Using Definite Integrals


 Darren Hubbard
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1 Chpter 6 Using Definite Integrls 6. Using Definite Integrls to Find Are nd Length Motivting Questions In this section, we strive to understnd the ides generted by the following importnt questions: How cn we use definite integrls to mesure the re between two curves? How do we decide whether to integrte with respect to x or with respect to y when we try to find the re of region? How cn definite integrl be used to mesure the length of curve? Introduction Erly on in our work with the definite integrl, we lerned tht if we hve nonnegtive velocity function, v, for n object moving long n xis, the re under the velocity function between nd b tells us the distnce the object trveled on tht time intervl. Moreover, bsed on the definition of the definite integrl, tht re is given precisely by b x = b. v(t) dt. Indeed, for ny nonnegtive function f on n intervl [, b], we know tht f (x) dx mesures the re bounded by the curve nd the xxis between x = nd Through our upcoming work in the present section nd chpter, we will explore how definite integrls cn be used to represent vriety of different physiclly importnt properties. In Preview Activity 6., we begin this investigtion by seeing how single definite integrl my be used to represent the re between two curves. 333
2 USING DEFINITE INTEGRALS TO FIND AREA AND LENGTH Preview Activity 6.. Consider the functions given by f (x) = 5 (x ) 2 nd g(x) = 4 x. () Use lgebr to find the points where the grphs of f nd g intersect. (b) Sketch n ccurte grph of f nd g on the xes provided, lbeling the curves by nme nd the intersection points with ordered pirs. (c) Find nd evlute exctly n integrl expression tht represents the re between y = f (x) nd the xxis on the intervl between the intersection points of f nd g. (d) Find nd evlute exctly n integrl expression tht represents the re between y = g(x) nd the xxis on the intervl between the intersection points of f nd g. (e) Wht is the exct re between f nd g between their intersection points? Why? Figure 6.: Axes for plotting f nd g in Preview Activity 6. The Are Between Two Curves Through Preview Activity 6., we encounter nturl wy to think bout the re between two curves: the re between the curves is the re beneth the upper curve minus the re below the lower curve. For the functions f (x) = (x ) 2 + nd g(x) = x + 2, shown in Figure 6.2, we see tht the upper curve is g(x) = x + 2, nd tht the grphs intersect t (, 2) nd (3, 5). Note tht we cn find these intersection points by solving the system of equtions given by y = (x ) 2 + nd y = x + 2 through substitution: substituting x + 2 for y in the first eqution yields x + 2 = (x ) 2 +, so x + 2 = x 2 2x + +, nd thus x 2 3x = x(x 3) =,
3 6.. USING DEFINITE INTEGRALS TO FIND AREA AND LENGTH g 4 2 g f 4 2 g f Figure 6.2: The res bounded by the functions f (x) = (x ) 2 + nd g(x) = x + 2 on the intervl [, 3]. from which it follows tht x = or x = 3. Using y = x + 2, we find the corresponding yvlues of the intersection points. On the intervl [, 3], the re beneth g is 3 while the re under f on the sme intervl is 3 Thus, the re between the curves is A = 3 (x + 2) dx (x + 2) dx = 2 2, [(x ) 2 + ] dx = 6. 3 [(x ) 2 + ] dx = = 9 2. (6.) A slightly different perspective is lso helpful here: if we tke the region between two curves nd slice it up into thin verticl rectngles (in the sme spirit s we originlly sliced the region between single curve nd the xxis in Section 4.2), then we see tht the height of typicl rectngle is given by the difference between the two functions. For exmple, for the rectngle shown t left in Figure 6.3, we see tht the rectngle s height is g(x) f (x), while its width cn be viewed s x, nd thus the re of the rectngle is A rect = (g(x) f (x)) x.
4 USING DEFINITE INTEGRALS TO FIND AREA AND LENGTH 6 4 g g(x) f (x) 2 g f x f x 2 3 Figure 6.3: The re bounded by the functions f (x) = (x ) 2 + nd g(x) = x + 2 on the intervl [, 3]. The re between the two curves on [, 3] is thus pproximted by the Riemnn sum A n (g(x i ) f (x i )) x, i= nd then s we let n, it follows tht the re is given by the single definite integrl A = 3 (g(x) f (x)) dx. (6.2) In mny pplictions of the definite integrl, we will find it helpful to think of representtive slice nd how the definite integrl my be used to dd these slices to find the exct vlue of desired quntity. Here, the integrl essentilly sums the res of thin rectngles. Finlly, whether we think of the re between two curves s the difference between the re bounded by the individul curves (s in (6.)) or s the limit of Riemnn sum tht dds the res of thin rectngles between the curves (s in (6.2)), these two results re the sme, since the difference of two integrls is the integrl of the difference: 3 3 g(x) dx f (x) dx = 3 (g(x) f (x)) dx. Moreover, our work so fr in this section exemplifies the following generl principle. If two curves y = g(x) nd y = f (x) intersect t (, g()) nd (b, g(b)), nd for ll x such tht x b, g(x) f (x), then the re between the curves is A = (g(x) f (x)) dx.
