MATH 253 WORKSHEET 24 MORE INTEGRATION IN POLAR COORDINATES. r dr = = 4 = Here we used: (1) The halfangle formula cos 2 θ = 1 2


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1 MATH 53 WORKSHEET MORE INTEGRATION IN POLAR COORDINATES ) Find the volume of the solid lying bove the xyplne, below the prboloid x + y nd inside the cylinder x ) + y. ) We found lst time the set of points in the plne lying inside the cylinder is D {r, θ) r cos θ}. Find fr, θ) describing the height of the solid bove ech such point. Solution: fr, θ) x + y r. b) Clculte the volume of the solid, tht is fr, θ) da. D Solution : Nturl to slice in rdil lines, tht is for ech θ integrte over 0 r cos θ. The bounding circle of D is tngent to the yxis t the origin, so θ rnges from π downwrd) to π upwrd). The integrl is thus θπ/ r cos θ θπ/ [ ] r r r cos θ r dr αθ θπ/ θ+π/ απ α π cos θ ) + cosθ) + cos α + cos α ) dα ) π 3 π. Here we used: ) The hlfngle formul cos θ + cosθ)) nd ) Tht the verge of cos α over full revolution is ero, while the verge of cos α is since cos α+sin α ). An lterntive to is using the hlfngle formul gin: cos α + cos α) nd now integrte from π to π. Solution : Let's slice in concentric circles. possible), so we cn lso write the integrl s r r 3 dr θ rccos r θ rccos r r/cos u 3 0 u0 The lrgest r vlue is when θ 0 this is r r 3 rccos dr ) uπ/ cos 3 u ) u) sin u) du by prts 8 [ u cos u ] uπ/ + 8 u0 8 π/ 0 cos u du uπ/ u0 cos u du md from now on we cn continue s bove since cos is even this is equl to +π/ π/ cos u du). ) In this problem we will nd the electricl eld due to sheet of chrge. Suppose we hve n innite conducting plte in the xy plne, contining σ units of chrge per unit re. The electricl eld due to the plte must point verticlly why?), nd cn only depend on the height bove the plte. Dte : //03.
2 ) Consider smll prt of the plte of re A ner the point x, y, 0). Wht is the chrge q in this smll prt? Solution : q σ A chrge/unit re re). ) By the inverse squre lw, the electricl eld t 0, 0, ) due to the chrge ner x, y, 0) is given by the vector k q v where v is the vector between the two points. Express the verticl v 3 component of this vector s function of x, y). Solution: We hve v x, y, so v x + y + nd the projection of v on the kσ verticl xis is. In other words, we hve E A. x +y + ) 3/ b) Express the electricl eld t 0, 0, ) by n integrl. Solution: Summing over the contributions from the whole plte, we get E c) Evlute the integrl. Solution: In polr coordintes R kσ r da x + y + 3/ ) r dr θπ kσ r + ) 3/ θ0 [ kσ π) ) r + ) / kσ π) πkσ. ] r Notes: ) For the whole plne we go over ll ngles 0 θ π) nd ll rdii 0 r < ). ) The derivtive of r + ) / is r which is our integrnd up to constnt 3) r + ) 3/ As r r 0. d) Cn you nd function φx, y, ) Electric potentil) such tht φ E? Solution: Ex, y, ) πkσ 0, 0, since the by rottionl symmetry the eld must point up or down. Since φ is perpendiculr to the level sets, we see tht the level sets must be plnes prllel to the xy plne, so φ cn depend on lone. Then φ 0, 0, φ nd if this is constnt we see tht φ must be propotionl to, speciclly tht φx, y, ) πkσ. 3) In this problem we will nd the re under the bell curve. Let I dx, nd let e x J e R x y da integrl over the whole plne). ) Using n iterted integrl in the xy coordintes relte J to I. Solution: R e x y da dx dye x y dxe x ) x dxe dye y dye y ) I. Note: reliing tht I wsn't esy. We re slowly working on dye y dxe x developing the needed mentl exibility. b) Switch to polr coordintes nd evlute J. u0 e u du π where we substituted u r Solution: J θπ θ0 r r dre r π nd used e u du. 0 c) Given σ > 0 nd number Z such tht Solution: to choose Z πσ. Z e x /σ ) dx ux/ σ Z /σ ) Z dx. e x σ e u du Z σ π πσ Z so we need
