Volume of Frustum (Truncated Cone)

[silvashadow]

If there is a frustum with base radii of 1.75 and 1.25 inches, and a height of 6 inches, what is the volume? I tried to use the V=|(b1*h1)/3-(b2*h2)/3)| from the Wikipedia page, but h2 is unknown. I get an answer of 42.41 in^3. Is this correct?

Please use basic calculus as that is all I have learned.

 

[Vid]

Think of the circles x^2+y^2 = (1.75)^2 z=0 and x^2+y^2 = (1.25)^2 z=6. Left y = 0. This gives us two points from our equations (1.75, 0, 0) and (1.25, 0 ,6). Now we make a parametric equations for the line going through these two points. v = (-.5,0,6)
r = Po + t*v
x = 1.75 - .5t
y = 0
z = 6t

We want x and y to be zero so we can solve for z. 1.75 = .5t t = 3.5

z = 6*(3.5) = 21. So the total height of the cone if it weren't missing anything would be 21. Find the volume for that cone then subtract out the volume of the cone with height 14 and base 1.25.

 

[silvashadow]

Think of the circles x^2+y^2 = (1.75)^2 z=0 and x^2+y^2 = (1.25)^2 z=6. Left y = 0. This gives us two points from our equations (1.75, 0, 0) and (1.25, 0 ,6). Now we make a parametric equations for the line going through these two points. v = (-.5,0,6)
r = Po + t*v
x = 1.75 - .5t
y = 0
z = 6t

We want x and y to be zero so we can solve for z. 1.75 = .5t t = 3.5

z = 6*(3.5) = 21. So the total height of the cone if it weren't missing anything would be 21. Find the volume for that cone then subtract out the volume of the cone with height 14 and base 1.25.

Thanks.

 

[Vid]

I just realized a much simpler solution using similar triangles.

6/.5 = x/1.75 x = 12*1.75 = 21.

 

[silvashadow]

I just realized a much simpler solution using similar triangles.

6/.5 = x/1.75 x = 12*1.75 = 21.

Yea, that what I ended up doing.
x/1.25=(x+6)/1.75

 

[ssd]

V=(h/3)*[A1+A2+sqrt(A1*A2)],
h = height, A1 and A2 are areas of the top and bottom circles.
It is not tough to find the formula using simple definite integral.