Surface integrals of the second kind (over oriented surfaces)

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Compute the following surface integral of the second kind: ∬Sx dy dz+y dx dz+z dx dy\displaystyle \iint_{S} x\, dy\, dz+y\, dx\, dz+z\, dx\, dy, where SS is the outer surface of the sphere x2+y2+z2=a2x^{2}+y^{2}+z^{2}=a^{2}.

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Compute the following surface integral of the second kind: ∬Sf(x) dy dz+g(y) dx dz+h(z) dx dy\displaystyle \iint_{S} f(x)\, dy\, dz+g(y)\, dx\, dz+h(z)\, dx\, dy, where f(x)f(x), g(y)g(y), h(z)h(z) are continuous functions and SS is the outer surface of the rectangular parallelepiped 0<x<a0<x<a, 0<y<b0<y<b, 0<z<c0<z<c.

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Compute the following surface integral of the second kind: ∬S(y−z) dy dz+(z−x) dx dz+(x−y) dx dy\displaystyle \iint_{S} (y-z)\, dy\, dz+(z-x)\, dx\, dz+(x-y)\, dx\, dy, where SS is the outer surface of the conical surface x2+y2=z2x^{2}+y^{2}=z^{2} (0⩽z⩽h)(0\leqslant z\leqslant h).

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Compute the following surface integral of the second kind: ∬Sdy dzx+dx dzy+dx dyz\displaystyle \iint_{S} \dfrac {dy\, dz}{x}+\dfrac {dx\, dz}{y}+\dfrac {dx\, dy}{z}, where SS is the outer surface of the ellipsoid x2a2+y2b2+z2c2=1\dfrac {x^{2}}{a^{2}}+\dfrac {y^{2}}{b^{2}}+\dfrac {z^{2}}{c^{2}}=1.

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Compute the following surface integral of the second kind: ∬Sx2 dy dz+y2 dx dz+z2 dx dy\displaystyle \iint_{S} x^{2}\, dy\, dz+y^{2}\, dx\, dz+z^{2}\, dx\, dy, where SS is the outer surface of the spherical shell (x−a)2+(y−b)2+(z−c)2=R2(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=R^{2}.

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Apply the Ostrogradsky formula to transform the following surface integral, where the smooth surface SS encloses the bounded volume VV, and cos⁡α\cos \alpha, cos⁡β\cos \beta, cos⁡γ\cos \gamma are the direction cosines of the outward normal of the surface SS: ∬Sx3 dy dz+y3 dx dz+z3 dx dy\displaystyle \iint_{S} x^{3}\, dy\, dz+y^{3}\, dx\, dz+z^{3}\, dx\, dy.

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Apply the Ostrogradsky formula to transform the following surface integral, where the smooth surface SS encloses the bounded volume VV, and cos⁡α\cos \alpha, cos⁡β\cos \beta, cos⁡γ\cos \gamma are the direction cosines of the outward normal of the surface SS: ∬Sxy dx dy+xz dx dz+yz dy dz\displaystyle \iint_{S} xy\, dx\, dy+xz\, dx\, dz+yz\, dy\, dz.

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Apply the Ostrogradsky formula to transform the following surface integral, where a smooth surface SS encloses a bounded volume VV, and cos⁡α\cos \alpha, cos⁡β\cos \beta, cos⁡γ\cos \gamma are the direction cosines of the outward normal of the surface SS: ∬Sxcos⁡α+ycos⁡β+zcos⁡γx2+y2+z2 dS\displaystyle \iint_{S} \dfrac {x\cos \alpha +y\cos \beta +z\cos \gamma }{\sqrt{x^{2}+y^{2}+z^{2}}}\, dS.

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Apply the Ostrogradsky formula to transform the following surface integral, where a smooth surface SS encloses a bounded volume VV, and cos⁡α\cos \alpha, cos⁡β\cos \beta, cos⁡γ\cos \gamma are the direction cosines of the outward normal of the surface SS: ∬S(∂u∂xcos⁡α+∂u∂ycos⁡β+∂u∂zcos⁡γ) dS\displaystyle \iint_{S} \left(\dfrac {\partial u}{\partial x}\cos \alpha +\dfrac {\partial u}{\partial y}\cos \beta +\dfrac {\partial u}{\partial z}\cos \gamma \right)\, dS.