5 6.. USING DEFINITE INTEGRALS TO FIND AREA AND LENGTH 337 Activity 6.. In ech of the following problems, our gol is to determine the re of the region described. For ech region, (i) determine the intersection points of the curves, (ii) sketch the region whose re is being found, (iii) drw nd lbel representtive slice, nd (iv) stte the re of the representtive slice. Then, stte definite integrl whose vlue is the exct re of the region, nd evlute the integrl to find the numeric vlue of the region s re. () The finite region bounded by y = x nd y = 4 x. (b) The finite region bounded by y = 2 2x 2 nd y = x 2 8. (c) The re bounded by the yxis, f (x) = cos(x), nd g(x) = sin(x), where we consider the region formed by the first positive vlue of x for which f nd g intersect. (d) The finite regions between the curves y = x 3 x nd y = x 2. Finding Are with Horizontl Slices At times, the shpe of geometric region my dictte tht we need to use horizontl rectngulr slices, rther thn verticl ones. For instnce, consider the region bounded by the prbol x = y 2 nd the line y = x, pictured in Figure 6.4. First, we observe tht by solving the second eqution for x nd writing x = y +, we cn eliminte vrible through substitution nd find tht y + = y 2, nd hence the curves intersect where y 2 y 2 =. Thus, we find y = or y = 2, so the intersection points of the two curves re (, ) nd (3, 2). We see tht if we ttempt to use verticl rectngles to slice up the re, t certin vlues of x (specificlly from x = to x =, s seen in the center grph of Figure 6.4), the curves tht govern the top nd bottom of the rectngle re one nd the sme. This suggests, s shown in the rightmost grph in the figure, tht we try using horizontl rectngles s wy to think bout the re of the region. For such horizontl rectngle, note tht its width depends on y, the height t which the rectngle is constructed. In prticulr, t height y between y = nd y = 2, the right end of representtive rectngle is determined by the line, x = y +, while the left end of the rectngle is determined by the prbol, x = y 2, nd the thickness of the rectngle is y. Therefore, the re of the rectngle is A rect = [(y + ) (y 2 )] y, from which it follows tht the re between the two curves on the yintervl [, 2] is
6 USING DEFINITE INTEGRALS TO FIND AREA AND LENGTH 2 x = y 2 2 x = y 2 2 x = y 2 y y = x y = x x = y + Figure 6.4: The re bounded by the functions x = y 2 nd y = x (t left), with the region sliced verticlly (center) nd horizontlly (t right). pproximted by the Riemnn sum A n [(y i + ) (yi 2 )] y. i= Tking the limit of the Riemnn sum, it follows tht the re of the region is A = y=2 y= [(y + ) (y 2 )] dy. (6.3) We emphsize tht we re integrting with respect to y; this is dictted by the fct tht we chose to use horizontl rectngles whose widths depend on y nd whose thickness is denoted y. It is strightforwrd exercise to evlute the integrl in Eqution (6.3) nd find tht A = 9 2. Just s with the use of verticl rectngles of thickness x, we hve generl principle for finding the re between two curves, which we stte s follows. If two curves x = g(y) nd x = f (y) intersect t (g(c), c) nd (g(d), d), nd for ll y such tht c y d, g(y) f (y), then the re between the curves is Activity 6.2. A = y=d y=c (g(y) f (y)) dy. In ech of the following problems, our gol is to determine the re of the region described. For ech region, (i) determine the intersection points of the curves, (ii) sketch
7 6.. USING DEFINITE INTEGRALS TO FIND AREA AND LENGTH 339 the region whose re is being found, (iii) drw nd lbel representtive slice, nd (iv) stte the re of the representtive slice. Then, stte definite integrl whose vlue is the exct re of the region, nd evlute the integrl to find the numeric vlue of the region s re. Note well: At the step where you drw representtive slice, you need to mke choice bout whether to slice verticlly or horizontlly. () The finite region bounded by x = y 2 nd x = 6 2y 2. (b) The finite region bounded by x = y 2 nd x = 2 2y 2. (c) The re bounded by the xxis, y = x 2, nd y = 2 x. (d) The finite regions between the curves x = y 2 2y nd y = x. Finding the length of curve In ddition to being ble to use definite integrls to find the res of certin geometric regions, we cn lso use the definite integrl to find the length of portion of curve. We use the sme fundmentl principle: we tke curve whose length we cnnot esily find, nd slice it up into smll pieces whose lengths we cn esily pproximte. In prticulr, we tke given curve nd subdivide it into smll pproximting line segments, s shown t left in Figure 6.5. To see how we find such definite integrl tht mesures rc length on y f h y x x x x 2 x 3 x L slice Figure 6.5: At left, continuous function y = f (x) whose length we seek on the intervl = x to b = x 3. At right, close up view of portion of the curve. the curve y = f (x) from x = to x = b, we think bout the portion of length, L slice, tht lies long the curve on smll intervl of length x, nd estimte the vlue of Lslice using wellchosen tringle. In prticulr, if we consider the right tringle with legs prllel to the coordinte xes nd hypotenuse connecting two points on the curve, s seen t right
8 USING DEFINITE INTEGRALS TO FIND AREA AND LENGTH in Figure 6.5, we see tht the length, h, of the hypotenuse pproximtes the length, L slice, of the curve between the two selected points. Thus, L slice h = ( x) 2 + ( y) 2. By lgebriclly rerrnging the expression for the length of the hypotenuse, we see how definite integrl cn be used to compute the length of curve. In prticulr, observe tht by removing fctor of ( x) 2, we find tht L slice = = ( x) 2 + ( y) 2 ( ) ( x) 2 + ( y)2 ( x) 2 + ( y)2 ( x) 2 x. Furthermore, s n nd x, it follows tht y tht L slice + f (x) 2 x. x dy dx = f (x). Thus, we cn sy Tking Riemnn sum of ll of these slices nd letting n, we rrive t the following fct. Given differentible function f on n intervl [, b], the totl rc length, L, long the curve y = f (x) from x = to x = b is given by Activity 6.3. L = + f (x) 2 dx. Ech of the following questions somehow involves the rc length long curve. () Use the definition nd pproprite computtionl technology to determine the rc length long y = x 2 from x = to x =. (b) Find the rc length of y = 4 x 2 on the intervl 2 x 2. Find this vlue in two different wys: () by using definite integrl, nd (b) by using fmilir property of the curve. (c) Determine the rc length of y = xe 3x on the intervl [, ]. (d) Will the integrls tht rise clculting rc length typiclly be ones tht we cn evlute exctly using the First FTC, or ones tht we need to pproximte? Why?
