Volumes as integrals of crosssections (Sect. 6.1) Volumes as integrals of crosssections (Sect. 6.1)


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1 Volumes s integrls of crosssections (ect. 6.1) Te volume of simple regions in spce Volumes integrting crosssections: Te generl cse. Certin regions wit oles. Volumes s integrls of crosssections (ect. 6.1) Te volume of simple regions in spce Volumes integrting crosssections: Te generl cse. Certin regions wit oles.
2 Te volume of simple regions in spce Remrk: Volumes of simple regions in spce re es to compute. Find te volume of rectngulr bo wit sides, b, nd c. c Te re of n oriontl crosssection is A = b. (Constnt long te verticl direction.) Te volume of te bo is b V = Ac. Remrk: We ve dded up long te verticl direction ec oriontl crosssection. V = c A() d = A c d V = Ac. Volumes s integrls of crosssections (ect. 6.1) Te volume of simple regions in spce Volumes integrting crosssections: Te generl cse. Certin regions wit oles.
3 Volumes integrting crosssections: Generl cse. Remrk: Tis interprettion of te clcultion bove is good definition of volume for rbitrr sped regions in spce. Definition A crosssection of 3dimensionl region in spce is te dimensionl intersection of plne wit te region. R Remrk: Like in te lst cse bove, te re of cross section is function of te direction norml to te crosssection. Volumes integrting crosssections: Generl cse. Definition Te volume of region in spce wit integrble crosssection re A() for, b] is given b V = b A() d. Find te volume of prmid wit squre bse side nd eigt. (1) Coose simple crosssections. Here, oriontl crosssections. () Coose coordinte sstem were te crosssection res ve simple epressions.
4 Volumes integrting crosssections: Generl cse. Find te volume of prmid wit squre bse side nd eigt. = m + b A() = () ] We must find nd invert / / () / () = m + b. = () = b, = (/) = m + m =. V = () = + () = ( ). ] ( ) ( ) 3 d = ] V = Te volume of simple regions in spce Find te volume of te tetredr: c b Te re of te tringulr cross section bove is A() = 1 ()(), () = b +, () = c b + c. A() = 1 V = c b b ( ) ( ( b) c ) ( b) = c b b b ( b). ( b) d = c ( b) 3 b 3 b ] V = 1 6 bc.
5 Volumes s integrls of crosssections (ect. 6.1) Te volume of simple regions in spce Volumes integrting crosssections: Te generl cse. Certin regions wit oles. Regions of revolution Definition A region of revolution is 3dimensionl region in spce obtined b rotting plne region bout n is in tt plne. f() f() A() = pi f()] Remrk: Te crosssections of region of revolution re disks: A = πr. In te emple, R() = f (). Terefore, A() = π f () ].
6 Regions of revolution Teorem Te volume of region of revolution defined b rotting te function vlues = f () for, b] bout te is is b f () ] d. Find te volume of spere of rdius R b rotting lf circle wit te sme rdius. f() = R R R R R f ()] d = π (R ) d R ( R R ] 3 R ]) R 3 R R 3 3 R3] V = 4 3 πr3. Regions of revolution Remrk: Te is of rottion could be n is in spce. Find te volume of cone wit bse of rdius R nd eigt. = (/R) + () We coose s te rottion is. () ] d. R () ince = R ( ), we get R ( ) d = π R ( ) 3 3 ] = π R 3 3 We conclude tt V = 1 3 πr.
7 Regions of revolution Find te volume of te region bounded b = + 4 for, ] wen it is rotted bout te is. 4 = 4 () () 4 () ] d. ince = + 4, we get 4 ( ) ( + 4) d = π 4 + (4) 4] = π ] We conclude tt V = 8π. Volumes s integrls of crosssections (ect. 6.1) Te volume of simple regions in spce Volumes integrting crosssections: Te generl cse. Certin regions wit oles.
8 Certin regions wit oles (wser metod) Definition A wser region is region of revolution wit ole, were te eterior nd interior surfces re obtined b rotting te function vlues = f et () nd = f int () long te is. ketc te wser region bounded b = + 4 nd = + 4 for, ], rotted bout te is. = 4 = = +4 = +4 = Certin regions wit oles (wser metod) Teorem Te volume of wser region bout te is wit eterior nd interior surfces generted b = f et () nd = f int () for, b], respectivel, is given b V = V et V int b ( fet () ] fint () ] ) d. Find te volume of te wser region in te previous emple. V = V p V c, V p = 8π, V c = 1 3 π( )(4) = 16 3 π. V = ( 1 1 3) 16π V = 8 3 π.
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