find the volume of the region bounded by y 5 y 036x42 and

Solution

At y=2 , x=4 , your volume is in range x=0 and x=4 V= Area h , if you make Little\'s h , called dx , between x=0 and x=4 , you obtain Little\'s dV dx = (4-0)/N , N the number of intervals , greater is better.- Suppose N=10 ,then dx = 4/10 = 0.40 .- the first number to evaluate is x=0 , the second is x+dx =0+0.40 =0.40 .- Between both you must take the average, ie, (0.40+0)/2 =0.20 .- Calculate dV1 =pi f(0.20)^2 (0.40) =0.2513 Calculate dV2 , (x+dx) +dx ( ie, the last x , +dx) = 0.40+0.40 =0.80 , the average is (0.80+0.40)/2 = 0.60 ( Or 0.20 +0.40 ( center to center of intervals) dV2= pi f(0.60)^2 (0.40) = 0.75396 x+dx = 0.80 +0.40 =1.20 Average = (1.20+0.80)/2 =1.0 ( Or 0.60+0.40) dV3= pi f(1)^2 (0.40) = 1.2566 x+dx = 1.20 +0.40=1.60 avg = (1.60+1.20)/2 =1.40 dV4= pi f(1.40)^2 (0.40) =1.7584 and so on until x+dx =4 ( The last limit) if you want to calculate the area of revolution you use the shell method, similar , you only change the formula A= 2pi f(x) ds ds = distance between points , sqrt( (f(x+dx) -f(x))^2 +(dx)^2 ) First point , x=0 , second x+dx= 0.40 , dx =0.40 ds= sqrt( (0.63-0)^2 +0.40^2) =0.748 A1= 2pi f(0.20) 0.748 = 2.10 x+dx = (0.40+0.40) =0.80 , avg = (0.80+0.40)/2 =0.60 ( use the same points that in V) A2= 2pi f(0.60) ds , ds = sqrt( (f(0.80)-f(0.40))^2 +0.40^2 ) =0.4748 A2= 2.310 .... after sum all dA and you are ready.-