Daniel B. answered • 11/15/21
A retired computer professional to teach math, physics
I suggest that you draw a picture of the situation.
Since both surfaces are symmetrical in x and y, on a plain piece of paper
you can draw the situation in the x-z plain.
Do that by setting y=0, which yields the two curves
x² + z² = 16
z = √(3x²) = |x|√3
The first is a circle of radius 4.
The second is a wedge whose edges are formed by the two lines
z = x√3 and z = -x√3 restricted to the first and second quadrant (i.e., z >= 0).
First we need to find the intersection of the circle and the lines.
Do that by substituting z = x√3 into x² + z² = 16 getting
x² + 3x² = 16
Solving
x = 2
z = 2√3
In the three dimensional space, the plane defined by z = 2√3
cuts the "ice cream cone" into
"ice cream", which is a part of the sphere, and "cone".
We find the volume of the two pieces separately.
Lets first do the "cone".
We can see its volume as sliced into many horizontal disks.
For each value of z between 0 and 2√3 we get a disk of radius x = z/√3,
and area π(z/√3)² = πz²/3.
Imagine that each disk has a very small thickness dz, so small that we can
ignore the difference in radius between the top and bottom.
The volume of each disk is then πz²/3 dz.
We can get the volume of the cone by adding the volumes of all the thin disks
∫πz²/3 dz
where the integral is between the bounds of 0 and 2√3.
The indefinite integral is πz³/9, so the definite integral is 8π√3/3.
Now let's consider the "ice cream" portion.
There z ranges from 2√3 to 4.
For each value of z we get a disk with radius √(16-z²) -- using Pythagorean theorem.
By the same reasoning as for the "cone", the volume of the "ice cream" is
∫π(16-z²)dz
where the integral is between the bounds 2√3 to 4.
The indefinite integral is πz(16-z²/3).
Therefore the definite integral is
4π(16-4²/3) - 2π√3(16-4×3/3) = 128π/3 - 24π√3
Thus the whole ice cream cone has the volume
8π√3/3 + 128π/3 - 24π√3 = 64π(2 - √3)/3
