pancakes. A typical pancake also appears in the sketch above. The pancake at height x (which is the fraction x of the total height of the cone) has
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1 Volumes One can epress volumes of regions in tree dimensions as integrals using te same strateg as we used to epress areas of regions in two dimensions as integrals approimate te region b a union of small, simple pieces wose volume we can compute and ten ten take te limit as te piece sie tends to ero. Often tis results in multivariable integrals tat are beond our present scope. But tere are some special cases in wic tis leads to integrals tat we can andle. Here are some eamples. Eample 1 (Cone) Find te volume of te circular cone of eigt and radius r. Solution. Here is a sketc of te cone. We ave called te vertical ais, just so tat we end up wit a d integral. To compute its volume, we slice it up into tin oriontal pancakes. A tpical pancake also appears in te sketc above. Te pancake at eigt (wic is te fraction of te total eigt of te cone) as r tickness 1 d and radius r and ence cross sectional area π ( r)2 and ence volume π ( r) 2 d As runs from to, te total volume is ( ) 2 π r πr 2 d = 2 2 d = πr2 2 [ ] r = 1 πr2 d Eample 1 1 Of course wat we reall do is pick a natural number n, slice te cone into n pancakes eac of tickness = n and ten take te limit n. In te integral is replaced b d. Knowing tat tis is wat is going to appen, we are just skipping a few steps. c Joel Feldman All rigts reserved. 1 Februar 4, 215
2 Eample 2 (Spere) Find te volume of te spere of radius r. Solution. We ll find te volume of te part of te spere in te first octant 2, sketced below. Tenwe ll multipl b8. To computetevolume, we sliceitupinto tinvertical pancakes = r 2 (, r 2 2, ) Eac pancake is one quarter of a circular disk. Te pancake a distance from te plane is sown in te sketc above. Te radius of tat pancake is te distance from te dot sown in te figure to te ais, i.e. te coordinate of te dot. To get te coordinates of te dot, observe tat it lies te plane and so as coordinate ero and tat it also lies on te spere, so tat its coordinates obe = r 2. So te pancake at distance from te plane as tickness d and radius r 2 2 cross sectional area 1 4 π( r 2 2) 2 and ence volume π 4( r 2 2) d As runs from to r, te total volume of te part of te spere in te first octant is r π r 4( 2 2) d = π ] r [r 2 = πr and te total volume of te wole spere is eigt times tat, wic is 4 πr, as epected. Eample 2 2 Te first octant is te set of all points (,,) wit, and. Yet again wat we reall do is pick a natural number n, slice te octant of te spere into n pancakes eac of tickness = r n and ten take te limit n. In te integral is replaced b d. Knowing tat tis is wat is going to appen, we are again just skipping a few steps. c Joel Feldman All rigts reserved. 2 Februar 4, 215
3 Eample (Pramid) Find te volume of te pramid wic as eigt and wose base is a square of side b. Solution. Here is a sketc of te part of te pramid tat is in te first octant. To compute its volume, we slice it up into tin oriontal square pancakes. A tpical pancake also (,,) (, b,) 2 ( b /2,,) (, b /2,) appears in te sketc above. Te pancake at eigt is te fraction of te distance from te peak of te pramid to its base. So it is a square of side b. As a ceck, note tat wen = te pancake as side b =, and wen = te pancake as side b = b. So te pancake as cross-sectional area ( b)2 and tickness d and ence volume ( b)2 d. Te volume of te wole pramid is ( b ) 2d = b 2 2 = b2 2 = b2 2 = 1 b2 [ t ( ) 2 d t 2 ( dt) ] wit t =, dt = d ] = [ b2 2 Eample Eample 4 (Napkin Ring) Suppose ou make two napkin rings b drilling oles wit different diameters troug two wooden balls one of radius r and te oter of radius R, wit R > r. Te ole diameters are cosen so tat bot napkin rings ave eigt 2. Wic ring as more wood in it? c Joel Feldman All rigts reserved. Februar 4, 215
4 2r 2 2R Solution. We ll compute te volume of te napkin ring wit radius R. Ten, to get te volume of te napkin ring of radius r, we just need to replace R b r. To compute te volume of te napkin ring of radius R, we slice it up into tin oriontal pancakes. Here is a sketc of te part of te napkin ring in te first octant sowing a tpical pancake. Te coordinates (, R 2 2,) (, R 2 2,) of te two points marked in te plane of tat figure are found b remembering tat te equation of te spere is = R 2. As te two points are in te plane, = for tem so tat = R 2 2. In particular, at te top of te napkin ring = so tat = R 2 2. Te pancake at eigt, sown in te sketc, is a waser a circular disk wit a circular ole cut in its center. Te outer radius of te waser is R 2 2 and te inner radius of te waser is R 2 2. So te cross sectional area of te waser is π ( R 2 2) 2 π ( R2 2) 2 = π( 2 2 ) Recall tat we pick sliced te napkin ring into tin oriontal pancakes. Te pancake at eigt as tickness d and cross sectional area π( 2 2 ) and ence volume π( 2 2 )d. So te total volume of wood in te napkin ring of radisu R is ] π( 2 2 )d = π [ 2 [ 2 = π 2 ) ] = ( 4π Tis is independent of R. Te napkin ring of radius r contains precisel te same volume of wood as te napkin ring of radius R! Eample 4 c Joel Feldman All rigts reserved. 4 Februar 4, 215
5 Eample 5 (Notc) A 45 notc is cut to te centre of a clindrical log aving radius 2 cm. One plane face of te notc is perpendicular to te ais of te log. See te sketc below. Wat volume of wood was removed? Solution 1. Slice te notc into rectangles as in te figure below. Suppose tat te base of te notc is in te plane. Ten te circular part of te boundar of te base of te notc as equation = 2 2. (We re putting te origin of te plane at te centre of te circle.) If our coordinate sstem is suc tat is constant on eac slice, ten te slice as widt 2 = and eigt (since te upper face of te notc is at 45 to te base). So te slice as cross sectional area and te volume is V = d Make te cange of variables u = 2 2 2, du = 2d. u /2 V = u( du) = 2 /2 = = 16, 2 Solution 2. Suppose tat te base of te notc is in te plane wit te skinn edge along te ais. Slice te notc into triangles parallel to te ais as in te figure below. c Joel Feldman All rigts reserved. 5 Februar 4, 215
6 Ten te circular part of te boundar of te base of te notc as equation = 2 2. Our coordinate sstem is suc tat is constant on eac slice, so tat te slice as bot base and eigt = (since te upper face of te notc is at 45 to te base). So te slice as cross sectional area 1 2( 22 2)2 and te volume is V = (2 2 2 )d = 2 (2 2 2 )d = ] [2 2 2 = 2 2 = 16, Eample 5 c Joel Feldman All rigts reserved. 6 Februar 4, 215
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