MTHE 7 Proble Set 5 Solutions (A Crdioid). Let C be the closed curve in R whose polr coordintes (r, θ) stisfy () Sketch the curve C. r = cos θ +. (b) Find pretriztion t (r(t), θ(t)), t [, b], of C in polr coordintes. As we hve seen in clss, for pth pretrized in polr coordintes s in prt (b), the rclength of the pth cn be coputed by the epression b ( dr dt ) + r ( dθ dt ) dt. () (c) Applying (), or otherwise, show tht the rclength of C is equl to 8. (The hlf-ngle forul cos( θ ) = +cos(θ) y coe in useful. You cn sve yourself fro deling with the bsolute vlue in the hlf-ngle forul by noticing soe syetry in the proble.) (d) Show tht the re enclosed by C is equl to 3π/. (e) Convert your pretriztion of C found in (b) to Crtesin coordintes (, y). (C is eber of fily of curves cut out in polr coordintes by r = (cos θ + ), with positive rel nuber, clled crdioids. The rclength of generl crdioid is 6, nd its re is 6π. The ne derives fro the Greek word krdi, which trnsltes to hert.) Solution. () To drw the pth, it is helpful to first drw the grph of + cos(θ), on the left. Grph of + cos(θ) (b) One possible pretriztion is π θ The Crdioid r = + cos θ For this pretriztion, we hve t (cos(t) +, t) = (r(t), θ(t)), dr dt = sin(t), dθ dt =. t [, π].
(c) We hve ( dr dt ) +r ( dθ dt ) = sin (t)+(+cos(t)) = sin (t)+cos (t)+ cos(t)+ = (+cos(t)), so tht ( dr dt ) + r ( dθ dt ) = The rclength integrl is + cos(t) = cos( t ). π cos( t ) dt = π cos( t ) dt + π π = [ sin( t π )] + [ sin( t π )] ( cos( t )) dt = ( sin( π ) sin()) + ( sin(π) + sin(π )) = 8. We could hve lso observed tht by syetry of cos(t) +, the rclength over the intervl t π is equl to the rclength over π t π, so the rclength is equl to π cos( t ) dt = 8. (d) Let R be the region enclosed by the crdioid. In polr coordintes, R = {(r, θ) r cos(θ) +, θ π}, so tht R is Type II region in polr coordintes. By Fubini s theore, π R da = = = = π π [ r π π = + + = 3π. cos(t)+ rdrdθ cos(t)+ ] dθ cos (t) + cos(t) + dθ cos(t) + cos(t) + 3 dθ 3(π ) using cos (t) = + cos(t) (e) The polr coordintes re relted to Crtesin ones by = r cos θ, y = r sin θ. Therefore, possible pretriztion of the crdioid in (, y) coordintes is (t) = r(t) cos(θ(t)) = (cos(t)+) cos(t), y(t) = r(t) sin(θ(t)) = (cos(t)+) sin(t), t [, π].
(Averges of Averges). Two coon wys of finding the verge of two nonnegtive rel nubers nd y re the rithetic en nd the geoetric en y. The two ens re relted by the (very useful) rithetic en-geoetric en inequlity, or AM-GM inequlity for short: y. Fi rel nuber >. If nd y re chosen independently nd uniforly t rndo fro the intervl [, ] (ening no vlue is ore likely thn ny other), the epected vlues of the rithetic nd geoetric ens re given by [,] [,] da nd [,] [,] y da, respectively. Copute the two integrls. Are your findings consistent with the AM-GM inequlity? Solution. Both of the integrls hve the squre [, ] [, ] s the region of integrtion, so by Fubini s theore, we hve [,] [,] da = = = [ ddy + y = ] dy = + y dy = [ y + y y= ] y= = (3 + 3 ) = 3 = 3
nd y [,] [,] da = = y ddy [ 3 3/ y / ] = = = 3 3/ y / dy = 3 3/ y / dy = 3 3/ [ y= 3 y3/ ] = 9 3 = 9. y= dy Becuse +y y for ny, y (nd ultipliction by the positive constnt / preserves the inequlity), we ust hve [,] [,] da y [,] [,] da Since / = /8 > /9, the epected vlues of AM nd GM coputed bove re indeed consistent with the AM-GM inequlity. 3 (Are of n Ellipse). Find the re of the region R bounded by the ellipse + y b =, by integrting the constnt function over R. Wht do you find in the cse when = b? You y ke use of the following without proof: t dt = t t + rcsin(t). Optionl Proble. Integrte t, by king the trigonoetric substitution t = sin θ. Solution. The ellipse is Type III region, so we cn integrte in either order. Let s rbitrrily choose s the outer vrible. Fro the stndrd for of the eqution of the ellipse ( + y b = ), we cn red off tht the ellipse looks like b The Ellipse + y b =
