The accompanying figure shows a spherical cap of radius ρ and height h cut from a sphere of radius r. Show that the volume of the spherical cap as below using the concept of integral calculus.

Question

[hint: revolve an appropriate portion of the circle x^2+y^2=r^2 about the y-axis.]

a.) V=1/3*πh²(3r-h)

b.) V=1/6*πh(3p²+h²)

Answer 1

One way you can derive either formula is by starting with an integral

The equation of the sphere if you center it at the origin is x² + y² + z² = r².

cross section at a given x is a circle with radius R(x) = √(r² - x²), giving a cross-sectional area π(r² - x²)

The cap is traced by the range r - h ≤ x ≤ r because it has height h.

We can now get the formula first formula.

V=∫^r_r-h π(r² - x²) dx = π(r²x - x³/3)|^r_r-h = π(r³ - r³/3) - π(r²(r-h) - (r-h)³/3)

= 2πr³/3 - π(r³ - r²h - r³/3 + r²h - rh² +h³/3)

= 2πr³/3 - π(2r³/3 - rh² +h³/3)

= 2πr³/3 - 2πr³/3 + πrh² - πh³/3

= πrh² - πh³/3

= πh²(3r - h)/3

For the second formula, note that ρ = R(r-h) = √(r² - (r-h)²) = √(2rh-h²).

So 2rh-h² = ρ², giving us r = (ρ² + h²) / (2h).

V = πh²(3(ρ² + h²) / (2h) - h)/3

= πh²(3ρ² + 3h²) / (2h) - 2h² / (2h))/3

= πh²(3ρ² + h²) / (2h))/3

= πh(3ρ² + h²) / 6