Given the function
[tex]y=\frac{1}{4}x[/tex]
the vertical line is given by the point
( x , 0 )
the area of the triangle is
[tex]At=\frac{b*h}{2}[/tex]
where
base= value of x
height=value of y
then
the expression of the area as a function of x is
[tex]At=\frac{x*\frac{1}{4}x}{2}[/tex][tex]At(x)=\frac{x^2}{8}[/tex]
Then when x=5
[tex]At(5)=\frac{5^2}{8}[/tex][tex]At(5)=\frac{25}{8}=3.125un^2[/tex]
The volume of the cone is given by
[tex]Vc=\frac{1}{3}\pi *r^2*h[/tex]
where
h= value of x
r= value of y
then the volume of the cone as a function of x is
[tex]Vc=\frac{1}{3}\pi *(\frac{1}{4}x)^2*x[/tex][tex]Vc=\frac{\pi x^3}{3*4^2}[/tex][tex]Vc(x)=\frac{\pi x^3}{48}[/tex]
The volume of the cone when x=16 is
[tex]Vc(16)=\frac{\pi(16)^3}{48}[/tex][tex]Vc(16)=\frac{256\pi}{3}[/tex][tex]Vc(16)=\frac{256\pi}{3}=268.082un^3[/tex]
