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16.8E: Ejercicios para la Sección 16.8

  • Page ID
    116677
    • Edwin “Jed” Herman & Gilbert Strang
    • OpenStax
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    Para los ejercicios 1 - 9, use un sistema algebraico por computadora (CAS) y el teorema de divergencia para evaluar la integral de superficie\(\displaystyle \int_S \vecs F \cdot \vecs n \, ds\) para la elección dada de\(\vecs F\) y la superficie límite\(S.\) Para cada superficie cerrada, asumir\(\vecs N\) es el vector normal de unidad exterior.

    1. [T]\(\vecs F(x,y,z) = x\,\mathbf{\hat i} + y\,\mathbf{\hat j} + z\,\mathbf{\hat k}\);\(S\) es la superficie del cubo\(0 \leq x \leq 1, \, 0 \leq y \leq 1, \, 0 < z \leq 1\).

    2. [T]\(\vecs F(x,y,z) = (\cos yz) \,\mathbf{\hat i} + e^{xz}\,\mathbf{\hat j} + 3z^2 \,\mathbf{\hat k}\);\(S\) es la superficie del hemisferio\(z = \sqrt{4 - x^2 - y^2}\) junto con el disco\(x^2 + y^2 \leq 4\) en el\(xy\) plano.

    Contestar
    \(\displaystyle \int_S \vecs F \cdot \vecs n \, ds = 75.3982\)

    3. [T]\(\vecs F(x,y,z) = (x^2 + y^2 - x^2)\,\mathbf{\hat i} + x^2 y\,\mathbf{\hat j} + 3z\,\mathbf{\hat k}; \)\(S\) es la superficie de las cinco caras del cubo unitario\(0 \leq x \leq 1, \, 0 \leq y \leq 1, \, 0 < z \leq 1.\)

    4. [T]\(\vecs F(x,y,z) = x\,\mathbf{\hat i} + y\,\mathbf{\hat j} + z\,\mathbf{\hat k}; \)\(S\) es la superficie de paraboloide\(z = x^2 + y^2\) para\(0 \leq z \leq 9\).

    Contestar
    \(\displaystyle \int_S \vecs F \cdot \vecs n \, ds = 127.2345\)

    5. [T]\(\vecs F(x,y,z) = x^2\,\mathbf{\hat i} + y^2 \,\mathbf{\hat j} + z^2 \,\mathbf{\hat k}\);\(S\) es la superficie de la esfera\(x^2 + y^2 + z^2 = 4\).

    6. [T]\(\vecs F(x,y,z) = x\,\mathbf{\hat i} + y\,\mathbf{\hat j} + (z^2 - 1)\,\mathbf{\hat k}\);\(S\) es la superficie del sólido delimitada por cilindro\( x^2 + y^2 = 4\) y planos\(z = 0\) y\(z = 1\).

    Contestar
    \(\displaystyle \int_S \vecs F \cdot \vecs n \, ds = 37.699\)

    7. [T]\(\vecs F(x,y,z) = x^2\,\mathbf{\hat i} + y^2 \,\mathbf{\hat j} + z^2 \,\mathbf{\hat k}\);\(S\) es la superficie delimitada arriba por esfera\(\rho = 2\) y abajo por cono\(\varphi = \dfrac{\pi}{4}\) en coordenadas esféricas. (Piense en\(S\) como la superficie de un “cono de helado”.)

    8. [T]\(\vecs F(x,y,z) = x^3\,\mathbf{\hat i} + y^3 \,\mathbf{\hat j} + 3a^2z \,\mathbf{\hat k} \, (constant \, a > 0)\);\(S\) es la superficie delimitada por cilindro\(x^2 + y^2 = a^2\) y planos\(z = 0\) y\(z = 1\).

    Contestar
    \(\displaystyle \int_S \vecs F \cdot \vecs n \, ds = \dfrac{9\pi a^4}{2}\)

    9. [T] Integral de superficie\(\displaystyle \iint_S \vecs F \cdot dS\), donde\(S\) está el sólido limitado por paraboloide\(z = x^2 + y^2\) y plano\(z = 4\), y\(\vecs F(x,y,z) = (x + y^2z^2)\,\mathbf{\hat i} + (y + z^2x^2)\,\mathbf{\hat j} + (z + x^2y^2) \,\mathbf{\hat k}\)

    10. Utilice el teorema de divergencia para calcular la integral de superficie\(\displaystyle \iint_S \vecs F \cdot dS\), donde\(\vecs F(x,y,z) = (e^{y^2} \,\mathbf{\hat i} + (y + \sin (z^2))\,\mathbf{\hat j} + (z - 1)\,\mathbf{\hat k}\) y\(S\) es el hemisferio superior\(x^2 + y^2 + z^2 = 1, \, z \geq 0\), orientada hacia arriba.

