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# Chapter 15.4-15.7 Notes MA 26100 - 300

Purdue

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This 8 page Class Notes was uploaded by Pawin Jiaravanon on Friday October 23, 2015. The Class Notes belongs to MA 26100 - 300 at Purdue University taught by David W Catlin in Fall 2015. Since its upload, it has received 30 views. For similar materials see Multivariate Calculus in Mathematics (M) at Purdue University.

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Date Created: 10/23/15

154 Double Integrals in Polar Coordinates Given a quotpolar rectangle how do we write it as a sum of many smaller polar rectangles alt9ltB aSer Subdivide anoltn1ltn2ltltnj1ltnjltltnmb Where nj nj1 An b am a90lt91lt92ltlt9k1lt9kltlt9nB Where 9k OH A9 B an AA T r12 Iii12 Ml27 AA T r Iii1 r rj1l Ml2n If we assume that rj1 z rj then AA z n rj Ar A9n on AA n An A9 Now imagine that the height of the box above the rectangle is fxy frj cos9k rj sin9k Then the volume of the solid region is V ij1frj cos9k rj sin9k nj An A9 Ifweletm ooand n9 IIR fxy dA uIBaIb fn cos9 n sin9 n dr d9 Eg Compute the integral L x2 y dA where R is the region in the first quadrant bounded by x2 y2 1 and x2 y2 4 gin21F x2 y r dr d6 gin21F n2 cosz6 r sin6 r dr dB This function is splittable X1 oIn21I2 r3 cosz6 dr d6 oInZ r44 n1n2 cosz6 dr d6 oInZ cos26 4 14 dr d6 154 oInZ 1 cos262 dr d6 158 rtZ 1516 sin26 low2 157t16 Y1 gin2112 r2 sin6 dr d6 oITE2l 133In1n2 sin6 dr d6 oInZ sin6 73 dr d6 73 cos6 low2 73 Therefore IR 151 16 73 Eg Find the volume of the solid bounded by z x2 y2 and 29 Volume ojznoP 9 r2 r dr d6 You can resolve the 6 integral right away 27 of 9r r2 dr 2T 9r22 r44o3 2n 992 814 814 2n 817t2 Now suppose D is bounded by D r9 a S 9 S B h19 S r S h29 UR fr cos6 r sin6 r dr d6 UR dA Area of D aIBh1eIh29 fr cos6 r sin6 r dr d6 In case it was unclear h1 is the lower bound of the function and h2 is the upper bound of the function Area OJB 12 h262 h162 A9 OJB r22 Irh1erh2e d9 Eg Use a double integral to find the area in a loop of r sin 49 A gin4 ojsinl m r dr d6 gin4 r22 ror quot 49 d6 gin4 sin24G2 d6 oft4 1 cos862 2 d6 rt16 sin8632 o 4 n16 Eg Find the volume of the shape between 2 Ix2 yz and x2 y2 z2 1 Z l1 x2 y2 x2 yz 1 x2 y2x2y2 9 1 r2 r29 2r2 1 9 r 12 Volume alkali 2 I1 r2 r r dr d6 Solve this for your answer 155 Density and Mass Consider a thin plate in the shape of a domain D The density of D may vary as a function of xy We define pxy lim AmAA 9 Am z pxy AA In the usual way we can approximate D by a union of rectangles RU each of area AA Ax Ay The mass of the rectangle RU is approximately pxiyj Ax Ay Therefore the total mass is approximately m 2L pxi Vj Ax Ay so m IID pxy dA Similarily if oxy charge density at xy then the total charge is Q In oxy dA And that concludes charge Forever Moments and center of mass In chapter 8 we defined Moment of D 2439 21 xi pxiy dA About the yaxis In x pxy dA My About the xaxis ID V pxy dA Mx x Mym V Mxm As before the center of mass of D is at x V where x Mym and V Mxm where m IID pxy dA Eg Find the center of mass of D if D is bounded by y x2 and y x 2 and pxy y Eg Find x V if D region in first quadrant bounded by x2 y2 1 and where density is p y m U0 V dA oInZ of r sin6 r dr d6 13 oInZ sin6 d6 13 cos6 H 