Volumes of solids Find the volume of the | Ch 7.4 - 45E

Problem 45EVolumes of solids Find the volume of the following solids.The region bounded by , y = 0, x = 1, and x = 2 is revolved about the x-axis.AnswerStep-1 In this problem we need to find the volume of the solid founded by rotating around in the region In order to find the volume, we will be using the following condition. If f is a function such that for all in the interval , then the volume of the solid generated by revolving, around the x axis, the region bounded by the graph of , the x axis (y = 0) and the vertical lines andis given by the integral Volume [ The radius for our cylinder would be the function f(x) and the height of our cylinder would be the distance of each disk ;dx The volume of each slice would be V = dx Adding the volumes of the disks with infinitely small dx would give us to the formula V = dx Here , f(x) = radius dx = height of each disk [a , b] = total height of a cylinder . ] Proper fraction definition ; In a rational fraction , if the degree of f(x) < the degree of g(x) , then the rational fraction is called a proper fraction. The sum of two proper fractions is a proper fraction. Example; Partial fractions Depending upon the nature of factors of Denominator ;1. When the denominator has non-repeated linear factors; A non - repeated linear factor (x-a) of denominator corresponds a partial fraction of the form , where A is a constant to be determined’ If g(x) = (x-a)(x-b)(x-c)............(x-n), then we assume that = ++ +...............+ Where A, B, C,............N are constants which can be determined by equating the numerator of L.H.Sto the numerator of R.H.S , and substituting x = a,b ,c ….n. Step-2 Consider We have to find the volume of the solid founded by rotating around x-axis ( i.e , y=0)in the region [1,2]Then Volume (V) The visual representation of the interval is given below. Consider = dx Thus … (1)To evaluate this integral let us use partial fractions.Step-3Let us first see about Partial Fractions. Consider = +…………….(2) = Thus , = Equating the numerator of L.H.S to the numerator of R.H.S Then , ( 1) = A(3-x) +B(x) 1 = (B-A)x+ (3A) B - A = 0 , and 3A =1.So , A = Then , B =A = Therefore , A = B = …………….(3) Therefore, from(2) and (3) = + = + , since from(2) = [+] Therefore , = [+] Step-4 Thus, from(1) becomes, V = = , = = [- ]+C = [ln(2) - ln(1) - ln(1)+ln(2)]+C = [2 ln(2)]+C (1.0471976)(1.3862943 1.45172413, since ln(2)= 0.69314718 , and = 3.14. Therefore , V = Thus volume of the solid generated is