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You can go slightly beyond the positive orthant. In the case of a 2-sphere in 3-dimensional space, you can render a cap with angular radius $45°$. This has a surface area of $2\pi(1-\sqrt{1/2})\approx(2\pi)\color{blue}{(0.293)}$ steradians, whereas the positive orthant has a surface area of $\pi/2=(2\pi)\color{blue}{(0.25)}$ steradians.

Although the cap is larger in area, the positive orthant cannot fit inside it. This is essentially a spherical analogue of the planar geometry result that a circle of unit diameter is larger than an equilateral triangle of unit side (or, for that matter, any regular odd-sided polygon whose longest chord measures one unit), but the triangle/odd-gon does not fit in the circle.

The counterpart to the cap also beats the positive orthant in spatial dimensions greater than 3; in fact the ratio between the two eventually becomes arbitrarily large with sufficiently many dimensions.

Let a $(d-1)$-sphere in $d$-dimensional space be defined by

$\sum\limits_{i=1}^d x_i^2=1$

The positive orthant is then given by $\min(x_i)\ge0$, while the cap may be defined by $x_1\ge\sqrt{1/2}$. In the latter case two antipodal points $(\sqrt{1/2},x_2,x_3,...x_d)$ and $(\sqrt{1/2},-x_2,-x_3,...-x_d)$ on the boubdary of the cap give the inner product

$1/2-\sum\limits_{i=2}^d x_i^2=0,$

where the equation for the sphere combined with defining $x_1=\sqrt{1/2}$ imply the second equality.

The fraction of the sphere covered by the positive orthant is then $2^{-d}$, while the fraction covered by the cap is given by a ratio of integrals:

$\dfrac{\int_{\sqrt{1/2}}^1(1-x^2)^{(3-d)/2}dx}{\int_{-1}^1(1-x^2)^{(3-d)/2}dx}.$

With $d=3$ this gives (to four decimal places) $0.1464$ for the cap versus $0.1250$ for the orthant. With four-dimensional space the corresponding fractions are $0.1051$ for the cap and $0.0625$ for the orthant. As $d\to\infty$, the integrals in the fraction for the cap become Laplace integrals and can be evaluated asymptotically by the appropriate method for such integrals; the result is that the fraction covered by the cap has the controlling $(\sqrt2)^{-d}$ versus $2^{-d}$ for the positive orthant. As $\sqrt2<2$, the cap ultimately becomes exponentially larger than the positive orthant.

You can go slightly beyond the positive orthant. In the case of a 2-sphere in 3-dimensional space, you can render a cap with angular radius $45°$. This has a surface area of $2\pi(1-\sqrt{1/2})\approx(2\pi)\color{blue}{(0.293)}$ steradians, whereas the positive orthant has a surface area of $\pi/2=(2\pi)\color{blue}{(0.25)}$ steradians.

Although the cap is larger in area, the positive orthant cannot fit inside it. This is essentially a spherical analogue of the planar geometry result that a circle of unit diameter is larger than an equilateral triangle of unit side (or, for that matter, any regular odd-sided polygon whose longest chord measures one unit), but the triangle/odd-gon does not fit in the circle.

The counterpart to the cap also beats the positive orthant in spatial dimensions greater than 3; in fact the ratio between the two eventually becomes arbitrarily large with sufficiently many dimensions.

Let a $(d-1)$-sphere in $d$-dimensional space be defined by

$\sum\limits_{i=1}^d x_i^2=1$

The positive orthant is then given by $\min(x_i)\ge0$, while the cap may be defined by $x_1\ge\sqrt{1/2}$. In the latter case two antipodal points $(\sqrt{1/2},x_2,x_3,...x_d)$ and $(\sqrt{1/2},-x_2,-x_3,...-x_d)$ on the boubdary of the cap give the inner product

$1/2-\sum\limits_{i=2}^d x_i^2=0,$

where the equation for the sphere combined with defining $x_1=\sqrt{1/2}$ imply the second equality.

The fraction of the sphere covered by the positive orthant is then $2^{-d}$, while the fraction covered by the cap is given by a ratio of integrals:

$\dfrac{\int_{\sqrt{1/2}}^1(1-x^2)^{(3-d)/2}dx}{\int_{-1}^1(1-x^2)^{(3-d)/2}dx}.$

With $d=3$ this gives (to four decimal places) $0.1464$ for the cap versus $0.1250$ for the orthant. With four-dimensional space the corresponding fractions are $0.1051$ for the cap and $0.0625$ for the orthant. As $d\to\infty$, the integrals in the fraction for the cap become Laplace integrals and can be evaluated asymptotically by the appropriate method for such integrals; the result is that the fraction covered by the cap has the controlling factor $(\sqrt2)^{-d}$ versus $2^{-d}$ for the positive orthant. As $\sqrt2<2$, the cap ultimately becomes exponentially larger than the positive orthant.

