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Calculus II

by: Ms. Ally Koelpin

Calculus II MATH 1352

Ms. Ally Koelpin
GPA 3.62


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This 21 page Class Notes was uploaded by Ms. Ally Koelpin on Thursday October 22, 2015. The Class Notes belongs to MATH 1352 at Texas Tech University taught by Staff in Fall. Since its upload, it has received 114 views. For similar materials see /class/226473/math-1352-texas-tech-university in Mathematics (M) at Texas Tech University.


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Date Created: 10/22/15
PROBLEM SET Practice Problems for Exam 1 Math 13527 Fall 2004 Oct 1 2004 ANSWERS Problem 1vLet R be the region bounded by the curves 1 y2 and y z A Find the volume of the solid generated by revolving the region R around the zaxis Answer Here s the picture of the region R 157 02 04 06 08 1 12 14 15 The intersections of the two curves are at 00 and 11 For revolution about the zaxis7 we use the method of washers The outside radius is the top curve I and the inside radius is the bottom curve 12 Thus the volume V is given by b V 7r outside radius2 7 inside radius2 dz a1 7r 12 7122 dz 0 1 7r 12 714dz 0 Wlxs3ir55l3 7r13715 7 27f 7 E Find the volume of the solid generated by revolving the region R around the y axis Answer First Solution Use the method of shells7 integrating in the zdirectionl The volume V is then given by V 27f bradiusheight dz 27r0 7 z2 dz 1 27r z2 7 z3 dz 0 27rz33 7 z44l DB7UM 7r6l Second Solution Use the method of washers7 integrating in the ydirectionl The outside radius is on the parabola Solving the equation y z2 for z gives us z W for the outside radius The inside radius is on the line7 and so the inside radius is z yr For the region R y ranges from 0 to ll Thus7 the volume is given by V77rOl m27y2wy W01y7y2dy f if 7r6l Problem 2 The base of a solid is the region in the zyplane bounded by the lines y 2z and z ll e cross sections of the solid perpendicular to the yaxis are squares Find the volume of the solid Answer The picture looks like this 2 y 227 157 E y 1 057 0 032 034 036 038 If we take a cross section at y the base of the square cross section is the segment labeled Zr To nd the length Z of this segment note that the lefthand endpoint is one the line y 21 so its zcoordinate is yZ The zcoordinate of the right hand endpoint is at z 1 so Z 1 7 yZi The area of the cross section at y is thus Ay l 7 yZ We want the cross sections to sweep out the solid so y should range from 0 to 2 Thus we have V 02Aydy 021y22dy To do this integral make the substitution u leg2i Then dy 72 dui When y 0 we have u 1 and when y 2 we have u 0 Thus we get V 10u272 du 1 2 quu 0 Problem 3 Find the area of one leaf of the fourleaf rose Whose polar equa tion is T sin 2 AnsweT The polar graph of the equation T sin2t9 looks like this Lets nd the area of the petal in the rst quadrant To see What range of 9 is required to draw this petal7 consider the rectangular coordinate graph of T sin2t97 Which looks like this Note that the 19axis is marked in units of 7r From this graph7 we see that the petal in the rst quadrant is drawn as 19 goes from 0 to 7r21 Thus the area A is given y 1 2 Aja f9 do 1 7r2 sin2219 d191 0 1 12 1 1 7 7 7 7 00546 d6 double angle formula 2 0 2 2 7r2 7 sin4 0 1 7r 1 7r 7 i 7 g slnlt4 gt 7 0 7 s1n0 7 I 7 8 Problem 41 Find the area of the region that is inside the cardioid 7 1 cos19 and outside the Circle 7 11 Answer 05 The rectangular plot of 7 l 0089 looks like this 2 0 in units ofvr From this we can see that if we start at 9 77r2 and proceed to 7r27 the part of the cardioid that is outside the Circle Will be drawn For a xed value of 9 in this range7 the outside curve Will be 7 l cost9 and the inside curve is 7 17 thus the area is given by 7r2 A iW21COS62 712dt9 7r2 1 2cost9 cos2t971dt9 2 42 1 7r2 7 2 cost9 cos2 d9 2 42 7r2 2 cost9 cos2 d9 cos is even 0 7r2 7r2 4 1 2 cost9 dt cos2t9 d9 0 0 For the rst integral in 411 we get 7r2 sin7r2 7 sin0 17 0 1 0 7r2 4 2 0 cost9 d6 sin6 