320 THEORY OF HEAT. [CHAP. VII. 



the section, is equal to the product of the coefficient k, of the area 



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dydzj of the element dt, and of the ratio -=- taken with the nega 

 tive sign. We must therefore take the integral k I dy I dz -= , 



from z = to z I, the half thickness of the bar, and then from 

 y = to y = I. We thus have the fourth part of the whole flow. 



The result of this calculation discloses the law according to 

 which the quantity of heat which crosses a section of the bar 

 decreases ; and we see that the distant parts receive very little 

 heat from the source, since that which emanates directly from it 

 is directed partly towards the surface to be dissipated into the air. 

 That which crosses any section whatever of the prism forms, if we 

 may so say, a sheet of heat whose density varies from one point 

 of the section to another. It is continually employed to replace 

 the heat which escapes at the surface, through the whole end of 

 the prism situated to the right of the section : it follows therefore 

 that the whole heat which escapes during a certain time from this 

 part of the prism is exactly compensated by that which penetrates 

 it by virtue of the interior conducibility of the solid. 



To verify this result, we must calculate the produce of the flow 

 established at the surface. The element of surface is dxdy, and v 

 being its temperature, hvdxdy is the quantity of heat which 

 escapes from this element during the unit of time. Hence the 



integral h\dx\dyv expresses the whole heat which has escaped 



from a finite portion of the surface. We must now employ the 

 known value of v in y t supposing z = 1, then integrate once from 

 y = QiQy = l, and a second time from x = x up to x = oo . We 

 thus find half the heat which escapes from the upper surface of 

 the prism ; and taking four times the result, we have the heat lost 

 through the upper and lower surfaces. 



If we now make use of the expression h Ida) I dz v, and give to 



y in v its value I, and integrate once from z = to z = l, and a 

 second time from x = to x = oo ; we have one quarter of the heat 

 which escapes at the lateral surfaces. 



The integral /? I dx \dy v, taken between the limits indicated gives 



