Find the volume of a solid generated when the region bounded by the graphs of y=sqrtx, y=0 and x=4 is revolved about x=4 using a horizontal and vertical component for each  delta y and delta x.

2nd problem is the same except it is rotated around y=2

1. The integral can be considered to be the sum of the volumes of discs parallel to the x axis. The radius of each disc is 4-x because the widest disc has radius 4, when x=0, reducing to zero as x approaches 4. For example, at x=1 the radius is 3. The thickness of each disc is dy so the integral is (pi)S((4-x)^2dy, where S denotes integral. y=sqrt(x) so y^2=x and the integral is (pi)S((4-y^2)^2dy) between limits for y of 0 to 2, written [0,2]. The integral in y becomes:

(pi)S[0,2]((4-y^2)^2dy)=(pi)S[0,2]((16-8y^2+y^4)dy)=

(pi)(16y-8y^3/3+y^5/5)[0,2]=

(pi)(32-64/3+32/5)=256(pi)/15=53.62 cu units approx.

2. This time the discs are vertical with thickness dx and radius 2-y, where x goes from 0 to 4 (y ranges from 0 to 2). The integral is (pi)S[0,4]((2-y)^2dx)=

(pi)S[0,4]((4-4y+y^2)dx)=

(pi)S[0,4]((4-4sqrt(x)+x)dx)=

(pi)(4x-2x^-(1/2)+x^2/2)[0,4]=

(pi)(16-1+8)=23(pi)=72.26 cu units approx.

by Top Rated User (695k points)