Question

Determine the areas in mm2 of the surfaces generated by revolving the bent wire ABC (shown) one revolution about the following axes. y 85 mm 102 mm B 68 mm X 255 mm (a) the x-axis mm2 (b) the y-axis mm2

Answer 1

There are many ways to solve this problem , here I have solved using simple geometry and standard results

Important note 1) Surface area of cone TRL AR √ R²242 RR- 2 ) Surface area olo frustum ob cone Ka dy a (&+R) L h to à Ce+R) / 2²+ (R-8)² 11 - R Scanned with CamScanner

When lines are revolve around either form a cone OR a axis it will frustum (7) for it. When line AB revolves around n-axis example will form cone Radius of base= 68 255 . Height of Its A. ... Surface aree AR R² +62 56350.27 mm² 255 () When line Al is revolved around_2-9xis it will form a cone Radied off bases 170 85. 190 Height of cone = Surface area = AR JR²442 = 101457.1123 mm? 85 R-170 &= 68 h=170 R=1761 iii) When line (B is revolve Ground x-axis it will som frustum of cone AB+r) F (R-8) +h² J (170-68)2+(170)2 148157. 957 ) {= 68 kl 265-85 = ñ (170768 CS scanned with CamScanner

-- Area of surface when lines are revolved around acaxi- 56 350.27 + 101451:1123 + 148151:951 305965. 3393_mom? 2) Similarly, when eine AB revelved around Y-axis. it will form a cone Area ARR4h² 211313.5135 mm? 255mm th=68mm А. is when line LA85 AC is revolved around X-axis. it will form a cone .: Area = TR JR²+h² ·50728.55617 mm? 1 170 When line BC is revolveel around Y-axis will form a frustum IZ 854 Area = A(&+R) JR-2)² + H2 KT 102 211654.2243 mm² 255 :: Area of Suface when lines tevolved around y-axis 50728. 55617 + 211654.2243 CS$canned with cams6173.696.2939 mm2, 211313.5135 +

Area of suface when wire is revolved around

X-axis = 305965.3393 mm2

Y-axis = 473696.2939 mm2