1. Use calculus to derive the formula for the area of a parallelogram of base b and height. y f(x)=mx+b

Transcription

1 Area and Volume Problems. Use calculus to derive te formula for te area of a parallelogram of base b and eigt. y f(x)=mxb b g(x)=mx Te area of te parallelogram is given by te integral of te dierence of te function f(x) describing te upper side minus te function g(x) describing te lower side: (f(x) ; g(x))dx = ((mx b) ; (mx))dx =. Derive te formula for te area of a triangle of base b and eigt. x bdx= b: Glue two suc triangles togeter to create a parallelogram wit base b and eigt. b Te area of te triangle is alf te area of te parallelogram namely, b. Here is anoter way: Te area of te triangle is te integral of te areas of te orizontal strips.

2 y b y x Eac strip as lengt y b and tickness dy. So te area is y bdy = b y # = b: 3. Derive te formula for te area of a trapezoid wit bases b b and eigt. Suppose b b. Cut te trapezoid into a parallelogram wit base b and eigt, and a triangle wit base b ; b and eigt. b b b-b Ten te area is b (b ; b ) = (b b ): Here is anoter way: Te area of te trapezoid is te integral of te areas of te orizontal strips.

3 y b b y x Eac strip as lengt b y (b ; b ) and tickness dy. So te area is (b y (b ; b ))dy = # "b y y (b ; b ) 4. Use calculus to derive te area of a circle of radius. = b (b ; b )= (b b ): To calculate te area using polar coordinates, remember tat te small unit of area is rdrd. r d θ r dr So te total area is rdrd = r d = d = = : 3

4 Here is anoter way: Construct a triangle wit base being a tiny piece of te circumference, dc, joined to te center of te circle. r dc So te eigt of te triangle is, and its area is da = dc. Integrating tis over te circumference of te circle gives te total area A = C = = : 5. Take te derivative of te formula for te area of a circle of radius. Wat formula do you get? Wy does tis appen? If A = ten da d =, wic is te formula for te circumference of te circle. Tis appens because wen te radius of te circle is increased by, ten te area increases by A C, were C is te circumference. C Δ 4

5 So A C, and in te limit, da d = C. 6. Use calculus to prove Cavalieri's principle: If two -dimensional regions ave te property tat all lines parallel to some xed line tat meet and do so in line segments aving equal lengts, wose endpoints are te boundary points of and,ten and ave te same area. (See te illustration on page 79 of te textbook.) Suppose as upper boundary f (x) andlower boundary g (x), and as upper boundary f (x) and lower boundary g (x). y f(x) g(x) f(x) g(x) a b x Ten and area( )= area( )= b a b a (f (x) ; g (x))dx (f (x) ; g (x))dx But by assumption, f (x) ; g (x) =f (x) ; g (x) for all x. So te two integrals are te same. 5

6 7. Two regions are similar if tere is a one-to-one correspondence between te points of and te points of, and a constant k, suc tat for every pair of points A B in and corresponding pair of points A B in,weave AB = ka B.If and are similar -dimensional regions, wat is te relationsip between te areas of and?(ifyou are not certain, try tis out wit simple sapes rst.) area( )=k area( ). Tis works for squares, so extends to areas of oter gures by calculus. 8. Use calculus to derive te formula for te volume of a pyramid wose base is a polygon of area B and wose eigt is. Calculate volumes by slices. B z Te orizontal slice at eigt z as area ( z ) B (by te previous problem) and tickness dz. So te volume is ( z ) Bdz= 3 z3 B = 3 B: 9. Use calculus to derive te formula for te volume of a cylinder of radius and eigt. We can use cylindrical coordinates. For a small cange in r,, and z, te unit of 6

7 rdrddz = volume is rdrddz. So te total volume is = = ri d dz i = dz = [ z] = : dz d dz. Derive te formula of te surface area of a cylinder of radius and eigt. Te top and te bottom of te cylinder ave te same area namely,. Cut o te top and bottom and unroll te side of te cylinder into a rectangle. Te eigt ofte rectangle is, and its lengt is te circumference of te cylinder,. So its area is. So te total area of te cylinder is =( ):. Use calculus to derive te formula for te volume of a cone of radius and eigt. Tis works exactly te same as te volume of a pyramid, were now B =.Sote total volume is 3 B = 3.. Derive te formula for te surface area of a cone of radius and eigt. z dc a ρ y x 7

8 Let a be te \slant eigt" of te cone, so a = p. Te base of te cone as area, so all we need to nd is te lateral area. One way istomake a tin triangle wit base being dc, a small piece of te circumference of te base, and eigt being a. Ten te triangle as area adc.integrating tis over te circumference gives te area ac = p. So te total area is p = ( p ): Anoter way to get te lateral surface area is to use sperical coordinates. Every point on te cone as te same value of. For a small cange in and, te unit of surface area is dd. So te total lateral area is a a dd = a = ia a a d d = a i = a = p : Here is yet anoter way to get te lateral surface area. Cut te side of te cone and unfold it to get a sector of a circle. a Te radius of te circle is a, te slant eigt of te original cone. Te arc lengt of te sector is C, te circumference of te base of te original cone. Tis is te fraction C a 8 C

