Bob

]]>volume of triangle inscribed in a sphere

triangle??

Do you mean cone?

Bob

]]>Let's see if I can 'talk' my way up this tower.

(i) At the bottom it has radius 2R and this tapers to a radius of R over a distance of height R

(ii) Then it becomes a cylinder. I agree with your formula for that.

(iii) At the top it is half a sphere, radius R.

So now for the formulas:

I'll do the radius at each stage. You can convert that to the area of a circle (they all have circular cross sections)

(i) This is a linear function as the taper is regular. I'll use y for the radius at any point

At x = 0, y = 2R. At x = R, y = R. So if you graphed this the gradient would be -1 (R across is R down * that's putting the tower on its side so that x is across as usual and y up. I've added a diagram to show this)

y = 2R -x

(iii) Here the x and y coordinates lie on the sphere so you can use Pythagoras to write an equation linking them:

(x-3R)^2 + y^2 = R^2

Re-arrange to make y the subject:

y = √(R^2 - [x-3R]^2)

Hopefully that should do it.

Bob

ps. Looks a bit like a Dalek.

]]>define the function.

i know that the function is (pie)*R^2 for the interval R≤ x≤ 3R

but what about the other intervals and yeah the domain of the function is is from 0 to 4R. i am facing problems with relating the area of the cross section with the radii and x of both the cone and semisphere help me please!]]>