For the second integral in 411 we have 7r2 7r2 cos2t9 d6 M d6 0 0 2 1 7r2 1 7r2 7 1dt97 cos2t9 dOdt 2 o 2 0 M2 0 SW 1 sin7r 7 sin0 070 1 l 4 mamama Na 43 Plugging the results of 412 and 413 into 4 17 we get 7r A 2 Problem 5 Find the arc length of the graph of 1 1 gzs 1171 on the interval from 12 This is Problem 9 on page 393 of the book Answer The formula for the arc length of a graph is b L x1f z2dzi In our case we have 1 7 1 fzz2ilz 2127Ei This gives us W 7 2 7 12 2H 7412 I4 i 161 We can then compute that 2 4 1 1 1fr71r 7 16z4 71 4 1 f 781416181 16x4 4z412 16x4 Thus7Weget 4z41 2 1 412 I m 1 WW So7 nally7 we have 2 L 1 wach 1 2 I2 4172 dz 1 Problem 6 A tank has the shape of a hemisphere see picture of radius 1 meter The tank is full of water7 which weights 9800 Nmgi How much work is required to pump all of the water to a point one meter above the top of the tank Answer Put in the y axis running upward from ground level Consider a slab of the water in the tank at height y and with thickness dyifor example7 the shaded slab in the picture belowi I II liiiiiil 1 05 We need to calculate the volume of this slab7 using the formula for the volume of a cylinder The height of the slab is dyi To gure out the radius of the slab7 take a crosssection of the picture7 as in the next gure 9 yams radius 1 slab at eumy X2y2t X75qm7yz 705 05 l The cross section of the tank is a semiCircle of the Circle of radius one The equation of the Circle of radius one centered at the origin is 12 y2 1 Solving this for 1 yields 1 i 1 7 y Thus as we see from the gure the radius of the slab is 1 7 y Thus the volume of the slab is given by 7rv17 92V dy 7 7r1 7 92 W Thus the weight of the slab is W1 7 y dy where w 9800 is the weight density of water The slab is a height y and we need to lift it to y 2 one meter above the top of the tank at y 1 so the distance to lift the slab is 2 7 yr Thus the total work for lifting this slab is 2 7 yW17 y dy 7 TWO 7 y17 92 W To get the total work for pumping out the tank we add the last quantity up over all the slabs in the tank Thus 1 Total Work awe y 1 y2 dy 0 1 w lt2iygtlt1iy2gtdy 0 1 Trw 272y27yy3dy 0 2 3 92 94 1 Wigy gy Zi0 7 137m 7 12 7 13749800 7 12 m 337353124 Joule 1 Problem 7 i The vertical cross sections of a tank are isosceles triangles7 point upwards The bottom of the tank is 4 feet across and it is 8 feet high If the tank is lled with water to a depth of 4 feet7 what is the total force on one end of the tank The weight density of water is w 62141b5 31 Answer Put in the coordinate system by letting the y axis run upwards from the center of the base of the tank Here s the picture 2153 43 7541 The light gray area is the part occupied by the water Take a strip at height y and thickness dy on the end of the tank shown as the dark gray rectangle Half the length of the strip is the distance labelled Z Next nd the equation of the line along the right side of the triangle This is goes through the points 20 and 08 so it s easy to calculate that the equation is y 8 7 41 If we solve this for z in terms of y we get I 2 7 y4i Thus Z 2 7 y4i From this we conclude that the length of the strip is 22 7 y4 4 7 yZi Since the thickness of the strip is dy we get Area of strip 4 7 yQ dyi The top of the water is at y 4 so we have Depth of strip 4 7 yr The pressure at this depth is Pressure on strip w4 7 The force on the strip is the pressure times the area so we have Force on strip w4 y4 yZ dy Now we have to add up on the force on all the little strips from y 0 to y 4 So we can calculate 4 Total force w4 y4 yZ dy 0 4 10 162y74y4ry22ldy 0 4 w 16 6yy22ldy 0 1 4 w 16y 7 3y2 7f 6 0 43 w 164 7 342 a 7 80 w 3 80 624 7 3 16641133 Problem 8 Let R be the region bounded by the curves y l 7 12 and y l 7 z A Find the area of Bi Answer The picture looks like this The points of intersection of the two curves are 01 and 10 The top curve is the parabola y 1 7 12 and the bottom curve is the line y1izso A0 17127171dz 1 171271zdz 0 011 7 12m 12 1 iiz Fig 71 1 Eig 71 g B Find the ycoordinate of the