Volumes by Slicing; Disks and Washers

Transcription

1 Volumes y Slicing; Disks nd Wshers Definite integrls hve een used to produce res. We now show tht they cn e used to produce volumes s well. Suppose tht solid etends long the -is nd is ounded on the left nd right y plnes perpendiculr to the -is t = nd =. A() - is

2 Suppose tht t ech point etween nd, we know the cross sectionl re A() of the solid. Theorem. If V is the volume of the solid, then V= Ad () A() - is

3 A ( ) i - is The reson for this is tht if we tke the Riemnn sum n A ( ) i i= 1 we get n pproimtion tht oviously tends to the volume of the solid s n tends to infinity.

4 The infinitesiml point of view Whenever we wnt to represent some physicl quntity s definite integrl, we cn show tht the Riemnn sum which converges to tht integrl is n pproimtion to the physicl quntity. This is wht we did ove. However it is often esier to set up the integrl y using n imprecise ut intuitive ide clled infinitesimls. An infinitesiml quntity is considered to e so smll tht it is essentilly zero. Then the integrl is used to dd the infinite numer of infinitesimls together. [There is ctully wy to mke infinitesimls mthemticlly rel, ut it is too complicted to go into here]

5 y - is dv = A()d d - is To return to the volume emple for minute, we cn regrd the red solid slice shown ove s n infinitesiml volume dv. Since the thickness, d, is regrded s essentilly 0, we think of dv s rectngulr solid whose volume is the cross sectionl re times the thickness, tht is A()d - we don t hve to think of the thin sides of the solid dv s hving slnted edges. We then dd them ll together y integrtion to get the entire volume V= Ad ()

6 We cn proceed in similr wy if volume is divided into verticl infinitesiml volumes. y - is d dy c dv = A(y)dy - is d V= c Aydy ( )

7 Proceed s follows: 1. Set up the infinitesiml quntity to e integrted first. It will involve some independent infinitesiml d, dy, etc depending on the physicl prolem we re ttempting to solve, nd vrious other quntities.. The independent infinitesiml d, dy, etc. determines the integrl, so ll other quntities must e epressed in terms of the corresponding vrile. Thus if we hve d, the we re going to do n integrl, nd every other quntity in the integrl must e epressed in terms of. On the other hnd, if we hve dy, then every other prt of the quntity we re integrting must e epressed in terms of y, nd so on.

8 Emple. A solid hs the unit circle + y = 1 s its se. Its cross sections, tken perpendiculr to the se nd to the - is re squres. Find the volume. y (, y) 1 1 is Solution. First we look t the se. For ech etween 1 nd 1 we see tht the se hs width y. Thus the re of the squre cross section t is 4y, the thickness is d, nd so the infinitesiml volume is dv = 4y d. Thus

9 (1 V= ) 4 16 y d= d= = Epress everything in the integrnd in terms of, since the epression involves d. Compute the typicl infinitesiml volume dv, then set up the integrl tht sums ll these infinitesimls.

10 Volumes of Revolution One of the most common methods of generting solid odies is y rotting n re out some is of revolution. For emple, suppose tht we look t the re under function f() from = to =, s shown elow, nd rotte this re 360 degrees out the is, generting the solid shown. y y=f() y

11 y y=f() y It is esy to see tht the cross sections of this solid re disks with thickness d nd ses tht re circles of rdius y. Thus the infinitesiml volume dv of the cross section is dv= πy d= π f() d Therefore, the totl volume is V= π f() d

12 Emple. Find the volume of the solid tht results when the region enclosed y the given curves is revolved out the - is. y= cos( ), y = 0, = π/4, = π/. Solution. The re is shown to the right. When revolved round the is, it mkes solid tht resemles the nose of ullet. y = f() The volume is π π π π 1 V= () ( cos( )) π f d= π d= π cos( d ) = π sin( ) = π 1 π π π π 4

13 If we now rotte the re etween two curves out the - is, we will in generl get cross sections tht re wshers, rther thn disks. y y=f() y y=g() The re of the cross section is π f() πg () = π f() g () so dv= π f() g () d. Thus the volume is V= f() g () π d.

14 Emple. Find the volume of the solid tht results when the region shown elow is rotted out the - is. Solution: Here the region is etween y = nd y = -, which meet t = 1. Thus nd dv= π d π = + d V= π 4 d π 4 38π + = + = ( )

15 It is just s likely tht the solid is formed y rotting n re out the y-is. Then the disks nd wshers re horizontl with thickness dy, nd they must e dded together y y-integrl. Of course this mens tht ll quntities ssocited with the re must e epressed in terms of y, nd d V= c Aydy ( )

16 Emple. Find the volume of the solid tht results when the region shown elow is rotted out the y - is. V Solution: Here the region is etween = 1/y nd =. Then y vries from 1/ to. Thus 1 1 dv= π dy= π 4 dy y y nd V= π 4 dy π 4y π 8 4 9π = + = 1 + = 1 y y

17 Emple. Find the volume of the solid tht results when the region enclosed y the given curves is revolved out the - is. y= e, y= 0, = 0, = 1 Solution. The re is shown to the right. The volume is π () 4 π 4 V= π f d π e d π e d e π 1 e 4 = = = =

18 Emple. Find the volume of the solid tht results when the region enclosed y the given curves is revolved out the y- is. y=, = y Solution. The re is shown to the right. The cross section is horizontl wsher whose thickness is dy, so every thing must e epressed in terms of y. y In terms of y, the left nd right curves re respectively = y nd = y. The infinitesiml volume is therefore dv y y = π dy π y y 4 = dy ( ) ( )

19 To find the limits, we solve the eqution y = y, or y 4 = y The only rel solutions of this re y = 0 nd y = 1, which must therefore e the limits. So The volume is y y V= π y y dy π y y dy π 3 π = = = the solid

20 Emple. Find the volume of the solid tht results when the sme region is revolved out the - is. Solution. The re is shown to the right. The cross section is horizontl wsher whose thickness is d, so every thing must e epressed in terms of. The curves re y=, y=. The infinitesiml volume is therefore dv = π d π 4 = d Thus the volume is the sme for this solid, nmely 3 π. 10 ( ) ( )

21 Prolem. Find the volume of the solid generted when the region enclosed y y=, y= 6 nd y = 0 is revolved out the - is. y= y= 6 To do this prolem, we need to rek the region into two prts s shown, then rotte oth prts nd dd the results. The curves meet when = 6 or 0= = = ( 4)( 9) tht is t = 4 nd = 9.

22 y= y= ( ) V1 d d π = π = π = = 8π u V 8 ( 6 ) d u ( du) u du π = π π = π = π = = Thus the totl integrl volume is 8 + 8π = 3π. 3 3