    Contestar
    \(\displaystyle \iint_S \vecs F \cdot dS = \dfrac{\pi}{3}\)

    11. Utilice el teorema de divergencia para calcular la integral de superficie\(\displaystyle \iint_S \vecs F \cdot dS\), donde\(\vecs F(x,y,z) = x^4\,\mathbf{\hat i} - x^3z^2\,\mathbf{\hat j} + 4xy^2z\,\mathbf{\hat k}\) y\(S\) es la superficie delimitada por cilindro\(x^2 + y^2 = 1\) y planos\(z = x + 2\) y\(z = 0\).

    12. Utilice el teorema de divergencia para calcular la integral de superficie\(\displaystyle \iint_S \vecs F \cdot dS\), cuando\(\vecs F(x,y,z) = x^2z^3 \,\mathbf{\hat i} + 2xyz^3\,\mathbf{\hat j} + xz^4 \,\mathbf{\hat k}\) y\(S\) es la superficie de la caja con vértices\((\pm 1, \, \pm 2, \, \pm 3)\).

    Contestar
    \(\displaystyle \iint_S \vecs F \cdot dS = 0\)

    13. Utilizar el teorema de divergencia para calcular la integral de superficie\(\displaystyle \iint_S \vecs F \cdot dS\), cuando\(\vecs F(x,y,z) = z \, \tan^{-1} (y^2)\,\mathbf{\hat i} + z^3 \ln(x^2 + 1) \,\mathbf{\hat j} + z\,\mathbf{\hat k}\) y\(S\) es una parte del paraboloide\(x^2 + y^2 + z = 2\) que se encuentra por encima del plano\(z = 1\) y se orienta hacia arriba.

    14. [T] Usar un CAS y el teorema de divergencia para calcular el flujo\(\displaystyle \iint_S \vecs F \cdot dS\), donde\(\vecs F(x,y,z) = (x^3 + y^3)\,\mathbf{\hat i} + (y^3 + z^3)\,\mathbf{\hat j} + (z^3 + x^3)\,\mathbf{\hat k} \) y\(S\) es una esfera con centro\((0, 0)\) y radio\(2.\)

    Contestar
    \(\displaystyle \iint_S \vecs F \cdot dS = 241.2743\)

    15. Utilice el teorema de divergencia para calcular el valor de la integral de flujo\(\displaystyle \iint_S \vecs F \cdot dS\), donde\(\vecs F(x,y,z) = (y^3 + 3x)\,\mathbf{\hat i} + (xz + y)\,\mathbf{\hat j} + \left(z + x^4 \cos (x^2y)\right)\,\mathbf{\hat k}\) y\(S\) es el área de la región delimitada por\(x^2 + y^2 = 1, \, x \geq 0, \, y \geq 0\), y\(0 \leq z \leq 1\).

    0, y>0, and z>0. A quarter of a cylinder is drawn with center on the z axis. The arrows have positive x, y, and z components; they point away from the origin." data-type="media">Un campo vectorial en tres dimensiones, con enfoque en el área con x 0, y>0 y z>0. Un cuarto de cilindro se dibuja con el centro en el eje z. Las flechas tienen componentes x, y y z positivos; apuntan lejos del origen." src="https://math.libretexts.org/@api/dek...16_08_202.jpeg">

    16. Usar el teorema de divergencia para calcular la integral de flujo\(\displaystyle \iint_S \vecs F \cdot dS\), where \(\vecs F(x,y,z) = y\,\mathbf{\hat j} - z\,\mathbf{\hat k}\) and \(S\) consists of the union of paraboloid \(y = x^2 + z^2, \, 0 \leq y \leq 1\), and disk \(x^2 + z^2 \leq 1, \, y = 1\), oriented outward. What is the flux through just the paraboloid?