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Apply the Ostrogradsky formula to transform the following surface integral, where a smooth surface SS encloses a bounded volume VV, and cos⁡α\cos \alpha, cos⁡β\cos \beta, cos⁡γ\cos \gamma are the direction cosines of the outward normal of the surface SS:

∬S[(∂R∂y−∂Q∂z)cos⁡α+(∂P∂z−∂R∂x)cos⁡β+(∂Q∂x−∂P∂y)cos⁡γ] dS. \begin{align} \iint _{S} \Bigg[\left(\dfrac {\partial R}{\partial y}-\dfrac {\partial Q}{\partial z}\right)\cos \alpha +\left(\dfrac {\partial P}{\partial z}-\dfrac {\partial R}{\partial x}\right)\cos \beta +\left(\dfrac {\partial Q}{\partial x}-\dfrac {\partial P}{\partial y}\right)\cos \gamma \Bigg]\, dS . \end{align}
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Using the Ostrogradsky formula, compute the following surface integral: ∬Sx2 dy dz+y2 dx dz+z2 dx dy\displaystyle \iint_{S} x^{2}\, dy\, dz+y^{2}\, dx\, dz+z^{2}\, dx\, dy, where SS is the outer surface of the boundary of the cube 0<x<a0<x<a, 0<y<a0<y<a, 0<z<a0<z<a.

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Using the Ostrogradsky formula, compute the following surface integral: ∬Sx3 dy dz+y3 dx dz+z3 dx dy\displaystyle \iint_{S} x^{3}\, dy\, dz+y^{3}\, dx\, dz+z^{3}\, dx\, dy, where SS is the outer surface of the sphere x2+y2+z2=a2x^{2}+y^{2}+z^{2}=a^{2}.

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Using the Ostrogradsky formula, compute the following surface integral: ∬S(x−y+z) dy dz+(y−z+x) dx dz+(z−x+y) dx dy\displaystyle \iint_{S} (x-y+z)\, dy\, dz+(y-z+x)\, dx\, dz+(z-x+y)\, dx\, dy, where SS is the outer surface of the surface ∣x−y+z∣+∣y−z+x∣+∣z−x+y∣=1\left|x-y+z\right|+\left|y-z+x\right|+\left|z-x+y\right|=1.

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Compute

∬S(x2cos⁡α+y2cos⁡β+z2cos⁡γ) dS, \iint _{S} (x^{2}\cos \alpha +y^{2}\cos \beta +z^{2}\cos \gamma )\, dS ,

where SS is the part of the conical surface x2+y2=z2x^{2}+y^{2}=z^{2} (0⩽z⩽h)(0\leqslant z\leqslant h), and cos⁡α\cos \alpha, cos⁡β\cos \beta, cos⁡γ\cos \gamma are the direction cosines of the outward normal of this surface.

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Find the flux of the vector r→\overrightarrow {r}: (1) through the lateral surface of the cone x2+y2⩽z2x^{2}+y^{2}\leqslant z^{2} (0⩽z⩽h)(0\leqslant z\leqslant h); (2) through the base of this cone.

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(a)
(b)

Find the flux of the vector a→=yzi→+zxj→+xyk→\overrightarrow {a}=yz\overrightarrow {i}+zx\overrightarrow {j}+xy\overrightarrow {k}: (a) through the lateral surface of the cylinder x2+y2⩽a2x^{2}+y^{2}\leqslant a^{2} (0⩽z⩽h)(0\leqslant z\leqslant h); (б) through the entire surface of this cylinder.

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(a)
(b)

Find the flux of the radius vector r→\overrightarrow {r} through the surface

z=1−x2+y2(0⩽z⩽1) z = 1-\sqrt{x^{2}+y^{2}} \qquad (0\leqslant z\leqslant 1)
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Find the flux of the vector a→=x2i→+y2j→+z2k→\overrightarrow {a}=x^{2}\overrightarrow {i}+y^{2}\overrightarrow {j}+z^{2}\overrightarrow {k} through the part of the sphere x2+y2+z2=1x^{2}+y^{2}+z^{2}=1, x⩾0x\geqslant 0, y⩾0y\geqslant 0, z⩾0z\geqslant 0, lying in the first octant.

