Use disks to find the volumes of the generated solid below.

Question

The curve y = √(r² -x²) (for any positive real number, r) is revolved around the x-axis. Use disks to find the volumes of the generated solid. What is the significance of this question?

If possible please explain the step by step process in a video.

Answer 1

V = π∫_-r^r(r² - x²)dx = 2π(r²x - x³/3)₀^r = 2π(r³ - r³/3) = 4πr³/3

Answer 2

The region under consideration is bounded by the circle y= sqrt(r^2-x^2) i.e. part of circle above x-axis intersecting x-axis at x=r & x=-r. It is revolved about x-axis. The solid formed will be sphere of radius r. The region is symmetrical about y-axis.Consider the region on the right of y-axis, bouned by x=0 to x=1. Consider the elementary strip of height y and thickness delta x (from x-axis to Circular boundary). Delta V= Pi(y^2.delta x)

Required Volume V =2 Int.(pi.y^2 )dx from x=0 to x=r

=2. Int. [pi.(r^2 -x^2)dx (x= 0 to r) =2. pi[r^2 x - (1/3)x^3](x=0 to r)

=2.pi.[r^3 - (1/3)r^3 -0 + 0] = (4/3).pi.r^3