50. Use symmetry to evaluate xx D is the region bounded by the square with vertices 5, Prove Property 11. y y CAS

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1 68 CHAPTE MULTIPLE INTEGALS 46. e da, 49. Evlute tn 3 4 da, where,. [Hint: Eploit the fct tht is the disk with center the origin nd rdius is smmetric with respect to both es.] 5. Use smmetr to evlute 3 4 da, where is the region bounded b the squre with vertices 5, 47. Prove Propert. nd, In evluting double integrl over region, sum of 5. Compute s da, where is the disk iterted integrls ws obtined s follows:, b first identifing the integrl s the volume of solid. f, da f, d d 3 3 f, d d Sketch the region nd epress the double integrl s n iterted integrl with reversed order of integrtion. CAS 5. Grph the solid bounded b the plne z nd the prboloid z 4 nd find its ect volume. (Use our CAS to do the grphing, to find the equtions of the boundr curves of the region of integrtion, nd to evlute the double integrl.).3 UBLE INTEGALS IN PLA CINATES Polr coordintes were introduced in Section 9.3. Suppose tht we wnt to evlute double integrl f, da, where is one of the regions shown in Figure. In either cse the description of in terms of rectngulr coordintes is rther complicted but is esil described using polr coordintes. + = + =4 + = FIGUE () =s(r, ) r, πd (b) =s(r, ) r, πd P(r, )=P(, ) ecll from Figure tht the polr coordintes r, of point re relted to the rectngulr coordintes, b the equtions r r r cos r sin FIGUE The regions in Figure re specil cses of polr rectngle r, r b, which is shown in Figure 3. In order to compute the double integrl f, da, where is polr rectngle, we divide the intervl, b into m subintervls r i, r i with lengths r i r i r i nd we divide the intervl, into n subintervls j, j with lengths j j j. Then the circles r r i nd the rs j divide the polr rectngle into the smll polr rectngles shown in Figure 4.

2 SECTIN.3 UBLE INTEGALS IN PLA CINATES 683 = j = j_ = r=b ij (r i *, j*) Î j r= å =å r=r i r=r i_ FIGUE 3 Polr rectngle FIGUE 4 ividing into polr subrectngles The center of the polr subrectngle hs polr coordintes ij r, r i r r i, j j r i * r i r i j* j j We compute the re of ij using the fct tht the re of sector of circle with rdius r nd centrl ngle is r. Subtrcting the res of two such sectors, ech of which hs centrl ngle, we find tht the re of is j ij A ij r i j r i j r i r i j r i r i r i r i j r i * r i j Although we hve defined the double integrl f, da in terms of ordinr rectngles, it cn be shown tht, for continuous functions f, we lws obtin the sme nswer using polr rectngles. The rectngulr coordintes of the center of ij re r i * cos j*, r i * sin j*, so tpicl iemnn sum is m n i j f r i * cos j*, r i * sin j* A ij m n i j f r i * cos j*, r i * sin j* r i * r i j If we write written s tr, rfr cos, r sin m n i j, then the iemnn sum in Eqution cn be tr i *, j* r i j which is iemnn sum for the double integrl b tr, dr d

3 684 CHAPTE MULTIPLE INTEGALS Therefore, we hve f, da lim m ri, j l m n i j lim m ri, j l m n i j f r i * cos j*, r i * sin j* A ij b f r cos, r sin rdrd tr i *, j* r i j b tr, dr d CHANGE T PLA CINATES IN A UBLE INTEGAL If f is continuous on polr rectngle given b r b,, where, then f, da b f r cos, r sin rdrd d da dr The formul in () ss tht we convert from rectngulr to polr coordintes in double integrl b writing r cos nd r sin, using the pproprite limits of integrtion for r nd, nd replcing da b rdrd. Be creful not to forget the dditionl fctor r on the right side of Formul. A clssicl method for remembering this is shown in Figure 5, where the infinitesiml polr rectngle cn be thought of s n ordinr rectngle with dimensions rd nd dr nd therefore hs re da rdrd. FIGUE 5 r r d EXAMPLE Evlute 3 4 da, where is the region in the upper hlfplne bounded b the circles nd 4. SLUTIN The region cn be described s,, 4 It is the hlf-ring shown in Figure (b), nd in polr coordintes it is given b r,. Therefore, b Formul, 3 4 da 3r cos 4r sin rdrd 3r cos 4r 3 sin dr d Here we use the trigonometric identit sin cos s discussed in Section 6.. [r 3 cos r 4 sin ] r r d 7 cos 5 sin d [7 cos 5 cos ] d 7 sin sin 5