9 6.. USING DEFINITE INTEGRALS TO FIND AREA AND LENGTH 34 (e) A moving prticle is trveling long the curve given by y = f (x) =.x 2 +, nd does so t constnt rte of 7 cm/sec, where both x nd y re mesured in cm (tht is, the curve y = f (x) is the pth long which the object ctully trvels; the curve is not position function ). Find the position of the prticle when t = 4 sec, ssuming tht when t =, the prticle s loction is (, f ()). Summry In this section, we encountered the following importnt ides: To find the re between two curves, we think bout slicing the region into thin rectngles. If, for instnce, the re of typicl rectngle on the intervl x = to x = b is given by A rect = (g(x) f (x)) x, then the exct re of the region is given by the definite integrl A = (g(x) f (x)) dx. The shpe of the region usully dicttes whether we should use verticl rectngles of thickness x or horizontl rectngles of thickness y. We desire to hve the height of the rectngle governed by the difference between two curves: if those curves re best thought of s functions of y, we use horizontl rectngles, wheres if those curves re best viewed s functions of x, we use verticl rectngles. The rc length, L, long the curve y = f (x) from x = to x = b is given by L = + f (x) 2 dx. Exercises. Find the exct re of ech described region. () The finite region between the curves x = y(y 2) nd x = (y )(y 3). (b) The region between the sine nd cosine functions on the intervl [ π 4, 3π 4 ]. (c) The finite region between x = y 2 y 2 nd y = 2x. (d) The finite region between y = mx nd y = x 2, where m is positive constnt. 2. Let f (x) = x 2 nd g(x) = x 2, where is n unknown positive rel number. For wht vlue(s) of is the re between the curves f nd g equl to 2? 3. Let f (x) = 2 x 2. Recll tht the verge vlue of ny continuous function f on n intervl [, b] is given by b f (x) dx.
10 USING DEFINITE INTEGRALS TO FIND AREA AND LENGTH () Find the verge vlue of f (x) = 2 x 2 on the intervl [, 2]. Cll this vlue r. (b) Sketch grph of y = f (x) nd y = r. Find their intersection point(s). (c) Show tht on the intervl [, 2], the mount of re tht lies below y = f (x) nd bove y = r is equl to the mount of re tht lies below y = r nd bove y = f (x). (d) Will the result of (c) be true for ny continuous function nd its verge vlue on ny intervl? Why?
11 6.2. USING DEFINITE INTEGRALS TO FIND VOLUME Using Definite Integrls to Find Volume Motivting Questions In this section, we strive to understnd the ides generted by the following importnt questions: How cn we use definite integrl to find the volume of threedimensionl solid of revolution tht results from revolving twodimensionl region bout prticulr xis? In wht circumstnces do we integrte with respect to y insted of integrting with respect to x? Wht djustments do we need to mke if we revolve bout line other thn the x or yxis? Introduction 2 y x 3 x x Figure 6.6: A right circulr cylinder. Just s we cn use definite integrls to dd the res of rectngulr slices to find the exct re tht lies between two curves, we cn lso employ integrls to determine the volume of certin regions tht hve crosssections of prticulr consistent shpe. As very elementry exmple, consider cylinder of rdius 2 nd height 3, s pictured in Figure 6.6. While we know tht we cn compute the re of ny circulr cylinder by the formul V = πr 2 h, if we think bout slicing the cylinder into thin pieces, we see tht ech is cylinder of rdius r = 2 nd height (thickness) x. Hence, the volume of representtive slice is V slice = π 2 2 x.
12 USING DEFINITE INTEGRALS TO FIND VOLUME Letting x nd using definite integrl to dd the volumes of the slices, we find tht V = 3 π 2 2 dx. Moreover, since 3 4π dx = 2π, we hve found tht the volume of the cylinder is 2π. The principl problem of interest in our upcoming work will be to find the volume of certin solids whose crosssections re ll thin cylinders (or wshers) nd to do so by using definite integrl. To tht end, we first consider nother fmilir shpe in Preview Activity 6.2: circulr cone. Preview Activity 6.2. Consider circulr cone of rdius 3 nd height 5, which we view horizontlly s pictured in Figure 6.7. Our gol in this ctivity is to use definite integrl to determine the volume of the cone. () Find formul for the liner function y = f (x) tht is pictured in Figure 6.7. (b) For the representtive slice of thickness x tht is locted horizontlly t loction x (somewhere between x = nd x = 5), wht is the rdius of the representtive slice? Note tht the rdius depends on the vlue of x. (c) Wht is the volume of the representtive slice you found in (b)? (d) Wht definite integrl will sum the volumes of the thin slices cross the full horizontl spn of the cone? Wht is the exct vlue of this definite integrl? (e) Compre the result of your work in (d) to the volume of the cone tht comes from using the formul V cone = 3 πr2 h. 3 y x 5 x x Figure 6.7: The circulr cone described in Preview Activity 6.2
13 6.2. USING DEFINITE INTEGRALS TO FIND VOLUME 345 The Volume of Solid of Revolution A solid of revolution is three dimensionl solid tht cn be generted by revolving one or more curves round fixed xis. For exmple, we cn think of circulr cylinder s solid of revolution: in Figure 6.6, this could be ccomplished by revolving the line segment from (, 2) to (3, 2) bout the xxis. Likewise, the circulr cone in Figure 6.7 is the solid of revolution generted by revolving the portion of the line y = x from x = to x = 5 bout the xxis. It is prticulrly importnt to notice in ny solid of revolution tht if we slice the solid perpendiculr to the xis of revolution, the resulting crosssection is circulr. We consider two exmples to highlight some of the nturl issues tht rise in determining the volume of solid of revolution. Exmple 6.. Find the volume of the solid of revolution generted when the region R bounded by y = 4 x 2 nd the xxis is revolved bout the xxis. Solution. First, we observe tht y = 4 x 2 intersects the xxis t the points ( 2, ) nd (2, ). When we tke the region R tht lies between the curve nd the xxis on this intervl nd revolve it bout the xxis, we get the threedimensionl solid pictured in Figure 6.8. y = 4 x 2 y x x Figure 6.8: The solid of revolution in Exmple 6.. Tking representtive slice of the solid locted t vlue x tht lies between x = 2 nd x = 2, we see tht the thickness of such slice is x (which is lso the height of the cylindershped slice), nd tht the rdius of the slice is determined by the curve y = 4 x 2. Hence, we find tht V slice = π(4 x 2 ) 2 x,
14 USING DEFINITE INTEGRALS TO FIND VOLUME since the volume of cylinder of rdius r nd height h is V = πr 2 h. Using definite integrl to sum the volumes of the representtive slices, it follows tht V = 2 2 π(4 x 2 ) 2 dx. It is strightforwrd to evlute the integrl nd find tht the volume is V = 52 5 π. For solid such s the one in Exmple 6., where ech crosssection is cylindricl disk, we first find the volume of typicl crosssection (noting prticulrly how this volume depends on x), nd then we integrte over the rnge of xvlues through which we slice the solid in order to find the exct totl volume. Often, we will be content with simply finding the integrl tht represents the sought volume; if we desire numeric vlue for the integrl, we typiclly use clcultor or computer lgebr system to find tht vlue. The generl principle we re using to find the volume of solid of revolution generted by single curve is often clled the disk method. If y = r(x) is nonnegtive continuous function on [, b], then the volume of the solid of revolution generted by revolving the curve bout the xxis over this intervl is given by V = πr(x) 2 dx. A different type of solid cn emerge when two curves re involved, s we see in the following exmple. Exmple 6.2. Find the volume of the solid of revolution generted when the finite region R tht lies between y = 4 x 2 nd y = x + 2 is revolved bout the xxis. Solution. First, we must determine where the curves y = 4 x 2 nd y = x + 2 intersect. Substituting the expression for y from the second eqution into the first eqution, we find tht x + 2 = 4 x 2. Rerrnging, it follows tht x 2 + x 2 =, nd the solutions to this eqution re x = 2 nd x =. The curves therefore cross t ( 2, ) nd (, ).
