Area of a Spherical Lune

Consider a two-dimensional view from the top of the spherical lune.

Let

L (α)

be the area of the spherical lune. It should be noted that if

α = 36 0^{\circ},

L (α)

is equal to the surface area of the sphere. Therefore, it is possible to relate the formula for the surface area of a sphere and

L (α)

for other angles writing a proportion.

\frac{L ( α )}{α} = \frac{4 π r ^{2}}{3 6 0}

This proportion can be solved to find the value of

L (α) .

\frac{L ( α )}{α} = \frac{4 π r ^{2}}{3 6 0}

▼

Solve for

L (α)

$LHS \cdot α = RHS \cdot α$

L (α) = \frac{4 π r ^{2}}{3 6 0} \cdot α

$\frac{a}{b} = \frac{a / 4}{b / 4}$

L (α) = \frac{π r ^{2}}{9 0} \cdot α

L (α) = \frac{π r ^{2} α}{9 0}

If the angle is given in radians the reasoning is the same, but there is a change. Instead of considering the total circumference angle as

36 0^{\circ},

the total circumference angle is

α = 2 π .

This changes the formula as follows.

\frac{L ( α )}{α} = \frac{4 π r ^{2}}{2 π}

$\frac{a}{b} = \frac{a / 2 π}{b / 2 π}$

\frac{L ( α )}{α} = 2 r^{2}

$LHS \cdot α = RHS \cdot α$

L (α) = 2 r^{2} α

{{ article.displayTitle }}

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount }}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount }}
	{{ 'ml-lesson-time-estimation' \| message }}