Indentify the radius and height of each shell and compute the volume: The region
ID: 3287685 • Letter: I
Question
Indentify the radius and height of each shell and compute the volume: The region bounded by x^2 + y^2 = 2y, revolved about y = 4. Please show all work, I know what the answer is, just not how to work: V = (integral)(0,2) 2 pi (4 - y) (sqrt(2y - y^2) dy.Explanation / Answer
The final answer is --- V= 6(pi)^2 --- There are not many calc problems which lead to (pi)^2 . One of them is the volume of a torus --- which is this problem . They want the volume when the area within the curve in the x y plane --- x^2 + y^2 =2y is revolved about y = 4 . We can complete the square on y for x^2 + y^2 =2y : x^2 + y^2 -2y =0 x^2 + y^2 -2y +1 =1 which is the same as : x^2 + (y-1)^2 = 1 . This is a circle of radius r = 1 with center at (0,1) , and its revolved about y = 4 . The solid generated when the area within a circle in a plane is revolved about a line ( in the plane ) which does not pass through the circle is a torus . Anyway , the limits on y are from --- y =1-1=0 to y =1+1 =2 . We can solve x^2 + y^2 =2y for x ----> x = + or - sqrt( 2y - y^2) . The radius of the cylindrical shell is : R=4-y ----- . The height of the cylindrical shell is H = sqrt(2y-y^2) - - sqrt(2y -y^2) = 2sqrt(2y -y^2) ---- . The differential of volume is --- dV =2(pi)RHdy = 2(pi)(4-y)(2sqrt(2y - y^2))dy ; or , we have : dV = 4(pi)(4-y)sqrt(2y - y^2)dy . We integrate from y = 0 to y =2 to find the volume : V = integral(0 to 2) 4(pi)[ 3 + (1-y)]sqrt (2y - y^2)dy (1) We can compute this integral in two parts --- V = V1 + V2 with : V1 = 12(pi)integral(0 to 2)sqrt(2y - y^2)dy and V2 = 4(pi)integral(0 to 2)(1-y)sqrt(2y - y^2)dy . We have V1 = 6(pi)^2 and V2 = 0 ------ email me at pk2712@aol.com and I will email you hand written solutions to these two integrals as pdf attachments to two emails .