See attachment
Yes, I have searched the forums and google but nothing quite yet what I was looking for.
Problem:
A water tank in the shape of a cone is underground with the distance above ground being called D, the base of the cone is horizontal with the ground ie pointing straight up, The height in feet of the tank is called H, which has a base radius r in feet, the tank is filled up to some point h. Set up the integral to calculate the total work done. No calculations.
My assessments
The equation to find the volume of my slice obviously would be:
The distance to move the slice is to go from the current level to the top of the tank to the distance above ground thus giving
D being distance above ground, H height of the tank, and h being current distance from top of the tank at the water level
Next I thought of my vertical being of h and the change in height was of dh, and thought of my horizontal axis r, radius. where 2r(diameter) is the width of the cone, this is perhaps where I confused my self not placing it on a x-y coordinate plan to make some sort of distinction. Even If i had placed it on a x-y plane the, apex, or peak would be at some value of y, I always noticed my radius could be at some point x or y, which usually results in the equation of the slope of the line r/h * x, but this when people place one of the cone's half horizontal to the x-axis, so the slope can be positive.
In general in need some help sorting out this problem, it's been bugging me such a simple concept can get so confusing.
Except you see, the cone is inverted, as the base is on the "x-axis" so as you lift each piece out of the cone the radius will increase as to where, in your picture as you lift each slice out the radius gets smaller and smaller, that's what threw me off. Perhaps I am over complicating things, and if even the cone is inverted the the function of the radius/height remains the same? even if radius increasing when I think about it