Section 8.16

The Ostrogradsky Formula

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Problem 4376

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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Problem 4377

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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Problem 4378

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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Problem 4379

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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Problem 4380

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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Problem 4381

Prove that if SS is a closed simple surface and l→\overrightarrow {l} is any fixed direction, then

∬Scos⁡(n→,l→) dS=0, \iint _{S} \cos (\overrightarrow {n},\overrightarrow {l})\, dS = 0 ,

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

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Problem 4382

Prove that the volume enclosed by the surface SS equals

V=13∬S(xcos⁡α+ycos⁡β+zcos⁡γ) dS, V = \dfrac {1}{3}\iint _{S} (x\cos \alpha +y\cos \beta +z\cos \gamma )\, dS ,

where cos⁡α\cos \alpha, cos⁡β\cos \beta, cos⁡γ\cos \gamma are the direction cosines of the outward normal of the surface SS.

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Problem 4383

Prove that the volume of the cone bounded by the smooth conical surface F(x,y,z)=0F(x,y,z)=0 and the plane Ax+By+Cz+D=0Ax+By+Cz+D=0 equals

V=13SH, V = \dfrac {1}{3}SH ,

where SS is the area of the base of the cone lying on the given plane, and HH is the height of the cone.

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Problem 4384

Find the volume of the solid bounded by the surfaces z=±cz=\pm c and

x=acos⁡ucos⁡v+bsin⁡usin⁡v, x = a\cos u\cos v+b\sin u\sin v , y=acos⁡usin⁡v−bsin⁡ucos⁡v, y = a\cos u\sin v-b\sin u\cos v , z=csin⁡u z = c\sin u

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Problem 4385

Find the volume of the body bounded by the surface

x=ucos⁡v,y=usin⁡v,z=−u+acos⁡v(u⩾0) x=u\cos v , \qquad y=u\sin v , \qquad z=-u+a\cos v \qquad (u\geqslant 0)

and the planes x=0x=0, z=0z=0 (a>0)(a>0).

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Problem 4385.1
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Problem 4386

Prove the formula

ddt{∭x2+y2+z2⩽t2f(x,y,z,t) dx dy dz}=∬x2+y2+z2=t2f(x,y,z,t) dS+∭x2+y2+z2⩽t2∂f∂t dx dy dz(t>0). \begin{align} \dfrac {d}{dt}\left\{ \iiint _{x^{2}+y^{2}+z^{2}\leqslant t^{2}} f(x,y,z,t)\, dx\, dy\, dz\right\} \\ & = \iint _{x^{2}+y^{2}+z^{2}=t^{2}} f(x,y,z,t)\, dS + \iiint _{x^{2}+y^{2}+z^{2}\leqslant t^{2}} \dfrac {\partial f}{\partial t}\, dx\, dy\, dz \qquad (t>0) . \end{align}
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Problem 4387

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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Problem 4388

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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Problem 4389

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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Problem 4390

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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Problem 4391

Prove the formula

∭Vdξ dη dζr=12∬Scos⁡(r→,n→) dS, \iiint _{V} \dfrac {d\xi \, d\eta \, d\zeta }{r} = \dfrac {1}{2}\iint _{S} \cos (\overrightarrow {r},\overrightarrow {n})\, dS ,

where SS is the closed surface enclosing the volume VV, n→\overrightarrow {n} is the outward normal at the moving point (ξ,η,ζ)(\xi ,\eta ,\zeta ) on the closed surface SS, and

r=(ξ−x)2+(η−y)2+(ζ−z)2, r = \sqrt{(\xi -x)^{2}+(\eta -y)^{2}+(\zeta -z)^{2}} ,

with r→\overrightarrow {r} the radius vector from the point (x,y,z)(x,y,z) to the point (ξ,η,ζ)(\xi ,\eta ,\zeta ).