centroid of Bi 0 Answer The yCoordinate of the centroid g is given by Q lacA7 Where A is the area and LE moment With respect to the zaxis is7 in general7 given by IT bfzl2 7 MW dry In our case the top curve is 1 7 12 and the bottom curve is 91 1711 us IE 011712271712dz 1 1 E17212z47172z12dz 0 1 1 E 17212z47121712dz 0 1 1 7 2173z2z4ldz 0 2 1 111 2 5 ii 710 From this we get r i y A Use the Theorem of Pappus to nd the volume of the solid generated When the region R is revolved around the zaxis Answer Pappus7 Theorem says that the volume of the solid of revolution is V As Where A is the area of the region and s is the distance traveled by the centroid In our case7 the centroid travels around a Circle of radius g so we have 3 67f 2 2 iA s 7ry 7r5 5 Thus the volume is VAsl6ll 6 5 5 Problem 9 Find the following integrals A z cos2z dzl Answer Use the integration by parts formula 96 uvdzuv7uvdz Set u z7 1 os2z7 so we have u l 17 v 00821 dz isin2zl Plugging this into 967 we get zcos2z dz z sin2z 7 sin2z dz z sin2z 7 sinez dz 1 l l 2z s1n2z7 E 7i cos2z C l l 2z sin2z 1 cos2z Cl 12 1111 dz Use the integration by parts formula with Answer u lnz7 1 z2 then Plugging into gives 12 111 11 W 1311 1 1 2 igzlnz7 z dz 1 113 igzlnz7 zc 1 1 gzln 7 613 C1 MATH 1352 Review Final Exam Time 150 minutes Total points 100 o This review is intended to highlight the topics being examined7 to make sure that all sections have the same information regarding the nal 0 The included sections are 617 65 717 77 817 88 and 917 94 0 Speci c topic omitted Surface Area in both cartesian and polar coordinates in Section 64 0 Only improper integrals of the type ff mdz will be required to be solved7 though the students should be aware of other types of improper integrals o If an instructor has not covered the material for a problem7 he might assign an alternate problem during the exam 0 Calculators are to be used at the discretion of the instructor H Solve the following problems i Identify the improper integrals among if any give a short reason lt1 dz ii f dz iii f02quottan m dz ii Is the following sum correct give a short mathematical reason 17r7r2 iii Do you agree with the following computation give a short mathematical reason 1 1 1 7 1 2 7w 7 f71Pdif71 dm 71 71 239 2 Solve one of the following two problems a Find the z co ordinate of the centroid of the region in the rst quadrant bounded by the curves y 2 and y b Find the point of intersection of the curves r 4 sin0 and r 2 ii Find the area enclosed by one loop of r 2 sin2 0 3 Solve one of the following two problems a According to Coulomb s law of physics7 two similarly charged particles repel each other with a force inversely proportional to the square of the distance between them Suppose the force is 12 dynes when they are 5 cm apart How much work is done in moving one particle from a distance of 10 cm to a distance of 8 cm from the other b Suppose that a ball is dropped from a height h and it rises 7 of the distance it had previously fallen If it travels a total distance of 21 ft7 what is h 4 Solve one of the following two problems a Find the solution of the differential equation z4 3z3y 5 that passes through 5 1 What is the value of y for z 71 00 b Find the sum of the series g1 5 Find the following integrals i f 11 39 ww281339 H 12i4mi4 11 f warm dm iii fsin dz A 6 Find whether the sequence quot converges or diverges 00 7 Does the series 2 71W1 Mil converge absolutely7 conditionally or diverges 8 Test the following series for convergence State the name of the tests that you have used to arrive at your conclusion M 0 00 3 111 Z w k1 k 00 1 1v 2 k k k2 00 2k 9 Find the interval of convergence for the power series 2 k1 10 Find the rst four terms of the Taylor series of the function i at c 5 11 Givenuj73kandv7ijk i Find u 1 ii Find the angle between u and 12 iii Find the area of the parallelogram determined by u and 1 iv Find a unit vector normal to u and v v If w i 7 2j 2k nd the scalar triple product of 741 and w


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