4 SECTIN.3 UBLE INTEGALS IN PLA CINATES 685 z (,, ) V EXAMPLE Find the volume of the solid bounded b the plne z nd the prboloid z. SLUTIN If we put z in the eqution of the prboloid, we get. This mens tht the plne intersects the prboloid in the circle, so the solid lies under the prboloid nd bove the circulr disk given b [see Figures 6 nd ()]. In polr coordintes is given b r,. Since r, the volume is FIGUE 6 V da d r r 3 dr r r rdrd r 4 4 If we hd used rectngulr coordintes insted of polr coordintes, then we would hve obtined V da s s d d which is not es to evlute becuse it involves finding 3 d. = r=h ( ) Wht we hve done so fr cn be etended to the more complicted tpe of region shown in Figure 7. It s similr to the tpe II rectngulr regions considered in Section.. In fct, b combining Formul in this section with Formul..5, we obtin the following formul. å =å r=h ( ) 3 then If f is continuous on polr region of the form r, f, da FIGUE 7 =s(r, ) å, h ( ) r h ( )d In prticulr, tking f,, h, nd h h in this formul, we see tht the re of the region bounded b,, nd r h is A, h r h h f r cos, r sin rdrd h da h r h rdrd d h d nd this grees with Formul

5 686 CHAPTE MULTIPLE INTEGALS FIGUE 8 (-)@+ = (or r= cos ) z V EXAMPLE 3 Find the volume of the solid tht lies under the prboloid z, bove the -plne, nd inside the clinder. SLUTIN The solid lies bove the disk whose boundr circle hs eqution or, fter completing the squre, (See Figures 8 nd 9.) In polr coordintes we hve r nd r cos, so the boundr circle becomes r r cos, or r cos. Thus the disk is given b r,, r cos nd, b Formul 3, we hve V da 4 cos4 d 8 cos r rdrd cos 4 d 8 r 4 cos 4 cos d d FIGUE 9 cos cos 4 d 3 [ 3 sin 8 sin 4] 3.3 EXECISES 4 A region is shown. ecide whether to use polr coordintes or rectngulr coordintes nd write f, da s n iterted integrl, where f is n rbitrr continuous function on Sketch the region whose re is given b the integrl nd evlute the integrl rdrd 6. 4 cos rdrd 7 Evlute the given integrl b chnging to polr coordintes. 7. da, where is the disk with center the origin nd rdius 3 8. da, where is the region tht lies to the left of the -is between the circles nd 4 9. cos da, where is the region tht lies bove the -is within the circle 9. s4 da, where, 4,. rctn da, where, 4,. e da, where is the region in the first qudrnt enclosed b the circle 5

6 SECTIN.3 UBLE INTEGALS IN PLA CINATES Use polr coordintes to find the volume of the given solid. 3. Under the cone z s nd bove the disk 4 4. Below the prboloid z 8 nd bove the -plne 5. A sphere of rdius 6. Inside the sphere z 6 nd outside the clinder 4 7. Above the cone z s nd below the sphere z 8. Bounded b the prboloid z nd the plne z 7 in the first octnt 9. Inside both the clinder 4 nd the ellipsoid 4 4 z 64. () A clindricl drill with rdius r is used to bore hole through the center of sphere of rdius r. Find the volume of the ring-shped solid tht remins. (b) Epress the volume in prt () in terms of the height h of the ring. Notice tht the volume depends onl on h, not on or. Use double integrl to find the re of the region.. ne loop of the rose r cos 3. The region enclosed b the curve r 4 3 cos 3 6 Evlute the iterted integrl b converting to polr coordintes r r 3 3 s9 sin d d s dd s d d s s d d 7. A swimming pool is circulr with 4-ft dimeter. The depth is constnt long est-west lines nd increses linerl from ft t the south end to 7 ft t the north end. Find the volume of wter in the pool. 8. An griculturl sprinkler distributes wter in circulr pttern of rdius ft. It supplies wter to depth of e r feet per hour t distnce of r feet from the sprinkler. () Wht is the totl mount of wter supplied per hour to the region inside the circle of rdius centered t the sprinkler? (b) etermine n epression for the verge mount of wter per hour per squre foot supplied to the region inside the circle of rdius. 9. Use polr coordintes to combine the sum s s d d s into one double integrl. Then evlute the double integrl. 3. () We define the improper integrl (over the entire plne where is the disk with rdius nd center the origin. Show tht (b) An equivlent definition of the improper integrl in prt () is where S is the squre with vertices,. Use this to show tht (c) educe tht (d) B mking the chnge of vrible t s, show tht (This is fundmentl result for probbilit nd sttistics.) 3. Use the result of Eercise 3 prt (c) to evlute the following integrls. () I lim l e d e da e da e da e da lim l S e d e d e d s e d s d d s s4 d d e d d e da (b) s e d