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Problem 4392

Compute the Gauss integral

I(x,y,z)=∬Scos⁡(r→,n→)r2 dS, I(x,y,z) = \iint _{S} \dfrac {\cos (\overrightarrow {r},\overrightarrow {n})}{r^{2}}\, dS ,

where SS is a simple closed smooth surface containing the volume VV, n→\overrightarrow {n} is the outward normal to the surface SS at the point (ξ,η,ζ)(\xi ,\eta ,\zeta ), r→\overrightarrow {r} is the radius vector connecting the point (x,y,z)(x,y,z) and the point (ξ,η,ζ)(\xi ,\eta ,\zeta ),

r=(ξ−x)2+(η−y)2+(ζ−z)2. r = \sqrt{(\xi -x)^{2}+(\eta -y)^{2}+(\zeta -z)^{2}} .

Consider two cases: (a) when the surface SS does not enclose the point (x,y,z)(x,y,z); (b) when the surface SS encloses the point (x,y,z)(x,y,z).

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Problem 4393

Prove that if

Δu=∂2u∂x2+∂2u∂y2+∂2u∂z2 \Delta u = \dfrac {\partial ^{2}u}{\partial x^{2}}+\dfrac {\partial ^{2}u}{\partial y^{2}}+\dfrac {\partial ^{2}u}{\partial z^{2}}

and SS is a smooth surface enclosing a bounded volume VV, then the following formulas hold:

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(a)

∬S∂u∂n dS=∭VΔu dx dy dz\displaystyle \iint_{S} \dfrac {\partial u}{\partial n}\, dS = \iiint_{V} \Delta u\, dx\, dy\, dz;

(b)

∬Su∂u∂n dS\displaystyle \iint_{S} u\dfrac {\partial u}{\partial n}\, dS

=∭V[(∂u∂x)2+(∂u∂y)2+(∂u∂z)2] dx dy dz+∭VuΔu dx dy dz, \begin{align} = \iiint _{V} \left[\left(\dfrac {\partial u}{\partial x}\right)^{2}+\left(\dfrac {\partial u}{\partial y}\right)^{2}+\left(\dfrac {\partial u}{\partial z}\right)^{2}\right]\, dx\, dy\, dz +\iiint _{V} u\Delta u\, dx\, dy\, dz , \end{align}

where uu and its partial derivatives up to second order are continuous functions in the domain V+SV+S, and ∂u∂n\dfrac {\partial u}{\partial n} is the derivative along the outward normal of the surface SS.

Problem 4394

Prove the second Green's formula in space

∭V∣ΔuΔvuv∣ dx dy dz=∬S∣∂u∂n∂v∂nuv∣ dS, \iiint _{V} \begin{vmatrix} \Delta u & \Delta v \\ u & v \end{vmatrix}\, dx\, dy\, dz = \iint _{S} \begin{vmatrix} \dfrac {\partial u}{\partial n} & \dfrac {\partial v}{\partial n} \\ u & v \end{vmatrix}\, dS ,

where the volume VV is enclosed by the surface SS, n→\overrightarrow {n} is the outward normal direction of the surface SS, and the functions u=u(x,y,z)u=u(x,y,z), v=v(x,y,z)v=v(x,y,z) are twice differentiable in the domain V+SV+S.

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Problem 4395

The function u=u(x,y,z)u=u(x,y,z) has continuous derivatives up to second order in some domain; if

Δu=∂2u∂x2+∂2u∂y2+∂2u∂z2=0, \Delta u = \dfrac {\partial ^{2}u}{\partial x^{2}}+\dfrac {\partial ^{2}u}{\partial y^{2}}+\dfrac {\partial ^{2}u}{\partial z^{2}} = 0 ,

then u(x,y,z)u(x,y,z) is called a harmonic function in this domain.

