Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 59

To verify the volume of a right circular cone, we consider the radius of the base (r) as an interval along the x-axis and height (h) as an interval along the y-axis. As shown in the attached image, a red line revolves about the y-axis to form a right circular cone. For the equation of the red line, we consider the points: (0,h) and (r,0) where: x_1= 0 , y_1=h , x_2=r , and y_2=0 .

The point (0,h) is a y-intercept point therefore it follows (0,b) then b =h in y=mx+b .
To solve for m, we follow m = ((y_2-y_1))/((x_2-x_1)) .
m= ((0-h))/((r-0)) = -h/r
Then plug-in m= -h/r and b = h, we get the equation of the red line as: y =-h/rx+h .
This can be rearrange into x = -(y-h)*r/h or x= ((h-y)r)/h .
Using the Disk Method, we consider a rectangular strip perpendicular to the axis of revolution.
For a horizontal rectangular strip with a thickness of "dy", we follow the formula for Disk Method as: V = int_a^b pi r^2 dy .
To determine the r, we consider the length of the rectangular strip = x_2-x_1 .
Then, r= ((h-y)r)/h - 0 = ((h-y)r)/h .
Boundary values of y: a=0 to b=h .
Plug-in the values on the formula: V = int_a^b pi r^2 dy , we get:
V = int_0^h pi (((h-y)r)/h)^2 dy
V = int_0^h pi (r^2/h^2)*(h-y)^2dy
Apply basic integration property: int c*f(y) dy = c int f(y) dy .
V =( pir^2)/h^2 int_0^h (h-y)^2 dy
To find the indefinite integral, we may apply u-substitution by letting u = h-y then du = -dy or (-1)du = dy .
V =( pir^2)/h^2 int (u)^2 *(-1)du
V =( -pir^2)/h^2 int (u)^2 du
Apply Power rule for integration: int y^n dy= y^(n+1)/(n+1) .
V =( -pir^2)/h^2* u^(2+1)/(2+1)
V =( (-pir^2)/h^2)* u^3/3
Plug-in y = h-y on (( pir^2)/h^2)* u^3/3 , we get:
V =(( -pir^2)/h^2)* (h-y)^3/3|_0^h
Apply definite integration formula: int_a^b f(y) dy= F(b)-F(a) .
V =((- pir^2)/h^2)* (h-h)^3/3-((- pir^2)/h^2)* (h-0)^3/3
V =(( -pir^2)/h^2)* (0)^3/3-(( -pir^2)/h^2)* (h)^3/3
V =0 -(( -pih^3r^2)/(3h^2))
V = 0 +(pih^3r^2)/(3h^2)
V =(pih^3r^2)/(3h^2)
V = (pihr^2)/3 or 1/3pir^2h

Note: Recall the Law of Exponent: y^n/y^m= y^((n-m))
then h^3/h^2= h^((3-2)) = h^ 1 or h .