15 6.2. USING DEFINITE INTEGRALS TO FIND VOLUME 347 When we tke the region R tht lies between the curves nd revolve it bout the xxis, we get the threedimensionl solid pictured t left in Figure 6.9. y R(x) x r(x) Figure 6.9: At left, the solid of revolution in Exmple 6.2. At right, typicl slice with inner rdius r(x) nd outer rdius R(x). Immeditely we see mjor difference between the solid in this exmple nd the one in Exmple 6.: here, the threedimensionl solid of revolution isn t solid in the sense tht it hs open spce in its center. If we slice the solid perpendiculr to the xis of revolution, we observe tht in this setting the resulting representtive slice is not solid disk, but rther wsher, s pictured t right in Figure 6.9. Moreover, t given loction x between x = 2 nd x =, the smll rdius r(x) of the inner circle is determined by the curve y = x + 2, so r(x) = x + 2. Similrly, the big rdius R(x) comes from the function y = 4 x 2, nd thus R(x) = 4 x 2. Thus, to find the volume of representtive slice, we compute the volume of the outer disk nd subtrct the volume of the inner disk. Since πr(x) 2 x πr(x) 2 x = π[r(x) 2 r(x) 2 ] x, it follows tht the volume of typicl slice is V slice = π[(4 x 2 ) 2 (x + 2) 2 ] x. Hence, using definite integrl to sum the volumes of the respective slices cross the integrl, we find tht V = 2 π[(4 x 2 ) 2 (x + 2) 2 ] dx. Evluting the integrl, the volume of the solid of revolution is V = 8 5 π. The generl principle we re using to find the volume of solid of revolution generted
16 USING DEFINITE INTEGRALS TO FIND VOLUME by single curve is often clled the wsher method. If y = R(x) nd y = r(x) re nonnegtive continuous functions on [, b] tht stisfy R(x) r(x) for ll x in [, b], then the volume of the solid of revolution generted by revolving the region between them bout the xxis over this intervl is given by Activity 6.4. V = π[r(x) 2 r(x) 2 ] dx. In ech of the following questions, drw creful, lbeled sketch of the region described, s well s the resulting solid tht results from revolving the region bout the stted xis. In ddition, drw representtive slice nd stte the volume of tht slice, long with definite integrl whose vlue is the volume of the entire solid. It is not necessry to evlute the integrls you find. () The region S bounded by the xxis, the curve y = x, nd the line x = 4; revolve S bout the xxis. (b) The region S bounded by the yxis, the curve y = x, nd the line y = 2; revolve S bout the xxis. (c) The finite region S bounded by the curves y = x nd y = x 3 ; revolve S bout the xxis. (d) The finite region S bounded by the curves y = 2x 2 + nd y = x 2 + 4; revolve S bout the xxis (e) The region S bounded by the yxis, the curve y = x, nd the line y = 2; revolve S bout the yxis. How does the problem chnge considerbly when we revolve bout the yxis? Revolving bout the yxis As seen in Activity 6.4, problem (e), the problem chnges considerbly when we revolve given region bout the yxis. Foremost, this is due to the fct tht representtive slices now hve thickness y, which mens tht it becomes necessry to integrte with respect to y. Let s consider prticulr exmple to demonstrte some of the key issues.
17 6.2. USING DEFINITE INTEGRALS TO FIND VOLUME 349 Exmple 6.3. Find the volume of the solid of revolution generted when the finite region R tht lies between y = x nd y = x 4 is revolved bout the yxis. Solution. We observe tht these two curves intersect when x =, hence t the point (, ). When we tke the region R tht lies between the curves nd revolve it bout the yxis, we get the threedimensionl solid pictured t left in Figure 6.. Now, it is prticulrly importnt y R(y) r(y) x Figure 6.: At left, the solid of revolution in Exmple 6.3. At right, typicl slice with inner rdius r(y) nd outer rdius R(y). to note tht the thickness of representtive slice is y, nd tht the slices re only cylindricl wshers in nture when tken perpendiculr to the yxis. Hence, we envision slicing the solid horizontlly, strting t y = nd proceeding up to y =. Becuse the inner rdius is governed by the curve y = x, but from the perspective tht x is function of y, we solve for x nd get x = y 2 = r(y). In the sme wy, we need to view the curve y = x 4 (which governs the outer rdius) in the form where x is function of y, nd hence x = 4 y. Therefore, we see tht the volume of typicl slice is V slice = π[r(y) 2 r(y) 2 ] = π[ 4 y 2 (y 2 ) 2 ] y. Using definite integrl to sum the volume of ll the representtive slices from y = to y =, the totl volume is V = y= y= π [ 4 y 2 (y 2 ) 2] dy. It is strightforwrd to evlute the integrl nd find tht V = 7 5 π.
