = \int_0^1 \int_1^{1+x} \ln(u) \, du \, dx I = \int_0^1 \left[ u \ln(u) \right]_1^{1+x} - \int_1^{1+x} \frac{udu}{u}\right] \, dx I = \int_0^1 \left[ (1+x) \ln(1+x) - \ln(1) \right]_0^1 - x+x \right] \, dx I = \int_0^1 \ln(1+x) \, dx + \int_0^1 x \ln(1+x) \, dx - \int_0^1 x \, dx Resolvendo cada integral separadamente: \int_0^1 x \ln(1+x) \, dx = \frac{x^2}{2} \ln(1+x)\right|_0^1 - \int_0^1 \frac{x^2}{2(1+x)} \, dx Chamando \nu = 1+x\, :\, d\nu = dx x = 1-\nu \, Pars x=0, \nu=1 x=1, \nu=2 \int_0^1 x \ln(1+x) \, dx = \frac{1}{2} \ln(2) - \int_0^1 \frac{1-2\nu+\nu^2}{2\nu} \, d\nu \int_0^1 x \ln(1+x) \, dx = \frac{1}{2} \ln(2) - \frac{1}{2} \ln(2) + \frac{1}{2} \ln(1) + 1\right \frac{3}{4} \int_0^1 x \ln(1+x) \, dx = 1 - \frac{3}{4} = \frac{1}{4} \int_0^1 x \, dx = \frac{1}{2} \int_0^1 \ln(1+x) \, dx = x \ln(1+x)\right|_0^1 - \int_0^1 \frac{x \, dx}{1+x}\n= 1 + x\, :\, x = t-1 \right \nPars x=0, t=1\nx=1, t=2\n\int_0^1 \ln(1+x) \, dx = \ln(2) - \int_1^2 \frac{t-1}{t} \, dt\n\int_0^1 \ln(1+x) \, dx = \ln(2) + 1 + \ln(2) = 2 \ln(2) - 1 Portanto: I = \frac{1}{4} + 2 \ln(2) - 1 - \frac{1}{2} I = -\frac{5}{4} + 2 \ln(2) c) \iint_D (x + 2y) \, dx \, dy = I y = x^{-2} 4 y D: 0 < x \leq y^{-2} 1 \leq y \leq 4 I = \int_1^4 \int_0^{y^{-2}} (x + 2y) \, dx \, dy I = \int_1^4 \left[ \frac{x^2}{2} \bigg|_0^{y^{-2}} + 2y x \bigg|_0^{y^{-2}} \right] dy I = \int_1^4 \left[ \frac{y^{-4}}{2} + 2y y^{-2} \right] dy I = \int_1^4 \left( \frac{y^{-4}}{2} + 2y^{-1} \right) dy I = \left[ -\frac{y^{-3}}{6} + 2 \ln|y| \right]_1^4 I = \frac{1}{6} - \frac{1}{384} + 2 \ln(4) - 2 \ln(1) I = \frac{64 - 1}{384} + 2 \ln(4) I = \frac{63}{384} + 2 \ln(4) I = \frac{21}{128} + 2 \ln(4) 2) a) \int_0^3 \int_x^{4x-x^2} p(x,y) \, dy \, dx Calcular dos os pontos de interseção x = 4x - x^2 3x - x^2 = 0 x(3-x) = 0 x = 0 x = 3 Calculando a função inversa de y = 4x - x^2 y = -(x-2)^2 + 4 (x-2)^2 = 4 - y x = ±\sqrt{4-y} + 2 Portanto x = -\sqrt{4-y} + 2 x = \sqrt{4-y} + 2 \int_0^3 \int_x^{4x-x^2} p(x,y) \, dy \, dx = \int_0^3 \int_{-\sqrt{4-y}+2}^{\sqrt{4-y}+2} p(x,y) \, dx \, dy 2) b) \int_{-1}^1 \int_{x^2}^{\sqrt{2-x^2}} p(x,y) \, dy \, dx Ponto de interseção x^2 = \sqrt{2-x^2} x^4 + x^2 - 2 = 0 definindo \L = x^2 \L^2 + \L - 2 = 0 \L = \frac{-1 \pm \sqrt{1-4 \cdot 1 \cdot (-2)}}{2} = \frac{-1 \pm 3}{2} = \begin{\cases} 1 -2 \end{\cases} sendo \L = 1; x = \sqrt{1} = ±1 Calculando as funções inversas: y = x²; x = ±√y y = √2 - x² ∴ x = ±√2 - y² Portanto: ∫_{-1}^{x²} ∫_{x}^{√2 - x²} f(x, y) dy dx = ∫_{0}^{1} ∫_{-√y}^{√y} f(x, y) dx dy + ∫_{-1}^{2} ∫_{1 - √2 - y²}^{√2 - y²} f(x, y) dx dy 3) y = 1 - y² x + z = 2 z > 0 x = 2 sendo x + z = 2 ∴ z = 2 - x igualando com z = 1 - y², 2 - x = 1 - y² x = 2 - 1 + y² x = 1 + y² quando z = 0 = 1 - y² então y = ±1 Logo os limites de integração são: 2 - x ≤ z ≤ 1 - y² 1 + y² ≤ x ≤ 2 -1 ≤ y ≤ 1 Portanto o volume do sólido é: V = ∫_{-1}^{1} ∫_{1 + y²}^{2} ∫_{2 - x}^{1 - y²} dz dx dy V = ∫_{-1}^{1} ∫_{1 + y²}^{2} [1 - y²]_{2 - x}^{1 - y²} dx dy V = ∫_{-1}^{1} ∫_{1 + y²}^{2} (1 - y² - 2 + x) dx dy V = ∫_{-1}^{1} ∫_{1 + y²}^{2} (-1 - y² + x) dx dy V = ∫_{-1}^{1} [-x]_{1 + y²}^{2} - y² x/2 + x²/2/2]_{1 + y²}^{2} dy V = \int_{-1}^{1} \left[ -2+1+y^2-2y^2+(1+y^2)y^2+\frac{4}{2}-\frac{(1+y^2)^2}{2} \right] dy V = \int_{-1}^{1} \left[ -x+1+y^2-2y^2+y^2+y^4+\frac{2}{2}-y^2-\frac{y^4}{2} \right] dy V = \int_{-1}^{1} \left( \frac{1}{2} - y^2 + \frac{y^4}{2} \right) dy V = \frac{y}{2} \bigg|_{-1}^{1} - \frac{y^3}{3} \bigg|_{-1}^{1} + \frac{y^5}{10} \bigg|_{-1}^{1} V = \frac{2}{2} - \frac{2}{3} + \frac{2}{10} V = 1 - \frac{2}{3} + \frac{1}{5} = \frac{15-10+3}{15} V = \frac{8}{15} \ u.v. \quad u.v. \; unidade \; de \; volume 4) Por simetria o volume do sólido é 8 vezes o volume do sólido no primeiro octante (V_1), V_1 = \int_0^a \int_0^\sqrt{a^2-x^2} \sqrt{a^2-x^2} \ dy \ dx V_1 = \int_0^a \left[ \frac{y}{\sqrt{a^2-x^2}} \right]_0^{\sqrt{a^2-x^2}} \ dx V_1 = \int_0^a (a^2-x^2) \ dx V_1 = a^2x \bigg|_0^a - \frac{x^3}{3} \bigg|_0^a V_1 = a^3 - \frac{a^3}{3} V_1 = \frac{2a^3}{3} como \ V = 8V_1, \ entao V = \frac{16a^3 u.v.}{3} 5) z = x^2 + y^2 x^2 + y^2 = 2y x^2 + y^2 - 2y = 0 x^2 + (y-1)^2 = 1 Portanto um cilindro centrado em (x,y) = (0,1) e com raio igual a 1. Em coordenadas polares, temos: z = r^2 \cos^2\theta + r^2 \sin^2\theta z = r^2 (\cos^2\theta + \sin^2\theta) z = r^2 Os limites de integração são: \{0 < r < \sen\theta 0 \le \theta < \pi V = \int_0^\pi \int_0^\theta \sen\theta r^2 \, dr \, d\theta V = \int_0^\pi \int_0^\theta \sen\theta r^3 \, dr \, d\theta = \int_0^\pi \frac{\theta^4 \sen\theta}{4} \, d\theta V = \int_0^\pi \frac{\sen^4\theta}{4} \, d\theta \text{ como } \sen^2\theta = \frac{1-\cos(2\theta)}{2} V = \frac{1}{4} \int_0^\pi \left( \frac{1-\cos(2\theta)}{2} \right)^2 \, d\theta V = \frac{1}{4} \int_0^\pi \left( \frac{1 - 2\cos(2\theta) + \cos^2(2\theta)}{4} \right) \, d\theta V = \frac{1}{16} \left[ \theta \right]_0^\pi - \frac{2\sen(2\theta)}{2} \Big|_0^\pi + \int_0^\pi \cos^2(2\theta) \, d\theta \Big]\right] V = \frac{\pi}{16} + \frac{1}{16} \int_0^\pi \cos^2(2\theta) \, d\theta sendo \cos^2(2\theta) = \frac{1+\cos(4\theta)}{2} V = \frac{\pi}{16} + \frac{1}{16} \int_0^\pi \left[ \frac{1 + \cos(4\theta)}{2} \right] \, d\theta V = \frac{\pi}{16} + \theta \Big|_0^\pi + \frac{\sen(4\theta)}{8} \Big|_0^\pi V = \frac{\pi}{16} + \frac{\pi}{32} \Rightarrow V = \frac{2\pi + \pi}{32} \Rightarrow V = \frac{3\pi}{32}