Prove that if uu is a harmonic function in the bounded closed domain VV enclosed by a smooth surface SS, then the following formulas hold:

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(a)

∬S∂u∂n dS=0\displaystyle \iint_{S} \dfrac {\partial u}{\partial n}\, dS = 0;

(b)

∭V[(∂u∂x)2+(∂u∂y)2+(∂u∂z)2] dx dy dz\displaystyle \iiint_{V} \left[\left(\dfrac {\partial u}{\partial x}\right)^{2}+\left(\dfrac {\partial u}{\partial y}\right)^{2}+\left(\dfrac {\partial u}{\partial z}\right)^{2}\right]\, dx\, dy\, dz

=∬Su∂u∂n dS, = \iint _{S} u\dfrac {\partial u}{\partial n}\, dS ,

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

Using formula (2), prove that a harmonic function in the domain VV is uniquely determined by its values on the boundary SS.

Problem 4396

Prove that if the function u=u(x,y,z)u=u(x,y,z) is a harmonic function in the bounded closed domain VV enclosed by a smooth surface SS, then

u(x,y,z)=14π∬S[ucos⁡(r→,n→)r2+1r∂u∂n] dS, u(x,y,z) = \dfrac {1}{4\pi }\iint _{S} \left[u\dfrac {\cos (\overrightarrow {r},\overrightarrow {n})}{r^{2}}+\dfrac {1}{r}\dfrac {\partial u}{\partial n}\right]\, dS ,

where r→\overrightarrow {r} is the radius vector drawn from the interior point (x,y,z)(x,y,z) of the domain VV to the moving point (ξ,η,ζ)(\xi ,\eta ,\zeta ) on the surface SS, and

r=(ξ−x)2+(η−y)2+(ζ−z)2, r = \sqrt{(\xi -x)^{2}+(\eta -y)^{2}+(\zeta -z)^{2}} ,

with n→\overrightarrow {n} the outward normal vector of the surface SS at the point (ξ,η,ζ)(\xi ,\eta ,\zeta ).

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Problem 4397

Prove that if u=u(x,y,z)u=u(x,y,z) is a harmonic function inside the ball SS of radius RR centered at the point (x0,y0,z0)(x_{0},y_{0},z_{0}), then

u(x0,y0,z0)=14πR2∬Su(x,y,z) dS u(x_{0},y_{0},z_{0}) = \dfrac {1}{4\pi R^{2}}\iint _{S} u(x,y,z)\, dS

(the mean value theorem).

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Problem 4398

Prove that a function u=u(x,y,z)u=u(x,y,z) that is continuous on a bounded closed domain VV and harmonic in its interior, if it is not constant, cannot attain its maximum and minimum values at interior points of the domain (the maximum principle).

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Problem 4399

A body VV is completely submerged in a liquid. Starting from Pascal's law, prove that the buoyant force of the liquid equals the weight of the liquid displaced by the body, directed vertically upward (Archimedes' principle).

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Problem 4400

Let StS_{t} be the moving sphere (ξ−x)2+(η−y)2+(ζ−z)2=t2(\xi -x)^{2}+(\eta -y)^{2}+(\zeta -z)^{2}=t^{2}, and let the function f(ξ,η,ζ)f(\xi ,\eta ,\zeta ) be continuous. Prove that the function

u(x,y,z,t)=14π∬Stf(ξ,η,ζ)t dSt u(x,y,z,t) = \dfrac {1}{4\pi }\iint _{S_{t}} \dfrac {f(\xi ,\eta ,\zeta )}{t}\, dS_{t}

satisfies the wave equation

∂2u∂x2+∂2u∂y2+∂2u∂z2=∂2u∂t2 \dfrac {\partial ^{2}u}{\partial x^{2}}+\dfrac {\partial ^{2}u}{\partial y^{2}}+\dfrac {\partial ^{2}u}{\partial z^{2}} = \dfrac {\partial ^{2}u}{\partial t^{2}}

together with the initial conditions u∣t=0=0u\big|_{t=0}=0, ∂u∂t∣t=0=f(x,y,z)\dfrac {\partial u}{\partial t}\bigg|_{t=0}=f(x,y,z).

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