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by: Jack Magann

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# Calculus 1 Chapter 7 notes MTH 161

Marketplace > University of Miami > Mathematics (M) > MTH 161 > Calculus 1 Chapter 7 notes
Jack Magann
UM
GPA 3.865
161 Calculus 1
Dr. Beverly

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## About this Document

This covers chapter 7.1, 7.2, 7.3 Topics include: area between curves, volumes, and the shell method
COURSE
161 Calculus 1
PROF.
Dr. Beverly
TYPE
Class Notes
PAGES
9
WORDS
CONCEPTS
Calculus 1 University of Miami
KARMA
25 ?

## Popular in Mathematics (M)

This 9 page Class Notes was uploaded by Jack Magann on Sunday February 15, 2015. The Class Notes belongs to MTH 161 at University of Miami taught by Dr. Beverly in Fall2014. Since its upload, it has received 76 views. For similar materials see 161 Calculus 1 in Mathematics (M) at University of Miami.

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Date Created: 02/15/15
Chapter 71 Distance area Between Curves Previously only found distance between curves and the XaXis One method for nding the distance between two curves is by using rectangles The negative area is subtracted from positive to nd the total area Represented as 2311 x gx2lt JA l l it A an Typiml rectangle A 11me gltx2ltgt1Ax A Lfmx gltxgt1dx so eX 1 Find the are of the region enclosed by the given curves y x2 y 2x x2 y x2 is a parabola that opens up and y 2x x2 is a parabola that opens down Find the interval enclosed by nding when the two equations are equal intersect x2 2x x2 2xx 1 0 x 1x 0 intersect at 00 11 Now the area formula looks like A f 01 f x gx dx Need to plug in a value in the interval into both equations to see which is below nd needs to be subtracted Now A f2x x2 x2 dx f2x 2x2 dx 2 fx x2 dx l 2 l 3 1 2 2 1 3 1 0 3 For these questions the area will be positive eX2 yx1 and9 x2 fromx2x 1 The rst is a line while the second is a parabola that opens down The parabola in this case is the higher function on the intervals so Af2 x29 x 1dxf2 x2 x8dx x3 x28x 10 4 69 These problems can also be looked at from right to left rather than curves that go up to down 117 These curves include sideways parabolas which are not functions unless they are put in terms of y rather than X W This are is represented as d A L fy 9ydy Here the left function is subtracted from the right eX 1 Find the area enclosed by the equations y x 1 and y2 2x 6 If we set these equations in terms of X then we would end up with three equations due to the square root that would result from the second equation Put them in terms of y to end up with two xy1 and x y2 3 Findtheinterval iy2 3y1 yz y 420 y 2y4 Plug in a number in this interval into both equations to see which is further right The resulting integral is then f 312 y 2 dy where Fy y3 y2 2y flt y2y 2gtdy lt 8gt 4gt118 1 IA 39 5k 1 These are examples of equations you would use top down integration Here is one for right left integration P I rquot Chapter 7 2 Volumes It is already known that for a sphere V 37W and A 47T I 2 It is clear that the Area is the derivative of the volume or the volume is the antiderivative of the area It is the same with the area and circumference of a 2D circle This suggests a relationship between area and volume There are volume formulas for speci c shapes like cubes and cones There is no such formula for more irregular shapes One such irregular shape is graphically represented To nd the area we can take many slices of areas and add them up This would then look like 394 4 l 1 1 1 d I 39I I quotI I 39I 1 u h Looking at a sphere Ii We can see that as Ax the width of the sections approaches 0 the volume is more accurately found The area can then be represented as V 1imlxsoo 21Ax2 Ax If Axdx Where Ax represents length and height and dx represents width Antiderivative should then be 3dimensional and equal the shape s volume ex 1 Find the volume of the solid obtained by rotating the region under the curve y xE from 0 to 1 about the X axis First try to visualize or draw the area you are looking for The equation will make a semicircle and then by rotating it it will become a half circle Here the area of each small section is represented as a circle so Ax m 2 radius is then the distance from the edge to the Xaxis which is xE Write the integral f 01 7T 2dx 2 7T fol xdx where F x x2 nfolxdx 7Tx12 0 2 1 2 ex 2 Find the volume of the solid obtained by rotating the region between the curves y x3 find the interval W 0 y0 Altygt m2 now no and y8x0 Put in terms ofy x W xamp x22 8 E 3 E V nfo ysdy where FCY E313 V n 83 0 967139 08 ex 3 y x and y x2 rotated about the Xaxis Intervalx2 x 0 x 0x 1 Rotating this area will create a washer type shape must subtract area created by the lower function from the area created by upper function N 39n I39I a x p N A A1 A2 nr12 7Tr22 7Tx2 7Tx22 1 1 1 2 V nfo x2 x4dx 7T 13 g15 0 1 5 2 14 2 ex 4 y 3x and y x about the XaX1s Interval 2x2 x2 19 4 0 11 Find which equation makes the outer shape y 19 4 x2 is the upper one 2 2 14 5 A 7tx2 7T x2 9 9 196 V 7tf11 Ex2 Ear dx where Fx 2 x x3 xS 81 9 81 81 27 405 Due to the symmetry of the shape the interval can be halved and a 2 put in front V27Tf01 Ex2 5 6x4 dx 271 E 13 xS 0 1232 81 9 81 81 27 405 405 Chapter 72 Find the area enclosed by the given curves y 1 2v y 1 x Area between two curves problem Interval 1 12x1x 02x 2x3 f0161 2amp 1 xdx 316 i az 0 2 Find the volume of the solid obtained by rotating the region bound by the curves 3 2 2x or x 312 and x 2 2y about the yaxis Volume problem Interval 1 51221 110 114 1 2 1 1 Ayn2y27I y2 n4y27IZy4n4y2Zy4 4 1 4 1 512 42 4 4345 2 nf0y 4y dy n 3 200 o 12 Chapter73 Previous chapter way slices 1 Find the volume of the solid obtained by rotating the region bound by the curves 2 2 3 3V 07 x y and y x or x 3 about the yaxis mowlt92 Interval O to 9 9 2 2 2 9 1 2187 HOW dyn lt92 E9gt5 o 10 Shell Method Rather than adding up individual slices you find a shell and accumulate the areas of each shell Think of it as f in Where you would make a rectangle you would rotate it to make a washer shape gt For infinite points a bunch of washers will add up to the area of the rotated figure A v w 4 This is shown in the bottom picture if the above graph where rotated about the yaXis The area for these washers is Area circumference x height gtlt change in X or dX The formula for the volume then is b b V 2 27txfxdx 2n xfxdx This method can be used rather then setting an equation with multiple x 5 equal to x For example y 2x2 x3 about the yaxis 2 2 y 3V and y 9 about the yaxis Interval O to 9 1 x2 E x3 A 271x 32 3 dx 27T32 dx V2 J93 x3d 2 69 94 0 21877t 0x 9x 5 36 10 All these notes are from a Calculus 1 class run by Dr on Beverly at the University of Miami A majority of the problems used by the teacher and copied in these notes are taken from the textbook Essential Calculus Second edition By James Stewart Citation Stewart James Essential Calculus 2nd ed Np Cangage Learning 2013 Print

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