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6.6 Surface integrals  (Page 13/27)

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∬ S F · d S = ∬ S − k ∇ T · d S .

Calculating heat flow

A cast-iron solid cylinder is given by inequalities x 2 + y 2 ≤ 1 , 1 ≤ z ≤ 4 . The temperature at point ( x , y , z ) in a region containing the cylinder is T ( x , y , z ) = ( x 2 + y 2 ) z . Given that the thermal conductivity of cast iron is 55, find the heat flow across the boundary of the solid if this boundary is oriented outward.

Let S denote the boundary of the object. To find the heat flow, we need to calculate flux integral ∬ S − k ∇ T · d S . Notice that S is not a smooth surface but is piecewise smooth, since S is the union of three smooth surfaces (the circular top and bottom, and the cylindrical side). Therefore, we calculate three separate integrals, one for each smooth piece of S . Before calculating any integrals, note that the gradient of the temperature is ∇ T = ⟨ 2 x z , 2 y z , x 2 + y 2 ⟩ .

First we consider the circular bottom of the object, which we denote S 1 . We can see that S 1 is a circle of radius 1 centered at point ( 0 , 0 , 1 ) , sitting in plane z = 1 . This surface has parameterization r ( u , v ) = ⟨ v cos u , v sin u , 1 ⟩ , 0 ≤ u < 2 π , 0 ≤ v ≤ 1 . Therefore,

t u = ⟨ − v sin u , v cos u , 0 ⟩ and t v = ⟨ cos u , v sin u , 0 ⟩ ,

and

t u × t v = ⟨ 0 , 0 , − v sin 2 u − v cos 2 u ⟩ = ⟨ 0 , 0 , − v ⟩ .

Since the surface is oriented outward and S 1 is the bottom of the object, it makes sense that this vector points downward. By [link] , the heat flow across S 1 is

∬ S 1 − k ∇ T · d S = −55 ∫ 0 2 π ∫ 0 1 ∇ T ( u , v ) · ( t u × t v ) d v d u = −55 ∫ 0 2 π ∫ 0 1 ⟨ 2 v cos u , 2 v sin u , v 2 cos 2 u + v 2 sin 2 u ⟩ · ⟨ 0 , 0 , − v ⟩ d v d u = −55 ∫ 0 2 π ∫ 0 1 ⟨ 2 v cos u , 2 v sin u , v 2 ⟩ · ⟨ 0 , 0 , − v ⟩ d v d u = −55 ∫ 0 2 π ∫ 0 1 − v 3 d v d u = −55 ∫ 0 2 π − 1 4 d u = 55 π 2 .

Now let’s consider the circular top of the object, which we denote S 2 . We see that S 2 is a circle of radius 1 centered at point ( 0 , 0 , 4 ) , sitting in plane z = 4 . This surface has parameterization r ( u , v ) = ⟨ v cos u , v sin u , 4 ⟩ , 0 ≤ u < 2 π , 0 ≤ v ≤ 1 . Therefore,

t u = ⟨ − v sin u , v cos u , 0 ⟩ and t v = ⟨ cos u , v sin u , 0 ⟩ ,

and

t u × t v = ⟨ 0 , 0 , − v sin 2 u − v cos 2 u ⟩ = ⟨ 0 , 0 , − v ⟩ .

Since the surface is oriented outward and S 1 is the top of the object, we instead take vector t v × t u = ⟨ 0 , 0 , v ⟩ . By [link] , the heat flow across S 1 is

∫ ∫ S 2 − k ∇ T • d S = −55 ∫ 0 2 π ∫ 0 1 ∇ T ( u , v ) • ( t v × t u ) d v d u = −55 ∫ 0 2 π ∫ 0 1 ⟨ 8 v cos u , 8 v sin u , v 2 cos 2 u + v 2 sin 2 u ⟩ • ⟨ 0 , 0 , v ⟩ d v d u = −55 ∫ 0 2 π ∫ 0 1 ⟨ 8 v cos u , 8 v sin u , v 2 ⟩ • ⟨ 0 , 0 , v ⟩ d v d u = −55 ∫ 0 2 π ∫ 0 1 v 3 d v d u = − 55 π 2 .

Last, let’s consider the cylindrical side of the object. This surface has parameterization r ( u , v ) = ⟨ cos u , sin u , v ⟩ , 0 ≤ u < 2 π , 1 ≤ v ≤ 4 . By [link] , we know that t u × t v = ⟨ cos u , sin u , 0 ⟩ . By [link] ,

∬ S 3 − k ∇ T • d S = −55 ∫ 0 2 π ∫ 1 4 ∇ T ( u , v ) • ( t v × t u ) d v d u = −55 ∫ 0 2 π ∫ 1 4 ⟨ 2 v cos u , 2 v sin u , cos 2 u + sin 2 u ⟩ • ⟨ cos u , sin u , 0 ⟩ d v d u = −55 ∫ 0 2 π ∫ 0 1 ⟨ 2 v cos u , 2 v sin u , 1 ⟩ • ⟨ cos u , sin u , 0 ⟩ d v d u = −55 ∫ 0 2 π ∫ 0 1 ( 2 v cos 2 u + 2 v sin 2 u ) d v d u = −55 ∫ 0 2 π ∫ 0 1 2 v d v d u = −55 ∫ 0 2 π d u = −110 π .

Therefore, the rate of heat flow across S is 55 π 2 − 55 π 2 − 110 π = −110 π .

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A cast-iron solid ball is given by inequality x 2 + y 2 + z 2 ≤ 1 . The temperature at a point in a region containing the ball is T ( x , y , z ) = 1 3 ( x 2 + y 2 + z 2 ) . Find the heat flow across the boundary of the solid if this boundary is oriented outward.

− 440 π 3

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Key concepts

  • Surfaces can be parameterized, just as curves can be parameterized. In general, surfaces must be parameterized with two parameters.
  • Surfaces can sometimes be oriented, just as curves can be oriented. Some surfaces, such as a Möbius strip, cannot be oriented.
  • A surface integral is like a line integral in one higher dimension. The domain of integration of a surface integral is a surface in a plane or space, rather than a curve in a plane or space.
  • The integrand of a surface integral can be a scalar function or a vector field. To calculate a surface integral with an integrand that is a function, use [link] . To calculate a surface integral with an integrand that is a vector field, use [link] .
  • If S is a surface, then the area of S is ∫ ∫ S d S .
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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