Therefore, runs fro (, ) to (, ) (you cn quickly check tht both of these points stisfy the eqution + y b = ). Then, solving the defining eqution for y in ters of, we hve y = b ( ), so tht the bounds on y re b y b. The re integrl is b / b / dyd = b + b = b d. = b u du = b [ u u + rcsin(u) = b ( + π/ = b π = bπ. When = b, we get the curve + y re π = π, s epected. d Letting u = /, du = d/, u= ] u= π/ ) (the rnge of rcsin is [ π/, π/]!) =, or + y =, which is circle of rdius, nd Solution of the Optionl Proble. Letting t = sin θ, dt = cos θ dθ, t dt = sin θ cos θ dθ = cos θ cos θ dθ = cos θ dθ = + cos(θ) dθ = θ + sin(θ) θ + sin(θ) cos(θ) = (using sin(θ) = sin(θ) cos(θ). The following right tringle, constructed so tht sin θ = t/ = t, lets us find cos θ: 5
t θ t By the Pythgoren theore, the length of the botto side is equl to t = t, nd cos θ = t = t. Therefore, t dt = θ + sin(θ) cos(θ) = rcsin(t) + t t.. () For ech of the following regions R, sketch R, nd set up the double integrl R f(, y) da s n iterted integrl (or possibly su of iterted integrls) in Crtesin coordintes. It is not necessry to evlute the integrl. (i) The region R to the left of the y-is nd inside the circle + y =. (ii) The region R bounded by the curves = y + 3 nd = y. (iii) The region R bounded by the prllelogr with vertices (, ), (3, 3), (5, ) nd (7, ). (b) For ech of the following, sketch the region of integrtion, switch the order of integrtion, nd evlute the integrl. (i) (ii) (iii) Solution. ( ( y rcsin() y dy) d e d) dy (Note: The ntiderivtive of e cnnot be written down in ters of sus, products nd powers of the usul functions, cos(), sin(), ep(), log(),.... It hs no ntiderivtive in eleentry ters.) / ( y ln d) dy + ( y ln d) dy. () (i) The region is Type III. Integrting with respect to y first, the bounds on y re y. 6
By Fubini s theore, the integrl is f(, y) dyd. (ii) The two curves intersect when y = y + 3, or 3y = 3, or y = ±. The region bounded by the two curves looks like This region is Type II, but not Type I. Therefore, we integrte with respect to first. The integrl is (pplying Fubini s theore, s usul) y+3 y f(, y) ddy. (iii) The region is Type III, but it is not esy to write the boundries s single eplicit function. Therefore, we brek it up into three prts, s indicted on the left digr. The equtions of the boundry lines re written down on the right digr. y = y = 3 y = +9 y = +3 3 5 7 7
The integrl is 3 (+3)/ f(, y) dyd + 3 5 (+9)/ (+3)/ f(, y) dyd + 5 7 (+9)/ 3 f(, y) dyd. (b) (i) The region of integrtion looks like π/ Therefore, the integrl with switched order of integrtion is π/ ( sin(y) Integrting by prts twice, we hve y d) dy = π/ y y sin(y) dy. y sin(y) dy = y cos(y)+ y cos(y) dy = y cos(y)+(y sin(y) sin(y) dy), so tht π/ y y sin(y) dy = [ y3 y=π/ 3 + y cos(y) y sin(y) cos(y)] (ii) The region of integrtion looks like = ( π3 + π ) ( + ) = π3 π +. y= 8
Therefore, the integrl with switched order of integrtion is ( e dy) d = (iii) The region of integrtion looks like e d = [e /] = = = e. y = y = / / Therefore, the integrl with switched order of integrtion is ( We hve, by prts, so tht ( / / ln dy) d = ln ln d = ln d. ln d = ln() d = ln(), ln dy) d = [ ln() ]= = = ( ln() + ) = ln(). 9