    Answer
    \(\displaystyle \iint_S \vecs F \cdot dS = -\pi\)

    17. Use the divergence theorem to compute flux integral \(\displaystyle \iint_S \vecs F \cdot dS\), where \(\vecs F(x,y,z) = x\,\mathbf{\hat i} + y\,\mathbf{\hat j} + z^4 \,\mathbf{\hat k}\) and \(S\) is a part of cone \(z = \sqrt{x^2 + y^2}\) beneath top plane \(z = 1\) oriented downward.

    18. Use the divergence theorem to calculate surface integral \(\displaystyle \iint_S \vecs F \cdot dS\) for \(\vecs F(x,y,z) = x^4\,\mathbf{\hat i} - x^3z^2\,\mathbf{\hat j} + 4xy^2 z\,\mathbf{\hat k}\), where \(S\) is the surface bounded by cylinder \(x^2 + y^2 = 1\) and planes \(z = x + 2\) and \(z = 0\).

    Answer
    \(\displaystyle \iint_S \vecs F \cdot dS = \dfrac{2\pi}{3}\)

    19. Consider \(\vecs F(x,y,z) = x^2\,\mathbf{\hat i} + xy\,\mathbf{\hat j} + (z + 1)\,\mathbf{\hat k}\). Let \(E\) be the solid enclosed by paraboloid \(z = 4 - x^2 - y^2\) and plane \(z = 0\) with normal vectors pointing outside \(E.\) Compute flux \(\vecs F\) across the boundary of \(E\) using the divergence theorem.

    In exercises 20 - 23, use a CAS along with the divergence theorem to compute the net outward flux for the fields across the given surfaces \(S.\)

    20. [T] \(\vecs F = \langle x,\, -2y, \, 3z \rangle; \) \(S\) is sphere \(\{(x,y,z) : x^2 + y^2 + z^2 = 6 \}\).

    Answer
    \(15\sqrt{6}\pi\)

    21. [T] \(\vecs F = \langle x, \, 2y, \, z \rangle\); \(S\) is the boundary of the tetrahedron in the first octant formed by plane \(x + y + z = 1\).

    22. [T] \(\vecs F = \langle y - 2x, \, x^3 - y, \, y^2 - z \rangle\); \(S\) is sphere \(\{(x,y,z) \,:\, x^2 + y^2 + z^2 = 4\}.\)

    Answer
    \(-\dfrac{128}{3} \pi\)

    23. [T] \(\vecs F = \langle x,y,z \rangle\); \(S\) is the surface of paraboloid \(z = 4 - x^2 - y^2\), for \(z \geq 0\), plus its base in the \(xy\)-plane.

    For exercises 24 - 26, use a CAS and the divergence theorem to compute the net outward flux for the vector fields across the boundary of the given regions \(D.\)

    24. [T] \(\vecs F = \langle z - x, \, x - y, \, 2y - z \rangle\); \(D\) is the region between spheres of radius 2 and 4 centered at the origin.

    Answer
    \(-703.7168\)

    25. [T] \(\vecs F = \dfrac{\vecs r}{\|\vecs r\|} = \dfrac{\langle x,y,z\rangle}{\sqrt{x^2+y^2+z^2}}\); \(D\) is the region between spheres of radius 1 and 2 centered at the origin.

    26. [T] \(\vecs F = \langle x^2, \, -y^2, \, z^2 \rangle\); \(D\) is the region in the first octant between planes \(z = 4 - x - y\) and \(z = 2 - x - y\).

    Answer
    \(20\)

    27. Let \(\vecs F(x,y,z) = 2x\,\mathbf{\hat i} - 3xy\,\mathbf{\hat j} + xz^2\,\mathbf{\hat k}\). Use the divergence theorem to calculate \(\displaystyle \iint_S \vecs F \cdot dS\), where \(S\) is the surface of the cube with corners at \((0,0,0), \, (1,0,0), \, (0,1,0), \, (1,1,0), \, (0,0,1), \, (1,0,1), \, (0,1,1)\), and \((1,1,1)\), oriented outward.

    28. Use the divergence theorem to find the outward flux of field \(\vecs F(x,y,z) = (x^3 - 3y)\,\mathbf{\hat i} + (2yz + 1)\,\mathbf{\hat j} + xyz\,\mathbf{\hat k}\) through the cube bounded by planes \(x = \pm 1, \, y = \pm 1, \) and \(z = \pm 1\).