9 of te total circumference of te circle. So te area of te sector is te same fraction of te area of te circle namely C a a = Ca = p. 3. Use calculus to derive te formula for te volume of a spere of radius. Using sperical coordinates, for a small cange in,, and te unit volume is sin ddd. So te total volume is sin ddd = = = = = 3 3 i = : 3 3 sin i 3 3 cos i 3 3 d 3 3 sin dd ; d d d d 4. Use calculus to derive te formula for te surface area of a spere of radius. Using sperical coordinates, all te points on te spere ave constant =. For a small cange in and, te unit area is d times sin d. So te total surface area is sin dd = ; cos i d = d = i =4 : 5. Take te derivative of te formula for te volume of a spere of radius. Wat formula do you get? Wy does tis appen? If V = ten dv d =4, wic is te formula for te surface area of te spere. Tis appens because wen te radius of te spere is increased by, ten te volume increases by V S, were S is te surface area. So V limit, dv d = S. S, and in te 6. Use calculus to prove Cavalieri's principle: If two 3-dimensional regions ave te property tat all planes parallel to some xed plane tat meet and do so in plane sections aving equal areas, wose boundaries lie in te boundaries of and,ten and ave te same volume. (See te illustration on page 79 of te textbook.) Suppose and are cut by orizontal planes, te lowest plane being at eigt z = a, and te igest plane being at eigt z = b. Let A (z) denote te area of te 9

10 cross-section of determined by te plane at eigt z, anda (z) denote te area of te cross-section of determined by te plane at eigt z. Ten and volume( )= volume( )= b a b a A (z)dz A (z)dx: But by assumption, A (z) =A (z) for all z. Sotetwo integrals are te same. 7. If and are similar 3-dimensional regions, wat is te relationsip between te volumes of and?between te surface areas of and? Suppose tere is a one-to-one correspondence between te points of and te points of, and a constant k, suc tat for every pair of points A B in and corresponding pair of points A B in,weave AB = ka B.Tenvolume( )=k 3 volume( ) and surface area( )=k surface area( ). 8. ecall tat if tat if OAB is a triangle in E wit coordinates O =( ), A =(x y ), B =(x y ), ten its area is x y x y Actually, it turns out tat if te angle 6 ;! OA to OB, ;! AOB is counterclockwise, measured from ray B O ten we can drop te absolute value above, and te area is in fact just equal to te determinant x y x y Now suppose tat you ave a convex polygon A A ::: A n in te plane E (listing its vertices in counterclockwise order) tat contains te origin in its interior. Suppose A

11 tat te vertex A i as coordinates (x i y i ), i = ::: n. Prove tat te area of te polygon is x y x y x y x 3 y 3 x n; y n; x n y n x n y n x y Actually, it turns out tat tis formula works even if te polygon is not convex, and even it does not contain te origin in its interior, as long as it does not cross itself. Use tis formula to calculate te areas of te following polygons: A4=(-,4) A3=(,3) A4=(,3)! O=(,) A=(4,) O=(,) A=(3,-) A3=(6,) A5=(-3,-) A=(,-) A5=(,-3) A=(5,-3) We can see wy te formula works by using te rst polygon and constructing triangles from every side of te polygon to te origin O. A4=(-,4) A3=(,3) O A=(4,) A5=(-3,-) A=(,-) Te triangle wit vertices ( ), (x i y i ), and (x i y i ) as area x i y i x i y i

12 Ten te total area of te polygon equals te sum of te areas of te triangles. Applying te formula to te two polygons, we get areas (8 6 ) = 8 and (496; 6 5) = Suppose a polygon (not necessarily convex, but not self-crossing) as te property tat eac of its vertices as integer coordinates. Let B be te number of integer points on its boundary, and I be te number of integer points in its interior. Consider some examples and conjecture a formula for te area of te polygon in terms of B and I. Tis is called Pick's Formula. Use tis formula to calculate te areas of te following polygons: Wit a bit of experimentation you can guess te formula A = B I ; : Applying tis formula to te two polygons, we get areas () 3 ; = 8 and () 4 ; = 9.. ecall tat S is te spere of radius centered at te origin. Suppose tat A and A are any two points on S tat are exactly opposite eac oter. Connect A and A wit two great alf-circles tat meet at A (and also at A ) at an angle A<8. Tese two alf-circles determine a region of te spere called a lune. Explain wy te area of tis lune is A. (See page 463 of te textbook.) 9 Te lune consists of a fraction of te total surface area S of te spere. Tis fraction is te same as te fraction of te total 36 A angle tat A consists of namely,.so 36 te area of te lune is A 36 S = A 36 4 = A 9 :

