Question

Determine the total surface area in mm2 and the volume in mm3 of the solid of revolution. (Let 50 mm be the difference between the inner and outer radii at the top of the solid.) 50 mm 45 mm 75 mm 25 mm A = mm 2 mm3

Answer 1

TO Sol: Determines the total surface area and the volume of the solid of revolution, som 450 Curved surface area of outer Frustum (TA, CT2+T4)) S da = √(re-r33"thr r2 = 50+ 45=95mm 75mm dos Jas-25)*7 75" - 102.39 mm 250m di = VE-Y3)'th Area = 1 = √(45-25) + 750 =177.620 mm - The Cran) 17h (e'tr't RX1) Volume = Curved Surface area of outer frustum % (102.59) (95-+25)) 19337.75 ma' Curved surface area of Inner =% (IT (77.62) (45+25)) = 185 34.764 mm) Frustum Shaded area: At top = [X THY = [10995.57 mm 2 s= atbre 2. a+bro 50+ 77.620+102:59 115.105 mm . 2 . Area of Triangle 3s-le (s-4-se) fle- 70-429) (s. 10 – 219) . 3746.75 mm Total surface area 19337.239+8534.764+10995527 3748.759 Area 42616.835 mm Joral Volume = % (TT(02+ ***+2x2x)) – Y Tu (r+r3+rir3) } ({ 11x75(95 T1X75 (951725785x25)) - (TTX,75 (45°+25°+(45825)) Total 323976736 mm 2 42616.835 mm Asea a volenne a 323976.736 mm