Volume of a Sphere

Cavalieri's Principle will be used to show that the formula for the volume of a sphere holds. For this purpose, consider a hemisphere and a right cylinder with a cone removed from its interior, each with the same radius and height.

Now, consider a plane that cuts the solids at a height

x

and is parallel to the bases of the solids.

The area of each cross-section will be calculated one at a time.

Finding the Hemisphere's Cross-Sectional Area

Draw a right triangle with height $x,$ base $y,$ and hypotenuse $r .$ Here, $x$ is the distance between the center of the base of the hemisphere and the center of the cross sectional circle, $y$ is the radius of the cross sectional circle, and $r$ is the radius of the hemisphere.

Using the Pythagorean Theorem, an expression for

y

can be found.

r^{2} = x^{2} + y^{2}

▼

Solve for

y

SubEqn

$LHS - x^{2} = RHS - x^{2}$

r^{2} - x^{2} = y^{2}

RearrangeEqn

Rearrange equation

y^{2} = r^{2} - x^{2}

SqrtEqn

$LHS = RHS$

y = \pm r^{2} - x^{2}

Since

y

is a distance, only the principal root is considered. Therefore, the area

A_{H}

of the circular cross-section can be found using the formula for area of a circle. For consistency,

y

will be used in place of

r

in the standard formula.

A_{H} = π y^{2}

Substitute

$y = r^{2} - x^{2}$

A_{H} = π \cdot (r^{2} - x^{2})^{2}

PowSqrt

$(a)^{2} = a$

A_{H} = π (r^{2} - x^{2})

This equation gives the area of the cross-section of the hemisphere at altitude

x .

Finding the Cylinder's Cross-Sectional Area

The area of the cross-section of the cylinder can be found similarly. The cross-section's area is equal to the area between two circles. Since the height and the radius of the cylinder are equal, an isosceles right triangle can be formed inside the cylinder. Therefore, the radius of the smaller circle is also

x .

Now that the radii of the circles are known, the area

A_{C}

of the cross-section can be calculated. It is the difference between the area

A_{G}

of the greater circle and the area

A_{S}

of the smaller circle.

A_{C} = A_{G} - A_{S}

SubstituteII

$A_{G} = π r^{2}$ , $A_{S} = π x^{2}$

A_{C} = π r^{2} - π x^{2}

FactorOut

Factor out $π$

A_{C} = π (r^{2} - x^{2})

The area of the cross-section of the cylinder at altitude

x

can be found by using the above equation.

Conclusion

It can be stated that both solids have the same cross-sectional area at every altitude.

A_{H} = π (r^{2} - x^{2}) = A_{C}

Moreover, they have the same height. By Cavalieri's Principle, the hemisphere and the cylinder with a cone removed from its interior have the same volume.