18 USING DEFINITE INTEGRALS TO FIND VOLUME Activity 6.5. In ech of the following questions, drw creful, lbeled sketch of the region described, s well s the resulting solid tht results from revolving the region bout the stted xis. In ddition, drw representtive slice nd stte the volume of tht slice, long with definite integrl whose vlue is the volume of the entire solid. It is not necessry to evlute the integrls you find. () The region S bounded by the yxis, the curve y = x, nd the line y = 2; revolve S bout the yxis. (b) The region S bounded by the xxis, the curve y = x, nd the line x = 4; revolve S bout the yxis. (c) The finite region S in the first qudrnt bounded by the curves y = 2x nd y = x 3 ; revolve S bout the xxis. (d) The finite region S in the first qudrnt bounded by the curves y = 2x nd y = x 3 ; revolve S bout the yxis. (e) The finite region S bounded by the curves x = (y ) 2 nd y = x ; revolve S bout the yxis Revolving bout horizontl nd verticl lines other thn the coordinte xes Just s we cn revolve bout one of the coordinte xes (y = or x = ), it is lso possible to revolve round ny horizontl or verticl line. Doing so essentilly djusts the rdii of cylinders or wshers involved by constnt vlue. A creful, welllbeled plot of the solid of revolution will usully revel how the different xis of revolution ffects the definite integrl we set up. Agin, n exmple is instructive. Exmple 6.4. Find the volume of the solid of revolution generted when the finite region S tht lies between y = x 2 nd y = x is revolved bout the line y =. Solution. Grphing the region between the two curves in the first qudrnt between their points of intersection ((, ) nd (, )) nd then revolving the region bout the line y =, we see the solid shown in Figure 6.. Ech slice of the solid perpendiculr to the xis of revolution is wsher, nd the rdii of ech wsher re governed by the curves y = x 2 nd y = x. But we lso see tht there is one dded chnge: the xis of revolution dds fixed length to ech rdius. In prticulr, the inner rdius of typicl slice, r(x), is given
19 6.2. USING DEFINITE INTEGRALS TO FIND VOLUME 35 y x Figure 6.: The solid of revolution described in Exmple 6.4. by r(x) = x 2 +, while the outer rdius is R(x) = x +. Therefore, the volume of typicl slice is V slice = π[r(x) 2 r(x) 2 ] x = π (x + ) 2 (x 2 + ) 2 x. Finlly, we integrte to find the totl volume, nd V = π (x + ) 2 (x 2 + ) 2 dx = 7 5 π. Activity 6.6. In ech of the following questions, drw creful, lbeled sketch of the region described, s well s the resulting solid tht results from revolving the region bout the stted xis. In ddition, drw representtive slice nd stte the volume of tht slice, long with definite integrl whose vlue is the volume of the entire solid. It is not necessry to evlute the integrls you find. For ech prompt, use the finite region S in the first qudrnt bounded by the curves y = 2x nd y = x 3. () Revolve S bout the line y = 2. (b) Revolve S bout the line y = 4. (c) Revolve S bout the line x =. (d) Revolve S bout the line x = 5. Summry
20 USING DEFINITE INTEGRALS TO FIND VOLUME In this section, we encountered the following importnt ides: We cn use definite integrl to find the volume of threedimensionl solid of revolution tht results from revolving twodimensionl region bout prticulr xis by tking slices perpendiculr to the xis of revolution which will then be circulr disks or wshers. If we revolve bout verticl line nd slice perpendiculr to tht line, then our slices re horizontl nd of thickness y. This leds us to integrte with respect to y, s opposed to with respect to x when we slice solid verticlly. If we revolve bout line other thn the x or yxis, we need to crefully ccount for the shift tht occurs in the rdius of typicl slice. Normlly, this shift involves tking sum or difference of the function long with the constnt connected to the eqution for the horizontl or verticl line; welllbeled digrm is usully the best wy to decide the new expression for the rdius. Exercises. Consider the curve f (x) = 3 cos( x3 4 ) nd the portion of its grph tht lies in the first qudrnt between the yxis nd the first positive vlue of x for which f (x) =. Let R denote the region bounded by this portion of f, the xxis, nd the yxis. () Set up definite integrl whose vlue is the exct rc length of f tht lies long the upper boundry of R. Use technology ppropritely to evlute the integrl you find. (b) Set up definite integrl whose vlue is the exct re of R. Use technology ppropritely to evlute the integrl you find. (c) Suppose tht the region R is revolved round the xxis. Set up definite integrl whose vlue is the exct volume of the solid of revolution tht is generted. Use technology ppropritely to evlute the integrl you find. (d) Suppose insted tht R is revolved round the yxis. If possible, set up n integrl expression whose vlue is the exct volume of the solid of revolution nd evlute the integrl using pproprite technology. If not possible, explin why. 2. Consider the curves given by y = sin(x) nd y = cos(x). For ech of the following problems, you should include sketch of the region/solid being considered, s well s lbeled representtive slice. () Sketch the region R bounded by the yxis nd the curves y = sin(x) nd y = cos(x) up to the first positive vlue of x t which they intersect. Wht is the exct intersection point of the curves?