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Prove that the flux of the vector a→\overrightarrow {a} through the surface SS given by the equation r→=r→(u,v)\overrightarrow {r}=\overrightarrow {r}(u,v) ((u,v)∈Ω)((u,v)\in \Omega ) equals

∬Sa→⋅n→ dS=∬Ω(a→ ∂r→∂u ∂r→∂v) du dv, \iint _{S} \overrightarrow {a}\cdot \overrightarrow {n}\, dS = \iint _{\Omega } \left(\overrightarrow {a}\ \dfrac {\partial \overrightarrow {r}}{\partial u}\ \dfrac {\partial \overrightarrow {r}}{\partial v}\right)\, du\, dv ,

where n→\overrightarrow {n} is the unit normal vector of the surface SS.

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Using the Gauss-Ostrogradsky theorem, calculate the integrals. ∬S(1+2x)dydz+(2x+3y)dzdx+(3y+4z)dxdy\iint_{S}(1+2 x) d y d z+(2 x+3 y) d z d x+(3 y+4 z) d x d y, where SS :

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(1)

the outer side of the surface of the pyramid x/a+y/b+z/c⩽1x / a+y / b+z / c \leqslant 1, x⩾0,y⩾0,z⩾0x \geqslant 0, y \geqslant 0, z \geqslant 0

(2)

the inner side of the surface ∣x−y+z∣+∣y−z+x∣+∣z−x+y∣=a\left|x-y+z\right|+\left|y-z+x\right|+\left|z-x+y\right| = a.

Using the Gauss-Ostrogradsky theorem, calculate the integrals. ∬Szdxdy+(5x+y)dydz\iint_{S} z d x d y+(5 x+y) d y d z, where SS :

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(1)

the outer side of the total surface of the cone x2+y2⩽z2x^{2}+y^{2} \leqslant z^{2}, 0⩽z⩽40 \leqslant z \leqslant 4

(2)

the inner side of the ellipsoid x2/4+y2/9+z2=1x^{2} / 4+y^{2} / 9+z^{2} = 1;

(3)

the outer side of the boundary of the region 1<x2+y2+z2<41 < x^{2}+y^{2}+z^{2} < 4.

Using the Gauss-Ostrogradsky theorem, calculate the integrals. ∬Sx2dydz+y2dzdx+z2dxdy\iint_{S} x^{2} d y d z+y^{2} d z d x+z^{2} d x d y, where SS :

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(1)

the inner side of the surface of the parallelepiped 0⩽x⩽a0 \leqslant x \leqslant a, 0⩽y⩽b,0⩽z⩽c;0 \leqslant y \leqslant b, 0 \leqslant z \leqslant c ;

(2)

the outer side of the total surface x2/a2+y2/b2⩽z2/c2,0⩽z⩽cx^{2} / a^{2}+y^{2} / b^{2} \leqslant z^{2} / c^{2}, 0 \leqslant z \leqslant c (cone).

Using the Gauss-Ostrogradsky theorem, calculate the integrals. ∬Sx3dydz+y3dzdx+z3dxdy\iint_{S} x^{3} d y d z+y^{3} d z d x+z^{3} d x d y, where SS :

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(1)

the outer side of the surface of the tetrahedron x+y+z⩽a,x⩾0x+y+z \leqslant a, x \geqslant 0, y⩾0,z⩾0y \geqslant 0, z \geqslant 0

(2)

the inner side of the sphere x2+y2+z2=R2x^{2}+y^{2}+z^{2} = R^{2}.

Using the Gauss-Ostrogradsky theorem, calculate the integrals. ∬Sx4dydz+y4dzdx+z4dxdy\iint_{S} x^{4} d y d z+y^{4} d z d x+z^{4} d x d y, where SS :

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(1)

the sphere x2+y2+z2=R2x^{2}+y^{2}+z^{2} = R^{2};

(2)

the outer side of the total surface of the hemisphere x2+y2+z2⩽R2,z⩾0x^{2}+y^{2}+z^{2} \leqslant R^{2}, z \geqslant 0.

References

[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Функции многих переменных. 2-е изд., ФИЗМАТЛИТ, 2003.