3 ) The electric potentil t point Z due to chrge q t the point X is Find the electricl. potentil t height bove the middle of squre plte of side length, if the chrge density is σ. Solution: Prmetrie the points on the plte s {x, y, 0) x, y }. Then the distnce to the point 0, 0, ) bove the middle of the plte is x + y +, so the integrl is kq XZ φ) [,] kσ da x + y + recll tht σ da is the innitesiml chrge t the re element da). We convert this integrl to polr coordintes. For this we divide the plte into eight tringles s in the left prt of the gure below. Ech of those hs the sme contribution since ny two dier only by chnging the sign of x or y or both, or by exchnging x nd y). So the totl potentil is eight times the potentil creted by the tringle mgnied in the righthnd side. We prmeterie the point x, y) insted by r, θ) s in the gure. Then x + y + ) / r + ) /. Also, we cn see tht r is t lest ero, nd t most the length of the hypotenuse in the right tringle below with ngle θ nd side. It follows tht the tringle cn be described s { r, θ) 0 θ π, 0 r } cos θ. The integrl is then 8kσ θ π θ0 r cos θ r dr 8kσ r + ) / 8kσ 8kσ θ π θ0 θ π θ0 θ π θ0 [ r + ) ] / r/ cos θ [ cos θ + ] cos θ + πkσ. Remrk. The solution up to here ws very dicult mny ides were needed, nd using polr coordintes on tringle would not occur to nyone), but within the scope of 53. Actully clculting the remining integrl is not. We now concentrte on the remining integrl. It seems nturl to write the integrnd s + cos θ cos θ + cos θ cos θ cos θ since cos θ d sin θ). Chnging vribles this wy we see tht θ π θ0 cos θ + u/ u0 3 + u u du.
4 We cn get rid of the squre root by setting u + sin α so tht + ) u + cos α, leding so the integrl equls For convenience set A trick + + ) u/ u0 u/ u0 u/ u0 + cos α cos α cos α sin cos α dα α cos α dα + ) sin α dα We then remove the cos α from the numertor, in preprtion for u/ u0 [ cos ] α A cos α A + A cos dα. α A We cn do the rst integrl, noting tht u 0 corresponds to α 0 nd u / corresponds to α rcsin so + ) u0 u/ u0 dα rcsin + ). For the second integrl, we divide through by cos α nd recll tht cos α + tn α setting t tn α we hve u/ A dα u/ cos α A dα cos α Now, This integrl therefore equls A A A u/ u0 [ log + t t u0 u/ u0 u/ u0 + + A cos α A + t ) dt A A. dt. t + t + ) t dt ] u/ u0. dtn α) dα, so Now t u 0 we hve α 0 nd thus t tn 0 0, t which point the logrithm vnishes. At u / we hve sin α so + ) cos α sin α + ) + +. Thus t u we hve t +.
5 In conclusion, u/ u0 nd the potentil is therefore + u u du rcsin φ) 8kσ rcsin rcsin + ) + log ) + log ) + kσ log πkσ. The following is entirely unrelted to MATH 53. Let's use Tylor expnsions + physics intuition to check this kind of complicted nswer. We'll exmine the behviour s 0 nd when ) nd check if it looks resonble. ) As 0, rcsin nd log + + re even functions of, so jointly hve Tylor + + ) expnsion kσ log + + C + for some C. It follows tht for smll, our potentil looks like + φ) kσ log πkσ + In this pproximtion, the closer we re to the plte the more it looks innite, nd indeed the nonconstnt term mtches the nswer to d). Wht bout the constnt term? Recll tht φ) in d) ws determined by its grdient, so we could put in it whtever constnt we wnted so in fct the nswers mtch, except for higherorder corrections terms in Tylor series). ) As, + ) ) / + Since the derivtive of rcsin x t x is / ), we mke liner pproximtion to see tht s, For the second term, [ π 8kσ rcsin + ) 8kσ ) / + ) / ) +. ] πkσ kσ ) +. to rst order in tht is, ignoring terms like or higher). Tking log we see tht + kσ log + + 8kσ log + ) 8kσ. Adding up everything we see tht s φ) πkσ kσ + 8kσ πkσ ) kσ σ) k. 5
6 Now ) is the re of the plte, so σ) is the totl chrge of the plte. In other words, s, φ) is pproximtely the potentil due to chrge equl to the totl chrge of the plte, plced t the origin. This is wht we expect from very fr wy, the plte relly looks like pointlike object). 6
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