21 6.2. USING DEFINITE INTEGRALS TO FIND VOLUME 353 (b) Set up definite integrl whose vlue is the exct re of R. (c) Set up definite integrl whose vlue is the exct volume of the solid of revolution generted by revolving R bout the xxis. (d) Set up definite integrl whose vlue is the exct volume of the solid of revolution generted by revolving R bout the yxis. (e) Set up definite integrl whose vlue is the exct volume of the solid of revolution generted by revolving R bout the line y = 2. (f) Set up definite integrl whose vlue is the exct volume of the solid of revolution generted by revolving R bout the x =. 3. Consider the finite region R tht is bounded by the curves y = + 2 (x 2)2, y = 2 x2, nd x =. () Determine definite integrl whose vlue is the re of the region enclosed by the two curves. (b) Find n expression involving one or more definite integrls whose vlue is the volume of the solid of revolution generted by revolving the region R bout the line y =. (c) Determine n expression involving one or more definite integrls whose vlue is the volume of the solid of revolution generted by revolving the region R bout the yxis. (d) Find n expression involving one or more definite integrls whose vlue is the perimeter of the region R.
22 DENSITY, MASS, AND CENTER OF MASS 6.3 Density, Mss, nd Center of Mss Motivting Questions In this section, we strive to understnd the ides generted by the following importnt questions: How re mss, density, nd volume relted? How is the mss of n object with vrying density computed? Wht is the center of mss of n object, nd how re definite integrls used to compute it? Introduction We hve seen in severl different circumstnces how studying the units on the integrnd nd vrible of integrtion enbles us to better understnd the mening of definite integrl. For instnce, if v(t) is the velocity of n object moving long n xis, mesured in feet per second, while t mesures time in seconds, then both the definite integrl nd its Riemnn sum pproximtion, v(t) dt n v(t i ) t, i= hve their overll units given by the product of the units of v(t) nd t: (feet/sec) (sec) = feet. Thus, v(t) dt mesures the totl chnge in position (in feet) of the moving object. This type of unit nlysis will be prticulrly helpful to us in wht follows. To begin, in the following preview ctivity we consider two different definite integrls where the integrnd is function tht mesures how prticulr quntity is distributed over region nd think bout how the units on the integrnd nd the vrible of integrtion indicte the mening of the integrl. Preview Activity 6.3. In ech of the following scenrios, we consider the distribution of quntity long n xis. () Suppose tht the function c(x) = 2 + e.x models the density of trffic on stright rod, mesured in crs per mile, where x is number of miles est of mjor interchnge, nd consider the definite integrl 2 (2 + e.x ) dx. i. Wht re the units on the product c(x) x?
23 6.3. DENSITY, MASS, AND CENTER OF MASS 355 ii. Wht re the units on the definite integrl nd its Riemnn sum pproximtion given by 2 n c(x) dx c(x i ) x? iii. Evlute the definite integrl 2 c(x) dx = 2 (2 + e.x ) dx nd write one sentence to explin the mening of the vlue you find. (b) On 6 foot long shelf filled with books, the function B models the distribution of the weight of the books, mesured in pounds per inch, where x is the number of inches from the left end of the bookshelf. Let B(x) be given by the rule B(x) =.5 +. (x+) 2 i= i. Wht re the units on the product B(x) x? ii. Wht re the units on the definite integrl nd its Riemnn sum pproximtion given by 36 n B(x) dx B(x i ) x? 2 iii. Evlute the definite integrl 72 B(x) dx = 72 (.5 + ) dx nd write (x+) 2 one sentence to explin the mening of the vlue you find. i= Density The mss of quntity, typiclly mesured in metric units such s grms or kilogrms, is mesure of the mount of quntity. In corresponding wy, the density of n object mesures the distribution of mss per unit volume. For instnce, if brick hs mss 3 kg nd volume.2 m 3, then the density of the brick is 3kg kg = 5.2m3 m 3. As nother exmple, the mss density of wter is kg/m 3. Ech of these reltionships demonstrte the following generl principle. For n object of constnt density d, with mss m nd volume V, d = m V, or m = d V. But wht hppens when the density is not constnt?
24 DENSITY, MASS, AND CENTER OF MASS If we consider the formul m = d V, it is reminiscent of two other equtions tht we hve used frequently in recent work: for body moving in fixed direction, distnce = rte time, nd, for rectngle, its re is given by A = l w. These formuls hold when the principl quntities involved, such s the rte the body moves nd the height of the rectngle, re constnt. When these quntities re not constnt, we hve turned to the definite integrl for ssistnce. The min ide in ech sitution is tht by working with smll slices of the quntity tht is vrying, we cn use definite integrl to dd up the vlues of smll pieces on which the quntity of interest (such s the velocity of moving object) re pproximtely constnt. For exmple, in the setting where we hve nonnegtive velocity function tht is not constnt, over short time intervl t we know tht the distnce trveled is pproximtely v(t) t, since v(t) is lmost constnt on smll intervl, nd for constnt rte, distnce = rte time. Similrly, if we re thinking bout the re under nonnegtive function f whose vlue is chnging, on short intervl x the re under the curve is pproximtely the re of the rectngle whose height is f (x) nd whose width is x: f (x) x. Both of these principles re represented visully in Figure 6.2. ft/sec y = v(t) y y = f (x) v(t) sec f (x) x t x Figure 6.2: At left, estimting smll mount of distnce trveled, v(t) t, nd t right, smll mount of re under the curve, f (x) x. In similr wy, if we consider the setting where the density of some quntity is not constnt, the definite integrl enbles us to still compute the overll mss of the quntity. Throughout, we will focus on problems where the density vries in only one dimension, sy long single xis, nd think bout how mss is distributed reltive to loction long the xis. Let s consider thin br of length b tht is situted so its left end is t the origin, where x =, nd ssume tht the br hs constnt crosssectionl re of cm 2. We let the function ρ(x) represent the mss density function of the br, mesured in grms per cubic centimeter. Tht is, given loction x, ρ(x) tells us pproximtely how much mss will be found in onecentimeter wide slice of the br t x.
