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# Calculus II MTH 252

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This 10 page Class Notes was uploaded by August Feeney on Monday October 19, 2015. The Class Notes belongs to MTH 252 at Portland Community College taught by Staff in Fall. Since its upload, it has received 49 views. For similar materials see /class/224648/mth-252-portland-community-college in Math at Portland Community College.

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Date Created: 10/19/15

Double integrals in rectangular coordinates an introductory example Suppose that we wanted to determine the volume of the solid shown in Figure 1 The lateral sides of this solid are formed by the planes x 0 y 0 and x y 6 The bottom of the solid lies on the xyplane and the top of the solid lies along the surface 2 xy 73y 25 In MTH 252 we estimated the area of planar regions by superimposing rectangles over the region and adding up the areas of these rectangles We can estimate the volume of our solid in a similar way we can superimpose rectangular parallelepipeds over the solid and add up the volumes of these In Figure 2 the yaxis of the region the solid projects onto the xyplane has been subdivided into 3 intervals of equal width Over each of these intervals a rectangle has been formed between the boundaries x0 and x67y using the values of x at the midpoint y to determine the position in which to draw the line segments determined by x 0 and x 67 y This is summarized in Table 1 Tabe1 k I x rectangular borders lmerva yquot x borders come from x 0 and x 6 7 ykquot 1 02 1 y0y2x0x5 2 24 L y2 y4 x0 x3 3 46 5 y4y6x0x1 Each of the rectangular regions in Table 1 has been further partitioned into 2 rectangles of equal length this partition happening parallel to the xaxis Rectangular parallelepipeds are now formed using the 6 rectangles in Figure 2 as bases The tops of the boxes are portions of the planes 2 fxxxy where 3611 is the midpoint of rectangle and fxyxy73y 25 The tops and bases of the boxes are shown in Figure 3 and a facsimile of the 6 boxes is shown in Figure 4 The top of the actual solid is shaded in Figure 5 Notice how each of the rectangular parallelepipeds has some volume that lies outside the actual solid while part of the actual solid does not lie inside any of the rectangular parallelepipeds The more balanced these two discrepancies the better our estimate Page 1 of4 Double integrals in rectangular coordinates an introductory example Let39s go ahead and use this scheme to estimate the volume of the solid Let39s use Figure 6 and Table 2 to help us come up with our estimate Table 1 mm ylnleml xfy fxy M F 1 1 DAY3 C0 078 23 90 1 15 l1 217 Ccle 31st gm mm C 039 Lg 1 v as 5 192 mm b Clan CW 52 any one C0rl 40 mg m mm 6 2 3 05 60 ngs 111 mm 1quot l L A xl i Figure6 13 XIX h C XI1 393 1 Vulvm 55 Srtxijji bnij a 553 335 L QIl Cams r 413m 5 4 1LS341U ll2 D Q1100 311 Eta 3 3 13 Page 2 of4 Double integrals in rectangular coordinates an introductory example 1 Let39s use Figure 6to help us set up a double integral of form fxydxdy that finds the exact R volume of our solid Let39s evaluate the integral by hand In this form the y Iimits of integration are points on the yaxis and the x Iimits of integration are boundary curves of form xLy and xRy f 3 37 4er VJ lin lg3 5 mayhem5 quot5 31 il 32 fat Jaly S 333 urva x j 5 x 1 If 36 Sci3 3 quotL J39quot quotquot 4 If 343 3 quot 4 L t J ahaquot 393 393 39 quotquotJ39 3 151 lit 4 376 l Lde our Lie x Buw MnY guau 3r 6 Figure 7 330 Page3of4 Double integrals in rectangular coordinates an introductory example Let39s use Figurei to help us set up a double integral of form fxydydx that finds the exact R volume of our solid Let39s evaluate the integral on our calculalors In this form the xIimits of integration are points on the x axis and the y Iimits of integration are boundary curves of form yBx and yTx c Lquotl la39uho 3 Xv371f Cum 0 3w Gquot j 76 ma Figure 8 A L XL D B of h BVJNDRLY 6 V 90 Page 4 of4 Double integrals in rectangular coordinates Z 5 Consider J 3 L ye 7 yZ dydx Let39s illustrate the region of integration on Figure 1 paylg close orienton to what are axis limit and what are boundary curve liarIs Let39s calculate the value of the integral by hand Let39s illustrate on Figure 2 what it is exactly that we just calculated of 393 Im Jrquot Curv a f B 39Hw Boa Joj owvc w 3935 l Figure1 53 W11 43439 1 l1 321 33 JV 3 2i 13 51920 11dova v 1 5 L 1 quot 6 3 7 4 3 J 239 50quot 37gt L3 ct H a Hem5 13 an S39sUp 43 um thu V9 31 3 Ej m l J V J kc UBIV 4 v quot Xy lut ARM Xy Plhc Page 1 of 6 Double integrals in rectangular coordinates a 367 Let39s use our calculator to find the exact value of J J xzydydx Let39s clearly label the 6 I region of integration on Figure 3 L quotBawgv cveuc j3 x 6 Hr h shdn f vl rg j 7 o L lL YL LS i lvl HL HL g 1quot Y 35 L Figure 3 a mix Let39s use our calculator to find the exact value of J J xZ ydydx after first reversing the 6 I order of integration Let39s clearly label the region of integration on Figure 4 Figure 4 3935 dim j xv Jy M If LQH Bth y21 Calai Q uL 36 d3L j 36quot zLj Page 2 of 6 2 The curves y 2x and y x7 are shown in figures 6 and 7 In Figure 5 These curves have been luTerully exTended up To The plane 22 We are going To seT up boTh of The double inTegruls ThaT calculaTe This volume We are also going To evaluaTe each double inTegral and hopefully geT The same value boTh Times I 5 9i 88C 1 a E x3 37 k2 Figure 6 J Ll no 4 golf 9 Jj ah Figure 7 ELLO Vs39l Page 3 of 6 Double integrals in rectangular coordinates On this page we are going to find both double integrals that calculate the volume of the tetrahedron formed in the first octant by the plane in Figure 8 We will then use technology to calculate each double integral and confirm that we get the same value t 39 gt 1 3v f J 1 L U 3i v j Figure 9 Figure 10 Page 4 of 6 Double integrals in rectangular coordinates The top of the solid outlined in Figure 11 is the plane 2 x The curved boundary in the xy plane is the function yth We are going to set up both double integrals that calculate the volume of the solid We are mercifully going to evaluate the double integrals using technology Figure 12 Figure 13 Page 5 of 6 Double integrals in rectangular coordinates 4 2 Finally we are going to calculate J I M e y2 dydx afler first reversing the order of inlegra 39on rho INT E Ul T 3quot BBC 3 1 Figure 14 1k quot x q j Que9 4 J L1 Sf C39SLJX J Figure 15 Page 6 of 6

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