Question

Question 5: [Total 12.5 Marks] a) A cylindrical tank of depth 20 feet and cross-sectional area A of radius 5 feet full of water (density of water 62 lb/ft?). i. Calculate the work done using the following definite integral rule: (2.5) W *wAh. dh ii. What is the work done if the cylinder is only half full? (1.5) iii. Interpret your result of part (i) and part (ii). What did you conclude from these two results? (0.5) b) Using the given function x2 = y3 and x3 = y2 ,determine the volume when the region is rotating on x-axis. (8.0)

Answer 1

cylinder. height= : 20 ft Radius= 5 ft 3 20. density = 621b/it Take a small strip of height dh done is dw work : nadh S dw = distance & X density xvolume distance 20-W volume T (2)~ dh: ( 59*dh +255 ch dw = (20-6).62 (257) dhe for total work 20 20 1) w= J dw [ (20-6) (62)(257) dw 0 0 20 20 -15507 % (20-6) da = 1550 7 (204-br] isson [ 20 x 20 - mon)-0 2 Jo W Full - isson CS Scanned with . CamScanner

3.100001 in if water upto half depth = loft work done (20-h] (62) (257).dh 10 0 15.50T [ (10x10 ~ 40) - Whalf =775001 W 4x 775007 = 4 x Whalf full 0 four times • The work done for full of water times more than the work done here for half full. 5) Trays ps=y"_0 x.x=yv a. yozgu x y = 1 25 -ای CS Scanned with CamScanner

in put a av=y3 ýr = y3 : ys=1 ye 1 XI 1. also to yat (1) is intersection point and · also put (010). thuis satisfies both equations then Drawing rough sketch from [011] 6 h we get of yax: 2/3 volume of region 3/2 1 A - y = x notated about n-anis is a oto 1 v=1*, -yo") da - ( 4 ) 4/3 3 X 2 T 23) de. V = 0 CS Scanned-with CamScanner

4 -4 v=T a + 4 +) V= x 7/3 73 V=T 137 [G 6-0 6 - V= 3 4 V = 55 28 50 is Volume S ef: Cs Scanned with CamScanner