df dx There is an infinite number of different paths from

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1 Integrl clculus line integrls Feb 7, 18 From clculus, in the cse of single vrible x1 F F x F x f x dx, where f x 1 x df dx Now, consider the cse tht two vribles re t ply. Suppose,, df M x y dx N x y dy x, y x y 1, 1 There is n infinite number of x, y to x, y different pths from 1 1 x 1 1 df x M x, y x dx N y, x y c y y dy Note how in the dx integrl we expressed y in terms of x nd in the dy integrl we expressed x in terms of y.

2 Exmple M x3 y, N 3x4y integrte long y x3 from (x,y) = (, 3) to (,7) c 7 y 3 df x x dx y dy x9dx y dy 3 8x 11 y 9 7 9x y

3 if df is n exct differentil, then c F F df dx dy F x y F x y c x y y x,, 1 1 The integrl is independent of the pth, i.e., only depends on the endpoints. if df not n exct differentil, the integrl depends on the pth The exmple given bove corresponds to n exct differentil. We lredy integrted this long the most direct pth. Now lets integrte from x = to, keeping y = 3, nd then from y = 3 to 7, keeping x = x y dx x y dy x 9x 6y y x dx y dy (49) sme s before (,3) (,7) (,3)

4 Exmple of n inexct differentil du dx xdy (,) Pth c y x 4 Pth b cb (y = x) dx xdy dx ydy dx xdy dx dy x y 6 (,) b b (,) pth is long the line y = x One cn hve line integrls with three or more independent vribles. c,,,,,, M xyzdx N xyzdy Pxyzdz

5 Exmple: du yzdx xzdy xydz This is n exct differentil (with u = xyz) Exct differentils re importnt in thermodynmics reversible processes proceed infinitely slowly cyclic processes: begin nd end in sme stte P, T, V P, T, V P, T, V ex A line integrl for cyclic process is written s du du if du is n exct differentil If du is not n exct differentil du obvious such process is irreversible, since going round cycle does not fully restore the system to its initil stte.

6 For gs (or liquid) d rev PdV This is not n exct differentil d not c rev rev rev Δ would imply tht it depends on the endpoints only Exmple: 1 mol of idel gs t T = 5 K, V =. l () expnd t constnt T to V = 4. l (b) cool t constnt volume to 3 K (c) compress to V = l t T = 3 K (d) het to T= 5 K t V = l rev 4 () PdV nrt dv V V JK 4 1. mol nrt n K n J V1 mol using PV nrt

7 Step (b) V, so work = nrt Step (c) rev PdV dv 179J V Step (d) rev since there is no volume chnge To go round the cycle, 1153J rev dq q c dq = het trnsferred in reversible process It is lso not n exct differentil du dq d U q so for cyclic process q = for the process worked out bove, het hd to be provided dq rev C dt, p dibtic process: dq du d d d constnt P nd no rection or phse chnge, C p = het cpcity d

8 Double integrls b I f x, y dxdy 1 b1 When doing the first integrl, tret the other vrible s constnt Exmple (where limits on integrls re just constnts) b x x y bx 4xy dydx 4xy 3 b b x dx bx b x 3 dx

9 Double integrls A somewht more complicted cse is when the upper limit of the inner integrl depends on the vrible used for the outer integrtion. Exmple 3x x xy y dydx 3 3x y xy xy dx x 3 3x 9x dx 1x dx x 4 4

10 Are squre b dydx b bdxb re given by D integrl or by 1D multiplied by height function Are semicircle of rdius x 1 dydx 1 1xdx Are circle = r Are semicircle = r Volume of cube of side dxdydz x y z 3 Or cn sy tht we hve height x re = 3 dx dy

11 A more chllenging exmple If upper surfce were flt (prllel to xy plne), with f = 3 4 x 4 dydx x dx x 1.3 x 4 3 If the upper surfce were flt with f = 6, the volume would be 63.9 The nswer for the ctul problem must be between these two numbers s the upper surfce vries from to 6 ccording to f = y y y V dzdydx y dydx y dx x 4 x 4 x 4 4 x 4 x 8x 16 x 4 dx x 8 dx x 6 3 x 6x 16 dx x 16x 1 3

12 In QM, for the H tom nd for prticle in 3 D box, we encounter triple integrls Prob b c b c x, y, z dxdydz Probbility of finding prticle in the specified volume (rectngulr box) A specil cse is when the integrl fctors b c b c f x g y h z dzdydx b c f xdx gydy hzdz 1 b1 c1 It my be esier to do n integrl in polr or sphericl coordintes The H tom wvefunction seprtes in r, φ, θ, but not in x, y, z

13 Integrls involving polr coordintes xy Be Be integrted over ll spce should give 1. B e dd B B e d Note, we pick up fctor of when integrting in polr coordintes Normlize B 1 B Note tht the bove integrl cn lso be esily done in Crtesin coordintes since the exponent seprtes into x nd y fctors This would not be true if we hd sqroot(x +y ) e x dx Similrly, for the integrl over y

14 f h1 if = height = if = height = h h h h 1 dd h h

15 Integrls in sphericl coordintes f r,, r sin drd d Here the integrl is over ll spce. is mx ngle for nd is the mx ngle for