You can go slightly beyond the positive orthant. In the case of a 2-sphere in 3-dimensional space, you can render a cap with angular radius $45°$. This has a surface area of $2\pi(1-\sqrt{1/2})\approx(2\pi)\color{blue}{(0.293)}$ steradians, whereas the positive orthant has a surface area of $\pi/2=(2\pi)\color{blue}{(0.25)}$ steradians.

Although the cap is larger in area, the positive orthant cannot fit inside it. This is essentially a spherical analogue of the planar geometry result that a circle of unit diameter is larger than an equilateral triangle of unit side (or, for that matter, any regular odd-sided polygon whose longest chord measures one unit), but the triangle/odd-gon does not fit in the circle.

The counterpart to the cap also beats the positive orthant in spatial dimensions greater than 3; in fact the ratio between the two eventually becomes arbitrarily large with sufficiently many dimensions.

Let a $(d-1)$-sphere in $d$-dimensional space be defined by

$\sum\limits_{i=1}^d x_i^2=1$

The positive orthant is then given by $\min(x_i)\ge0$, while the cap may be defined by $x_1\ge\sqrt{1/2}$. In the latter case two antipodal points $(\sqrt{1/2},x_2,x_3,...x_d)$ and $(\sqrt{1/2},-x_2,-x_3,...-x_d)$ on the boubdary of the cap give the inner product

$1/2-\sum\limits_{i=2}^d x_i^2=0,$

where the equation for the sphere combined with defining $x_1=\sqrt{1/2}$ imply the second equality.

The fraction of the sphere covered by the positive orthant is then $2^{-d}$, while the fraction covered by the cap is given by a ratio of integrals:

$\dfrac{\int_{\sqrt{1/2}}^1(1-x^2)^{(3-d)/2}dx}{\int_{-1}^1(1-x^2)^{(3-d)/2}dx}.$

With $d=3$ this gives (to four decimal places) $0.1464$ for the cap versus $0.1250$ for the orthant. With four-dimensional space the corresponding fractions are $0.1051$ for the cap and $0.0625$ for the orthant. As $d\to\infty$, the integrals in the fraction for the cap become Laplace integrals and can be evaluated asymptotically by the appropriate method for such integrals; the result is that the fraction covered by the cap has the controlling $(\sqrt\pi)^{-d}$ versus $2^{-d}$ for the positive orthant. As $\sqrt\pi<2$, the cap ultimately becomes exponentially larger than the positive orthant.

You can go slightly beyond the positive orthant. In the case of a 2-sphere in 3-dimensional space, you can render a cap with angular radius $45°$. This has a surface area of $2\pi(1-\sqrt{1/2})\approx(2\pi)\color{blue}{(0.293)}$ steradians, whereas the positive orthant has a surface area of $\pi/2=(2\pi)\color{blue}{(0.25)}$ steradians.

Although the cap is larger in area, the positive orthant cannot fit inside it. This is essentially a spherical analogue of the planar geometry result that a circle of unit diameter is larger than an equilateral triangle of unit side (or, for that matter, any regular odd-sided polygon whose longest chord measures one unit), but the triangle/odd-gon does not fit in the circle.

The counterpart to the cap also beats the positive orthant in spatial dimensions greater than 3; in fact the ratio between the two eventually becomes arbitrarily large with sufficiently many dimensions.

Let a $(d-1)$-sphere in $d$-dimensional space be defined by

$\sum\limits_{i=1}^d x_i^2=1$

The positive orthant is then given by $\min(x_i)\ge0$, while the cap may be defined by $x_1\ge\sqrt{1/2}$. In the latter case two antipodal points $(\sqrt{1/2},x_2,x_3,...x_d)$ and $(\sqrt{1/2},-x_2,-x_3,...-x_d)$ on the boubdary of the cap give the inner product

$1/2-\sum\limits_{i=2}^d x_i^2=0,$

where the equation for the sphere combined with defining $x_1=\sqrt{1/2}$ imply the second equality.

The fraction of the sphere covered by the positive orthant is then $2^{-d}$, while the fraction covered by the cap is given by a ratio of integrals:

$\dfrac{\int_{\sqrt{1/2}}^1(1-x^2)^{(3-d)/2}dx}{\int_{-1}^1(1-x^2)^{(3-d)/2}dx}.$

With $d=3$ this gives (to four decimal places) $0.1464$ for the cap versus $0.1250$ for the orthant. With four-dimensional space the corresponding fractions are $0.1051$ for the cap and $0.0625$ for the orthant. As $d\to\infty$, the integrals in the fraction for the cap become Laplace integrals and can be evaluated asymptotically by the appropriate method for such integrals; the result is that the fraction covered by the cap has the controlling $(\sqrt2)^{-d}$ versus $2^{-d}$ for the positive orthant. As $\sqrt2<2$, the cap ultimately becomes exponentially larger than the positive orthant.