13 . Complete te discussion on page 464 of te textbook to obtain a formula for te area of a sperical triangle. eferring to te book for te denitions of regions K, I, II,andIII,we can rst see tat K [ I is te lune determined by anglea, K [ II is te lune determined by angle C, and K [ III is te lune determined by angle B. So for areas we ave Adding tese togeter gives K I = A 9 K II = C 9 K III = B 9 3K I II III = (A B C): 9 We can also observe tatk [ I [ II [ III occupies exactly alf te total area of te spere tis is most easily seen by examining a pysical model. So K I II III = 4 =: Subtracting tis equation from te previous one yields K = (A B C) ; 9 K = (A B C) ; 8 (A B C ; 8): = 8 So te total area is proportional to te amount bywic te angle sum of te triangle exceeds 8.. Suppose a spere is inscribed in a cylinder wose radius is te radius of te spere, and wose eigt is te diameter of te spere. 3

14 Let be any region on te spere. Project tis region out orizontally onto te cylinder to get te region on te cylinder. Use calculus to prove tat and ave te same areas. Tis is one way tomake maps of te eart tat accurately represent areas of countries, altoug te sapes of te countries are distorted. 4

15 r r sin φ dθ r d θ r sin φ dθ r sin φ d φ rdφ r d φ φ Using sperical coordinates, te unit of area on te spere is (r sin d) (rd)= r sin dd, wereas te projection of tis unit of area onto te cylinder gives a unit of area (rd) (r sin d)=r sin dd. Since tese units of area are identical in size, integrating tem over regions and will yield te same result. 3. Assume tat OABC is a tetraedron wit coordinates O =( ), A =(x y z ), 5

16 B =(x y z ), C =(x 3 y 3 z 3 ). Give two examples to sow tattevolume is 6 x y z x y z x 3 y 3 z 3 Explain wy tevolume of te tetraedron is t of te parallelepiped determined by 6 A, B, andc. Tere are many examples to coose from. Here is a simple one: A =( ), B = ( ), C =( 3). According to te formula, te volume sould be 6 3 =: Te base triangle 4OAB as area, and te eigt of te tetraedron is 3, so its volume is baseeigt= 3=. 3 3 D F C E B G O You can cut te parallelepiped into six tetraedra as sown above: OABC, ABCD, ACDE, ABDG, ADEG, DEF G. Eac of tese tetraedra ave te same volume, as can be demonstrated bysowing tat certain pairs of tem ave bases of common area, and equal eigts. For example, OABC as base 4OBC and eigt OA, wileabcd as base 4BCD and eigt OA. So tese two tetraedra ave te same volume. 4. Try to nd a formula analogous to tat of problem (8) tat applies to a convex 3- dimensional polyedron, all of wose faces are triangles. Wat could you do if not all faces are triangles? 6 A

17 Suppose tat te vertices of te polyedron ave coordinates P i =(x i y i z i ), i = ::: n. If te structure contains te origin in its interior,ten you can construct a tetraedron from eac triangular face to te origin and add up te volumes of te tetraedra. For any given face wit vertices P j P k P`, tevolume of te tetraedron will be 6 xj yj zj x k y k z k x` y` z` Adding up tese volumes, one per face, gives te volume of te entire polyedron. It turns out tat if te vertices of eac face are listed in te determinant in counterclockwise order as viewed from te outside, ten te determinant will be automatically positive, and te absolute value can be dropped. Once tis is done, it can also be proved tat te formula works even if te polyedron is not convex, and even if does not contain te origin in its interior, so long as it is a closed structure and not selfintersecting. If not all faces are triangles, te nontriangular faces can be be rst subdivided into triangles by drawing some diagonals. Ten te formula can be applied. Example: an octaedron. Te vertices are A =( ), B =(; ), C =( ), D =( ; ), E =( ), F =( ;). Te faces, wit vertices listed in counterclockwise order, are ACE, CBE, BDE, DAE, CAF, BCF, DBF, ADF. E : D C B A F 7

18 Te volume is V = 6 B@ ; ; ; ; ; ; = 6 ()= 4 3 : ; ; ; ; ; ; Example: acube. Te vertices are A = ( ), B =(; ), C =( ; ), D =(; ; ), E =( ;), F =(; ;), G =( ; ;), H =(; ; ;). Cut eac of te six squares wit a diagonal. Te triangles, wit vertices listed in counterclockwise order, are GAC, AGE, EBA, BEF, DGC, GDH, BHD, HBF, BCA, CBD, FGH, GF E. D B CA C A H F G E 8

19 Te volume is V = 6 B@ ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; = 6 ( )=8: ; ; ; ; ; ; CA ; ; ; ; ; 9