    Answer
    \(\displaystyle \iint_S \vecs F \cdot dS = 8\)

    29. Let \(\vecs F(x,y,z) = 2x\,\mathbf{\hat i} - 3y\,\mathbf{\hat j} + 5z\,\mathbf{\hat k}\) and let \(S\) be hemisphere \(z = \sqrt{9 - x^2 - y^2}\) together with disk \(x^2 + y^2 \leq 9\) in the \(xy\)-plane. Use the divergence theorem.

    30. Evaluate \(\displaystyle \iint_S \vecs F \cdot \vecs n \, dS\), where \(\vecs F(x,y,z) = x^2 \,\mathbf{\hat i} + xy\,\mathbf{\hat j} + x^3y^3\,\mathbf{\hat k}\) and \(S\) is the surface consisting of all faces except the tetrahedron bounded by plane \(x + y + z = 1\) and the coordinate planes, with outward unit normal vector \(\vecs N.\)

    A vector field in three dimensions, with arrows becoming larger the further away from the origin they are, especially in their x components. S is the surface consisting of all faces except the tetrahedron bounded by the plane x + y + z = 1. As such, a portion of the given plane, the (x, y) plane, the (x, z) plane, and the (y, z) plane are shown. The arrows point towards the origin for negative x components, away from the origin for positive x components, down for positive x and negative y components, as well as positive y and negative x components, and for positive x and y components, as well as negative x and negative y components.

    Contestar
    \(\displaystyle \iint_S \vecs F \cdot \vecs n \, dS = \dfrac{1}{8}\)

    31. Encuentre el flujo neto de campo hacia afuera\(\vecs F = \langle bz - cy, \, cx - az, \, ay - bx \rangle\) a través de cualquier superficie lisa y cerrada en\(R^3\) donde\(a, \, b,\) y\(c\) son constantes.

    32. Utilizar el teorema de divergencia para evaluar\(\displaystyle \iint_S ||\vecs R||\vecs R \cdot \vecs n \, ds,\) dónde\(\vecs R(x,y,z) = x\,\mathbf{\hat i} + y\,\mathbf{\hat j} + z\,\mathbf{\hat k}\) y\(S\) es esfera\(x^2 + y^2 + z^2 = a^2\), con constante\(a > 0\).

    Contestar
    \(\displaystyle \iint_S ||\vecs R||\vecs R \cdot \vecs n \, ds = 4\pi a^4\)

    33. Utilice el teorema de divergencia para evaluar\(\displaystyle \iint_S \vecs F \cdot dS,\) dónde\(\vecs F(x,y,z) = y^2 z\,\mathbf{\hat i} + y^3\,\mathbf{\hat j} + xz\,\mathbf{\hat k}\) y\(S\) es el límite del cubo definido por\(-1 \leq x \leq 1, \, -1 \leq y \leq 1\), y\(0 \leq z \leq 2\).

    34. Dejar\(R\) ser la región definida por\(x^2 + y^2 + z^2 \leq 1\). Usa el teorema de la divergencia para encontrar\(\displaystyle \iiint_R z^2 \, dV.\)

    Contestar
    \(\displaystyle \iiint_R z^2 dV = \dfrac{4\pi}{15}\)

    35. Dejar\(E\) ser el sólido delimitado por el\(xy\) -plano y paraboloide de\(z = 4 - x^2 - y^2\) manera que\(S\) sea la superficie de la pieza paraboloide junto con el disco en el\(xy\) -plano que forma su fondo. Si\(\vecs F(x,y,z) = (xz \, \sin(yz) + x^3) \,\mathbf{\hat i} + \cos (yz) \,\mathbf{\hat j} + (3zy^2 - e^{x^2+y^2})\,\mathbf{\hat k}\), encontrar\(\displaystyle \iint_S \vecs F \cdot dS\) usando el teorema de la divergencia.

    Un campo vectorial en tres dimensiones con todas las flechas apuntando hacia abajo. Parecen seguir el camino del paraboloide dibujado abriéndose hacia abajo con vértice en el origen. S es la superficie de este paraboloide y el disco en el plano (x, y) que forma su fondo.

    36. Let

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