25 6.3. DENSITY, MASS, AND CENTER OF MASS 357 x Figure 6.3: A thin br of constnt crosssectionl re cm 2 with density function ρ(x) g/cm 3. x If we now consider thin slice of the br of width x, s pictured in Figure 6.3, the volume of such slice is the crosssectionl re times x. Since the crosssections ech hve constnt re cm 2, it follows tht the volume of the slice is x cm 3. Moreover, since mss is the product of density nd volume (when density is constnt), we see tht the mss of this given slice is pproximtely mss slice ρ(x) g cm 3 x cm3 = ρ(x) x g. Hence, for the corresponding Riemnn sum (nd thus for the integrl tht it pproximtes), n ρ(x i ) x ρ(x) dx, i= we see tht these quntities mesure the mss of the br between nd b. (The Riemnn sum is n pproximtion, while the integrl will be the exct mss.) At this point, we note tht we will be focused primrily on situtions where mss is distributed reltive to horizontl loction, x, for objects whose crosssectionl re is constnt. In tht setting, it mkes sense to think of the density function ρ(x) with units mss per unit length, such s g/cm. Thus, when we compute ρ(x) x on smll slice x, the resulting units re g/cm cm = g, which thus mesures the mss of the slice. The generl principle follows. For n object of constnt crosssectionl re whose mss is distributed long single xis ccording to the function ρ(x) (whose units re units of mss per unit of length), the totl mss, M of the object between x = nd x = b is given by Activity 6.7. M = ρ(x) dx. Consider the following situtions in which mss is distributed in nonconstnt mnner. () Suppose tht thin rod with constnt crosssectionl re of cm 2 hs its mss
26 DENSITY, MASS, AND CENTER OF MASS distributed ccording to the density function ρ(x) = 2e.2x, where x is the distnce in cm from the left end of the rod, nd the units on ρ(x) re g/cm. If the rod is cm long, determine the exct mss of the rod. (b) Consider the cone tht hs bse of rdius 4 m nd height of 5 m. Picture the cone lying horizontlly with the center of its bse t the origin nd think of the cone s solid of revolution. i. Write nd evlute definite integrl whose vlue is the volume of the cone. ii. Next, suppose tht the cone hs uniform density of 8 kg/m 3. Wht is the mss of the solid cone? iii. Now suppose tht the cone s density is not uniform, but rther tht the cone is most dense t its bse. In prticulr, ssume tht the density of the cone is uniform cross cross sections prllel to its bse, but tht in ech such cross section tht is distnce x units from the origin, the density of the cross section is given by the function ρ(x) = 4 + 2, mesured in +x 2 kg/m 3. Determine nd evlute definite integrl whose vlue is the mss of this cone of nonuniform density. Do so by first thinking bout the mss of given slice of the cone x units wy from the bse; remember tht in such slice, the density will be essentilly constnt. (c) Let thin rod of constnt crosssectionl re cm 2 nd length 2 cm hve its mss be distributed ccording to the density function ρ(x) = 25 (x 5)2, mesured in g/cm. Find the exct loction z t which to cut the br so tht the two pieces will ech hve identicl mss. Weighted Averges clss grde grde points credits chemistry B clculus A history B psychology B Tble 6.: A college student s semester grdes. The concept of n verge is nturl one, nd one tht we hve used repetedly s prt of our understnding of the mening of the definite integrl. If we hve n vlues, 2,..., n, we know tht their verge is given by n, n
27 6.3. DENSITY, MASS, AND CENTER OF MASS 359 nd for quntity being mesured by function f on n intervl [, b], the verge vlue of the quntity on [, b] is b f (x) dx. b As we continue to think bout problems involving the distribution of mss, it is nturl to consider the ide of weighted verge, where certin quntities involved re counted more in the verge. A common use of weighted verges is in the computtion of student s GPA, where grdes re weighted ccording to credit hours. Let s consider the scenrio in Tble 6.. If ll of the clsses were of the sme weight (i.e., the sme number of credits), the student s GPA would simply be clculted by tking the verge = 3.. But since the chemistry nd clculus courses hve higher weights (of 5 nd 4 credits respectively), we ctully compute the GPA ccording to the weighted verge = 3.6. The weighted verge reflects the fct tht chemistry nd clculus, s courses with higher credits, hve greter impct on the students grde point verge. Note prticulrly tht in the weighted verge, ech grde gets multiplied by its weight, nd we divide by the sum of the weights. In the following ctivity, we explore further how weighted verges cn be used to find the blncing point of physicl system. Activity 6.8. For quntities of equl weight, such s two children on teetertotter, the blncing point is found by tking the verge of their loctions. When the weights of the quntities differ, we use weighted verge of their respective loctions to find the blncing point. () Suppose tht shelf is 6 feet long, with its left end situted t x =. If one book of weight lb is plced t x =, nd nother book of weight lb is plced t x 2 = 6, wht is the loction of x, the point t which the shelf would (theoreticlly) blnce on fulcrum? (b) Now, sy tht we plce four books on the shelf, ech weighing lb: t x =, t x 2 = 2, t x 3 = 4, nd t x 4 = 6. Find x, the blncing point of the shelf. (c) How does x chnge if we chnge the loction of the third book? Sy the loctions of the lb books re x =, x 2 = 2, x 3 = 3, nd x 4 = 6.
28 DENSITY, MASS, AND CENTER OF MASS (d) Next, suppose tht we plce four books on the shelf, but of vrying weights: t x = 2lb book, t x 2 = 2 3lb book, nd x 3 = 4 lb book, nd t x 4 = 6 lb book. Use weighted verge of the loctions to find x, the blncing point of the shelf. How does the blncing point in this scenrio compre to tht found in (b)? (e) Wht hppens if we chnge the loction of one of the books? Sy tht we keep everything the sme in (d), except tht x 3 = 5. How does x chnge? (f) Wht hppens if we chnge the weight of one of the books? Sy tht we keep everything the sme in (d), except tht the book t x 3 = 4 now weighs 2 lbs. How does x chnge? (g) Experiment with couple of different scenrios of your choosing where you move the loction of one of the books to the left, or you decrese the weight of one of the books. (h) Write couple of sentences to explin how djusting the loction of one of the books or the weight of one of the books ffects the loction of the blncing point of the shelf. Think crefully here bout how your chnges should be considered reltive to the loction of the blncing point x of the current scenrio. Center of Mss In Activity 6.8, we sw tht the blncing point of system of pointmsses (such s books on shelf) is found by tking weighted verge of their respective loctions. In the ctivity, we were computing the center of mss of system of msses distributed long n xis, which is the blncing point of the xis on which the msses rest. For collection of n msses m,..., m n tht re distributed long single xis t the loctions x,..., x n, the center of mss is given by x = x m + x 2 m 2 + x n m n m + m m n. Wht if we insted consider thin br over which density is distributed continuously? If the density is constnt, it is obvious tht the blncing point of the br is its midpoint. But if density is not constnt, we must compute weighted verge. Let s sy tht the In the ctivity, we ctully used weight rther thn mss. Since weight is computed by the grvittionl constnt times mss, the computtions for the blncing point result in the sme loction regrdless of whether we use weight or mss, since the grvittionl constnt is present in both the numertor nd denomintor of the weighted verge.
