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...

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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` .