a < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1


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1 Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the region under grph y = f(x) nd bove n intervl [, b] on the xxis, we slice up the intervl into n increments (smll prts) of width x = b n, with division points: < + x < +2 x < < +n x = b, nd we tke smple points x,..., x n, one in ech increment. This slices the re into increments A,..., A n, ech pproximtely rectngle: The re of n increment is pproximtely: A i (height) (width) = f(x i ) x, so the totl re is the Riemnn sum (see 4.,I): A = A i f(x i ) x. Tking the limit of very mny, very thin slices defines the integrl, which computes the exct re: A = lim n f(x i ) x = f(x) dx. This only gives pproximte numericl nswers to our question, though, unless we cn determine the limit, which is very difficult directly ( 4.,II). So fr, there is no relly surprising ide here: for thousnds of yers, mthemticins used similr methods to compute res, with very limited success. The genius of Newton shines in three ides which solve the problem, esily computing lmost ny re ( 4.3). The first ide is to mke re into function: A(x) = x f(t) dt is the re over vrible intervl t [, x]. Second, we find tht A (x) = f(x): mzingly, the re function is n ntiderivtive of f(x), becuse the rte of chnge of re is the height of the grph. Tht is, s x moves rightwrd, A(x) increses quickly or slowly depending on how high f(x) is: this is the First Fundmentl Theorem. Notes by Peter Mgyr
2 The third nd clinching ide is the lgebr of differentition. If we cn reverse the derivtive rules to find nother ntiderivtive, known function F (x) with F (x) = f(x), then the Second Fundmentl Theorem tells us tht the two ntiderivtives re the sme, except for dding some constnt: A(x) = x f(t) dt = F (x) + C. Since 0 = A() = F () + C, we must hve C = F (), so A(x) = F (x) F (), nd: A = A(b) = f(x) dx = F (b) F (). Tht is, if f(x) is the rte of chnge of F (x), so tht F i = f(x i ) x re the smll chnges (increments) in F (x), then the integrl is the totl chnge in F (x) over the intervl [, b]. These three brillint ides, working perfectly together, mke it esy to compute the res of most shpes, provided we re skillful enough t reversing derivtive rules to find indefinite integrls. exmple: Find the re of the leftmost region enclosed below the curve y = x cos(x 2 ) nd bove the line y = x. The left corner of the region is t x = 0. To find the right corner, we set x cos(x 2 ) = x nd get x(+ cos(x 2 )) = 0, whose smllest positive solution is x = π.77. To compute the re, we consider thin slices from the top curve to the bottom line, with height x cos(x 2 ) ( x) = x cos(x 2 ) + x, nd tke limit of Riemnn sums to get: A = π 0 ( x cos(x 2 ) + x ) dx = π 0 x cos(x 2 ) dx + π 0 x dx. We cn evlute the first term using the substitution u = x 2, du = 2x dx, so x cos(x 2 ) dx = 2 cos(x 2 ) 2x dx = 2 cos(u) du = 2 sin(u) = 2 sin(x2 ). Thus: [ A = 2 sin(x2 ) + 2 x2] x= π = 2 π.57. x=0 The indefinite integrl f(x) dx = F (x) + C denotes the generl ntiderivtive function, lso clled the primitive function. F (b) F () is number. By contrst, the definite integrl f(x) =
3 Tht looks bout right from the picture. Not result you could get from elementry geometry!
4 Solids of revolution. The integrl is very powerful tool to compute the totl of mny smll prts, such s the thin slices which fill up n re. A similr method llows us to compute volumes. We strt with solid of revolution which is formed by rotting region in the plne round the x xis, sweeping out kind of brrel shpe. The totl volume is the sum of n disk slices t smple points x,..., x n in [, b], ech with rdius f(x i ) nd thickness x = b n. The volume of ech slice is pproximtely: V i (circle re) (thickness) = πf(x i ) 2 x, nd tking the limit s n gives the exct volume s n integrl: V = lim n πf(x i ) 2 x = πf(x) 2 dx. Once we express our nswer s n integrl, we no longer consider its geometric motivtion: finding n ntiderivtive nd determining the vlue is purely lgebric problem. This nlysis should not be surprising: we expect our nswer to converge to n integrl s soon s we cn slice up the problem into smll increments.
5 exmple: Find the volume of the trumpet solid obtined by rotting round the xxis the region defined by: 0 y x nd x 2. This is the solid of revoltion of the curve y = x over the intervl x [, b]. Applying our formul: V = 2 π ( ) 2 2 x dx = πx 2 dx = [ πx ] x=2 x= = ( π2 ) ( π ) = π 2. exmple: Find the volume of the solid obtined by rotting the region defined by x 2 y 2 nd x, round the verticl xis x =. V = The setup is different from the xxis rottion, so we must repet our slice nlysis. Since we rotte round the verticl line x =, we should tke horizontl slices positioned by their height y, nd we should write our region s lying next to the intervl y [, 2] between the curves x = nd x = y. Our slices re thin horizontl disks t smple heights y,..., y n in the intervl y [, 2]. The thickness of ech disk is y. The rdius t height y i is the horizontl distnce from the xis x = to the curve x = y; this distnce is y i. The volume of ech disk is V i π( y i ) 2 y, nd: 2 π( y ) 2 dy = 2 π(y 2 y+) dy = [ )] y=2 ( π( 2 y2 4 3 y3/2 +y = π 23 y= ).
6 Solids with specified cross sections. Let us consider solid rther like the sil of the mnofwr jellyfish. It hs bse outlined by the circle x 2 + y 2 =, nd its verticl cross section over ech line x = c is n isosceles right tringle, with height equl to hlf its bse. The slices re defined over x [, ], with ( ech slice )( t x = ) x i hving thickness x nd re 2 (bse) (height) = 2 2 x 2 i x 2 i. Thus: V = 2 ( 2 )( ) x 2 x 2 dx = x 2 dx = [x 3 x3] x= x= = 4 3. Method of slice nlysis to compute size. We hve now seen severl cses where we compute the size or bulk of geometric objects. Let S denote ny mesure of the size of n object: length, re, volume, mss, etc.. Cut the object into slices whose position is determined by some vrible x [, b]. 2. Mrk off the intervl [, b] into n increments of width x = b n, ech with smple point x i. This splits the object into n slices, nd summing up their sizes gives the totl size: S = n S i. 3. Becuse the slice t x i is so thin, we cn find good pproximtion of its size by some simple formul of the form S i f(x i ) x. 4. Tking n nd x 0, the pproximtions become exct: S = f(x i ) x = f(x) dx. lim n 5. Hving expressed S = f(x) dx, we evlute this integrl by lgebric or numericl techniques. chllenge problem: Consider the solid of revolution round the xxis of the curve y = f(x) over the intervl x [, b]. Explin the following formul for the surfce re of this solid: Hint: S = ( x) 2 + ( y) 2 = 2πf(x) +(f (x)) 2 dx. + ( y x )2 x + ( dy dx )2 x.
We divide the interval [a, b] into subintervals of equal length x = b a n
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