Question

please help

Determine the surface area in in?, volume in in, and weight in pounds of the brass doorknob shown. Brass has a specific weight of 0.315 lb/in. 0.96 in -0.96 in

Answer 1

y=064 0.96 0.26 - - - - - Region I Given doorknob can be generated by Rotaling Carme (4B about x axis Equation of Cune (4 Regint (x - 2)² + (y=0-692 (09672 equation of Curre AB (X-896)2+(9-10)? = (696)2 Consider two seperate regions I, I Surface Area (1) Consider a disk of length de at a distance x from the origin Outer Surface area of disk = Perimeter x da - 200 de - 2 1176.94)? (-0-96)2 + 0 Toral Surface Area of region I A = 2 0962 - (-:98) + 16 ) dx +0261.6 da = 21 x 2.25982 147 451964T) (mus

I Similarly hatten Region & A = 21 fr da. + $0.96) 2 - x & Jon. -0.96 = 25 x 1.33822 A= 2.67644T Total Surface Area of dark knob A A A I A = (4.51964 + 2-67646) (A = 7.19608 TT (inches) T (2) Volume Volume for Region II (Disk method). V ST 12 dx. 0.96 (1090) (2 0 332 +15) de = T 5.36366 - 5.36366 T ( inches

Volume for I = T region. I T 12 de - o.ge T P (ocht! (0.96)? x2 je de -0.96 - π 1.90953 V = 1.909535 (in Toral Volume of doorlends = VIIVI = 1.509537 + 5.36366 V = 4.27319 T (ing weight of door knon W = Specific weight = 0.315 is x 7.273191 (w = 2.29105 485 us in 3

equation of line AC 10.6 5.3 2 y = 2x equation of line 36 q-o) = 10'6-0 / y = (-1)(x - 15:9) y tn= 15.9 Consider a Cylindrical shell of thickness dy at a distance y from the origin. Radius (r)=y height = (15.9 -9) - Volume of shell over lenfee Area & dy - 215rh dy or= 24 y (159 - 39 dy

Total volume of region Ve dr 5.3 of try (159 - 2g, dy = 2 ( 148.877) = 297.754 π T in- y = 2 Consider a disk of thickness dy at a distance 2= (15.gg) y from dingin Required volume of dite ar= T (12 - 13 dy. T (15.9-7 ² - ( 18 day Total volume of solid VETS (15.9-y? Gdy. = T (930.481) IV = 930.481T in