29 6.3. DENSITY, MASS, AND CENTER OF MASS 36 function ρ(x) tells us the density distribution long the br, mesured in g/cm. If we slice the br into smll sections, this enbles us to think of the br s holding collection of djcent pointmsses. For slice of thickness x t loction x i, note tht the mss of the slice, m i, stisfies m i ρ(x i ) x. Tking n slices of the br, we cn pproximte its center of mss by x x ρ(x ) x + x 2 ρ(x 2 ) x + + x n ρ(x n ) x. ρ(x ) x + ρ(x 2 ) x + + ρ(x n ) x Rewriting the sums in sigm nottion, it follows tht x ni= x i ρ(x i ) x ni=. (6.4) ρ(x i ) x Moreover, it is pprent tht the greter the number of slices, the more ccurte our estimte of the blncing point will be, nd tht the sums in Eqution (6.4) cn be viewed s Riemnn sums. Hence, in the limit s n, we find tht the center of mss is given by the quotient of two integrls. For thin rod of density ρ(x) distributed long n xis from x = to x = b, the center of mss of the rod is given by x = x ρ(x) dx. ρ(x) dx Note prticulrly tht the denomintor of x is the mss of the br, nd tht this quotient of integrls is simply the continuous version of the weighted verge of loctions, x, long the br. Activity 6.9. Consider thin br of length 2 cm whose density is distributed ccording to the function ρ(x) = 4 +.x, where x = represents the left end of the br. Assume tht ρ is mesured in g/cm nd x is mesured in cm. () Find the totl mss, M, of the br. (b) Without doing ny clcultions, do you expect the center of mss of the br to be equl to, less thn, or greter thn? Why? (c) Compute x, the exct center of mss of the br. (d) Wht is the verge density of the br? (e) Now consider different density function, given by p(x) = 4e.2732x, lso for br of length 2 cm whose left end is t x =. Plot both ρ(x) nd p(x) on
30 DENSITY, MASS, AND CENTER OF MASS the sme xes. Without doing ny clcultions, which br do you expect to hve the greter center of mss? Why? (f) Compute the exct center of mss of the br described in (e) whose density function is p(x) = 4e.2732x. Check the result ginst the prediction you mde in (e). Summry In this section, we encountered the following importnt ides: For n object of constnt density D, with volume V nd mss m, we know tht m = D V. If n object with constnt crosssectionl re (such s thin br) hs its density distributed long n xis ccording to the function ρ(x), then we cn find the mss of the object between x = nd x = b by m = ρ(x) dx. For system of pointmsses distributed long n xis, sy m,..., m n t loctions x,..., x n, the center of mss, x, is given by the weighted verge x = ni= x i m i ni= m i. If insted we hve mss continuously distributed long n xis, such s by density function ρ(x) for thin br of constnt crosssectionl re, the center of mss of the portion of the br between x = nd x = b is given by x = x ρ(x) dx. ρ(x) dx In ech sitution, x represents the blncing point of the system of msses or of the portion of the br. Exercises. Let thin rod of length hve density distribution function ρ(x) = e.x, where x is mesured in cm nd ρ in grms per centimeter. () If the mss of the rod is 3 g, wht is the vlue of?
31 6.3. DENSITY, MASS, AND CENTER OF MASS 363 (b) For the 3g rod, will the center of mss lie t its midpoint, to the left of the midpoint, or to the right of the midpoint? Why? (c) For the 3g rod, find the center of mss, nd compre your prediction in (b). (d) At wht vlue of x should the 3g rod be cut in order to form two pieces of equl mss? 2. Consider two thin brs of constnt crosssectionl re, ech of length cm, with respective mss density functions ρ(x) = +x 2 nd p(x) = e.x. () Find the mss of ech br. (b) Find the center of mss of ech br. (c) Now consider new cm br whose mss density function is f (x) = ρ(x) + p(x). i. Explin how you cn esily find the mss of this new br with little to no dditionl work. ii. Similrly, compute x f (x) dx s simply s possible, in light of erlier computtions. iii. True or flse: the center of mss of this new br is the verge of the centers of mss of the two erlier brs. Write t lest one sentence to sy why your conclusion mkes sense. 3. Consider the curve given by y = f (x) = 2xe.25x + (3 x)e.25(3 x). () Plot this curve in the window x =... 3, y =... 3 (with constrined scling so the units on the x nd y xis re equl), nd use it to generte solid of revolution bout the xxis. Explin why this curve could generte resonble model of bsebll bt. (b) Let x nd y be mesured in inches. Find the totl volume of the bsebll bt generted by revolving the given curve bout the xxis. Include units on your nswer (c) Suppose tht the bsebll bt hs constnt weight density, nd tht the weight density is.6 ounces per cubic inch. Find the totl weight of the bt whose volume you found in (b). (d) Becuse the bsebll bt does not hve constnt crosssectionl re, we see tht the mount of weight concentrted t loction x long the bt is determined by the volume of slice t loction x. Explin why we cn think bout the function ρ(x) =.6π f (x) 2 (where f is the function given t the strt of the problem) s being the weight density function for how the weight of the bsebll bt is distributed from x = to x = 3.
32 DENSITY, MASS, AND CENTER OF MASS (e) Compute the center of mss of the bsebll bt.
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