2 13 My ID x y dA of of r cos6 r sin6 r dr d6 of sin6 cos64 d6 M sin2G2 H 2 18 Therefore Y Mym 1813 38 MX Id y2 dA of 011r sin62 r dr d6 Orr2 sin26 d6 18 Orr2 1 cos2e d6 68 sin2e16 lat2 7116 Therefore V Mxm Tc16 3 37116 Yx2andyx2intersectwhenx2x20rx2 x 209x 2x109x12 Mass ID y dA 112 XXIX2 y dy dx 4J2 y22 xxX2 X y dX m 112 X 222 X42 dx Solve this for mass Mvm MyIIDXvdA XI 1I2 m X v dv dX 1I2 X v22 yxxyx2 dX 1I2 X X 222 X52 dX My 21I2x x 22 x5 dx V Mxm M H y2 dA 1I2 m y2 dv dX 1I2 v33 yxxyx2 dX MX 13 1I2x 23 x6 dx 156 Surface Area The surface 2 fxy AAxifxAyk BijfyAyk Ax f x Ax a Compute area of parallelogram i j k fxAxAyi fyAxijAxAyk Ax 0 fX Ax My 0 fyAv Area fx2 fy2 1 2 Ax Ay Therefore area of S above D is A ID 1 fx2 fy212 dA Eg Find surface area of paraboloid z x2 y2 below 2 4 x2y2z49 r2 z x2 y2 fxy fX 2x fy 2y ID 1 4x2 4y212 dA SA ofquot of 1 4r212 r dr d6 2n80I214r2128rdr u14r2 du8rdr Tt4 1I17u12 du Tt4 23 u32 I117 7T5 1732 1 Eg Find the surface area of ax by cz 1 above the xyplane in the 1st octant Z 1 ax byc ofox ac ofoy bc A Ho 1 ac2 bc212 dx dy c2 a2 b212c ID dA c2 a2 b212c 12 a b c2 a2 b2122 a b c 157 Triple Integrals GivenaboxBxyz anSbcSySdeSsz We divide B into subboxes of width Ax Ay and A2 where Ax b al Ay d cm and A2 f en We define the triple Riemann Sum by 21 21 Zk1fxi V 2 Ax Ay Az where xi yj 2k is in the box Bijk xi1xi x Vj1yj x zk1zk The x s between the s to the left are cross products If we let m 9 M then HID fxyz dV limit of Riemann sum as lmn 9 This is called the triple integral of f over the box B To calculate it we use Fubini s Theorem IIID fXIyIZ eIf CId aIb fXyZ dX dZ Integrate first with respect to x then y then 2 or in any order that involves integrating one variable at a time Eg fBxyz 05x5215y5205251 CalculatejjjBxyde 0I21I20I1xy2zdzdydx olz 1I2 X V2 222 zoZ1 dV dX oI21I2Xv22 dv dX oI2Xv36 y1y2dx of2 MB x6 dx of2 x2 dx x24 Io2 1 A solid region E is said to be of type I if it lies between the graphs of 2 functions of x and y over a region D E xvz w D U1xv S 2 S uzlxv IIIE fXIyIZ IID IU1XYUZXIY fXIyIZ dz If D is a type II region then IIIE fXIyIZ CId Ih1Yh2Y IU1XYUZXIY fXItyIz dz dX Or if D is of type I then IIIE fXIyIZ aIb I81Xg2X IU1XYUZXIY fXIyIZ dz dX Eg Let D solid region bounded by 2 x2 2y 1 and z y 2 in the first octant Find the volume X22y1y29y1 x2 Volume 1 1 dV all 0139XXIXX2y1V2 1 dz dy dx oll OIL Z zxx2y12quot2 dv dX oll of 139 Y 2 X2 2V 1 dv dX oll oll39xx 1 X2 Y dv dX of y 1 x2 y22 IVOV139XX dx of 12 1 x22 dx 12 oil 1 2x2 x4 dx 12 x 2X3l3 X55 xox1 415 You are always trying to project the integral into a lower dimension Eg Let R triangular region in the xyplane between yx and y1 for 0 S x S 1 and let E solid region between the surfaces 2 2 and z 1 x2 for xy S R Evaluate ID 2139XX x 1 dz dy dx of J1 x 1 z 2139XX dy dx of XI1x 11 x2 2 dy dx of XI1x3 x2 3x 3 dy dx 0I139X3 x2 3x 3 y yxv391 dx 011x4 4x2 3 dx 2815 Eg Suppose that a tetrahedron is bounded by x 2y 22 2 x 0 y O and z 0 and the density is p 22 Calculate the mass m Setz09x2y2gty1 x2 0I20I139X2 1X2V2OIVOIX0I20I139X2 VX212dVdX 012 y xz 133 y0v1gtlt2 dx 012 1 x2 x2 133 x2 133 dx Various applications Vol HIE dV HIE 1 dV Center of mass iii where Myzm Y szm V and Mxym Ewhere MyZ II x p dV sz U y p dV and MXy 1 2 p dV

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