Yoplit with Ares nd Volumes Yoplit yogurt comes in two differently shped continers. One is truncted cone nd the other is n ellipticl cylinder (see photos below). In this exercise, you will determine the volume of the Yoplit truncted cone in three different wys: (1) using geometry, () using the Disk Method from Clculus, nd (3) using the Shell Method. You will then determine the re formul for n ellipse using clculus, nd then compute the volume of the cylinder. Ech continer clims to contin six ounces of yogurt, which is equivlent to 10.83 cubic inches. I mesured the continers to the nerest tenth of n inch. In the ellipticl cylinder, the mjor xis ws.5 inches, the minor xis ws.1 inches, nd it ws.6 inches tll. In the truncted cone, the dimeter of the smller circle (top) ws inches, the dimeter of the lrger circle (bottom) ws.4 inches, nd it ws.8 inches tll. In the ltter, the height is deceiving becuse the continer is recessed on the bottom; the.8 inches is the height of the portion of the continer tht holds the yogurt the ctul continer is over three inches tll.
(1) The Yoplit Truncted Cone Continer Using geometry to find its volume First extend the sides of the continer until it forms cone, s pictured in the digrm below. Let h = the height of the truncted cone. Let x = the height of the new cone tht is formed by extending the sides. Solve for x. Now solve for the volume of the truncted cone by subtrcting the volumes of the two cones. How does this nswer compre to 10.83 cubic inches (the six ounces of yogurt)? () The Yoplit Truncted Cone Continer Using the Disk Method to find its volume In the digrm in (1), let R represent the positive x-xis, nd h + x represent the positive y-xis. Let the intersection of those two segments represent the origin. Now solve for the eqution of the line representing the lterl edge of the cone. Now determine the volume obtined when rotting tht line bout the y-xis from y = 0 to y =.8. How does this nswer compre to the nswer obtined in (1)?
(3) The Yoplit Truncted Cone Continer Using the Shell Method to find its volume Now determine the volume of the truncted cone by using the Shell Method. Note tht this must be done in two steps. Find the volume from x = 0 to x = 1, nd then find the volume from x = 1 to x = 1.. The height of the cylindricl shells from x = 0 to x = 1 is constnt, but the heights diminish from x = 1 to x = 1.. How does this nswer compre to the nswers obtined in (1) nd ()? (4) The Yoplit Ellipticl Cylinder Continer -- Find the re of n ellipse Use clculus to determine the re of n ellipse with mjor xis nd minor xes b. x y Use the formul 1s the formul for the ellipse, nd integrte the function in b the first qudrnt from x = 0 to x =, nd then multiply by four. You will need to use the method of Trigonometric Substitution nd some trig identities. (5) The Yoplit Ellipticl Cylinder Continer -- Find the volume Now, using the formul obtined in (4), multiply by the height to get the volume of the cylinder. How does this nswer compre to the nswers obtined in (1), (), nd (3)?
Answer Key 1. Use similr tringles to solve for x: x.8 x 1 1. 1.x.8 x.x.8 x 14 Now determine the volumes of the two cones nd subtrct: 1 1 Volume of the truncted cone = R H r h 3 3 1 1 = 1. 16.8 1 14 3 3 = 5.33-14.66 = 10.673 cubic inches. First, solve for the eqution of the lterl edge of the cone on the right hnd side. Note tht it intersects the y-xis t (0, 16.8) nd tht it intersects the x-xis t (1., 0). Therefore, its slope is m = - 14, nd the eqution is y = - 14x + 16.8. Now, we wnt to rotte this line bout the y-xis from y = 0 to y =.8 (the top of the truncted cone). Note tht the disks hve thickness dy. Before setting up the integrl for the volume, you must first solve the eqution of the line for x in terms of y. y x 1. 14 yd y.8 y V g( y) dy 1. dy 10.673 cubic inches 14 yc y 0
3. In the Shell Method, rottion is still bout the y-xis but the function is in terms of x, nd thickness is therefore dx. Also, note tht this must be done in two steps becuse the heights of the shells re not found in the sme wy. From x = 0 to x = 1, the height of the shells is constnt.8 ; but from x = 1 to x = 1., the heights of the shells re from the x-xis to the line y = -14x + 16.8. xb xc V xy dx xy dx x x b x1 x1. V x.8 dx x 14x 16.8 dx x0 x1 V 8.796459 1.87657 10.673 cubic inches (4) To determine the re of n ellipse with mjor xis nd minor xes b, use the formul x y 1s the formul for the ellipse, nd integrte the function in the first b qudrnt from x = 0 to x =, nd then multiply by four. x y You must first solve the eqution 1for y in terms of x in the first qudrnt: b b y b x Now, find the re of the ellipse by integrting this function from x = 0 to x = : x b A f ( x) dx b x dx x0 x x0 x x0 x A b 1 dx
Now use the Method of Trig Substitution to integrte: Let x sin Then dx cos d x sin A b 1 cos d x0 x A b 1sin cos d x0 x A b cos d x0 To integrte, use the trig identity clled the double ngle formul: cos 1 cos x x 1 cos 1 1 A b d b cos d x0 x0 1 x sin sin cos 1 1 1 1 A b b Sin 4 4 0 0 1 1 x 1 x x A b Sin 1 A b 0 b0 0 1 b 4 Now multiply by four, to obtin the formul for the re of n ellipse: A b 0
(5) Now to determine the volume of the ellipticl cylinder, multiply the re of the bse by the height of the cylinder. V V V bh 1.51.05.6 10.7 cubic inches Six ounces of yogurt is equivlent to 10.83 cubic inches. In computtions #1,, nd 3, we obtined 10.673 cubic inches. In computtion #4 nd 5, we obtined 10.7 cubic inches. The error probbly lies in the mesurements tken, but ll computtions were close.