Question

Determine the surface areas in in2 and the volumes in in of the bodies of revolution obtained by revolving the bright у 3.6 in 1.8 in X 3.6 in 3.6 in (a) the y-axis in 2 in V- (b) the x-axis A- in 2 V - in3

Answer 1

3.6 Esch L 7,2 Esch 1.8 inch calculate we recolve a one way to solve it is to find out the equation of the above fegure e integrate it. But as this figure is consist of a rectangle and semicurcle, we can Area and volume manually lets revolve this figure along Y-Y axis, As we know that, if rectangle, et will create a cylinder & semiurcle will create sphere. Dimension of Generated cylinder raduis (ri & Height (h) = 3-6 inch & Dimension of sphere So A = Area of cylinder - Area of sphere -Area of a ů so, 7-2 5 ch ů 1.8 Sch. semisphere 2 2 gar (ren) kar? sar? 2*** 7-2 (1-2+3.6) - 4***1-8?- 3x^x 1.82 417.32 6 27 7 A

va Volume of cylinder ar²n volume of sphere - volume of Shemisphere 3. 3 460 ar Ar 6 3 3 ax7.2²3.6 ax1.83 X7X **** 118 U V = 577-135 s 3 2) when we revolve it around x-x it will create a hollow cylinder of inner rodias 1.8cm , outer radius 5.4 inch 4 two hemisphere of raduis 1.8 in. Height of cylinder is 712 inch [an ho (Roth) - da Ri (Ri+n)]- Bari -sar? [20X5-465-4472) - 20x1.8 (1-877-2)] – 2*3*** 1.99 A = 2 A 2 A = 264.64 u 3 2 R; ²) h v [ * (Ro - Ri 3 3 A xps - A Ars a (s.y?-1.8°) 1.2 - fxxuos- a tolva)'] 2 3 v z 561.86 in 2 A tinal Aus V a) y-axis 417.32 577.135 in 3 b) x-arů A = 264-64 264-64 & 2 561-86 å 3 V