You can go slightly beyond the positive orthant. In the case of a 2-sphere in 3-dimensional space, you can render a cap with angular radius $45°$. This has a surface area of $2\pi(1-\sqrt{1/2})\approx(2\pi)\color{blue}{(0.293)}$ steradians, whereas the positive orthant has a surface area of $\pi/2=(2\pi)\color{blue}{(0.25)}$ steradians.

Although the cap is larger in area, the positive orthant cannot fit inside it. This is essentially a spherical analogue of the planar geometry result that a circle of unit diameter is larger than an equilateral triangle of unit side (or, for that matter, any regular odd-sided polygon whose longest chord measures one unit), but the triangle/odd-gon does not fit in the circle.

The counterpart to the cap also beats the positive orthant in spatial dimendions greater than 3; in fact the ratio between the two eventually becomes arbitrarily large with sufficiently many dimensions.

Let a $(d-1)$-sphere in $d$-dimensional space be defined by

$\sum\limits_{i=1}^d x_i^2=1$

The positive orthant is then given by $\min(x_i)\ge0$, while the cap may be defined by $x_1\ge\sqrt{1/2}$. In the latter case two antipodal points $(\sqrt{1/2},x_2,x_3,...x_d)$ and $(\sqrt{1/2},-x_2,-x_3,...-x_d)$ on the boubdary of the cap give the inner product

$1/2-\sum\limits_{i=2}^d x_i^2=0,$

where the equation for the sphere combined with defining $x_1=\sqrt{1/2}$ imply the second equality.

The fraction of the sphere covered by the positive orthant is then $2^{-d}$, while the fraction covered by the cap is given by a ratio of integrals:

$\dfrac{\int_{\sqrt{1/2}}^1(1-x^2)^{(3-d)/2}dx}{\int_{-1}^1(1-x^2)^{(3-d)/2}dx}.$

With $d=3$ this gives (to four decimal places) $0.1464$ for the cap versus $0.1250$ for the orthant. With four-dimensional space the corresponding fractions are $0.1051$ for the cap and $0.0625$ for the orthant. As $d\to\infty$, the integrals in the fraction for the cap become Laplace integrals and can be evaluated asymptotically by the appropriate method for such integrals; the result is that the fraction covered by the cap has the controlling $(\sqrt\pi)^{-d}$ versus $2^{-d}$ for the positive orthant. As $\sqrt\pi<2$, the cap ultimately becomes exponentially larger than the positive orthant.

You can go slightly beyond the positive orthant. In the case of a 2-sphere in 3-dimensional space, you can render a cap with angular radius $45°$. This has a surface area of $2\pi(1-\sqrt{1/2})\approx(2\pi)\color{blue}{(0.293)}$ steradians, whereas the positive orthant has a surface area of $\pi/2=(2\pi)\color{blue}{(0.25)}$ steradians.

Although the cap is larger in area, the positive orthant cannot fit inside it. This is essentially a spherical analogue of the planar geometry result that a circle of unit diameter is larger than an equilateral triangle of unit side (or, for that matter, any regular odd-sided polygon whose longest chord measures one unit), but the triangle/odd-gon does not fit in the circle.

The counterpart to the cap also beats the positive orthant in spatial dimensions greater than 3; in fact the ratio between the two eventually becomes arbitrarily large with sufficiently many dimensions.

Let a $(d-1)$-sphere in $d$-dimensional space be defined by

$\sum\limits_{i=1}^d x_i^2=1$

The positive orthant is then given by $\min(x_i)\ge0$, while the cap may be defined by $x_1\ge\sqrt{1/2}$. In the latter case two antipodal points $(\sqrt{1/2},x_2,x_3,...x_d)$ and $(\sqrt{1/2},-x_2,-x_3,...-x_d)$ on the boubdary of the cap give the inner product

$1/2-\sum\limits_{i=2}^d x_i^2=0,$

where the equation for the sphere combined with defining $x_1=\sqrt{1/2}$ imply the second equality.

The fraction of the sphere covered by the positive orthant is then $2^{-d}$, while the fraction covered by the cap is given by a ratio of integrals:

$\dfrac{\int_{\sqrt{1/2}}^1(1-x^2)^{(3-d)/2}dx}{\int_{-1}^1(1-x^2)^{(3-d)/2}dx}.$

With $d=3$ this gives (to four decimal places) $0.1464$ for the cap versus $0.1250$ for the orthant. With four-dimensional space the corresponding fractions are $0.1051$ for the cap and $0.0625$ for the orthant. As $d\to\infty$, the integrals in the fraction for the cap become Laplace integrals and can be evaluated asymptotically by the appropriate method for such integrals; the result is that the fraction covered by the cap has the controlling $(\sqrt\pi)^{-d}$ versus $2^{-d}$ for the positive orthant. As $\sqrt\pi<2$, the cap ultimately becomes exponentially larger than the positive orthant.