Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do I calculate the surface area of the unit sphere above the plane $z=\frac12$?

EDIT: I have been attempting things and I am thinking about parameterizing this... While I know that surface area is given by the double integral of the cross products of partial derivatives of the new parameters, I don't know what to set them to.. (sorry I'm not good with the fancy notation)

share|improve this question
If you just want a formula, Wikipedia has it. MathWorld has a derivation as well. –  Rahul Jun 3 '12 at 22:15
add comment

4 Answers

The circumference of an infinitesimal ring of the unit sphere between $z$ and $z+\mathrm dz$ is $2\pi\sqrt{1-z^2}$, and its width is $\mathrm dz/\sqrt{1-z^2}$. Thus its surface area is $2\pi\,\mathrm dz$. That is, the surface area of a slab of the unit sphere between two $z$ coordinates (or in fact between any two parallel planes) is simply $2\pi$ times the difference of the $z$ coordinates (or, generally, the distance between the two planes). Thus the surface area of the slab of the unit sphere between $z=1/2$ and $z=1$ is $2\pi\cdot(1-1/2)=\pi$.

share|improve this answer
add comment

So if this is your paramterization $$X\left(u,v\right)=\left(\begin{array}{c} r\sin u\cos v\\ r\sin u\sin v\\ r\cos u \end{array}\right)$$ these are the elements of tangent space (partial derivatives wrt $u$ and $v$ respectively): $$X_{u}=\left(\begin{array}{c} r\cos u\cos v\\ r\cos u\sin v\\ -r\sin u \end{array}\right)$$ $$X_{v}=\left(\begin{array}{c} -r\sin u\sin v\\ r\sin u\cos v\\ 0 \end{array}\right)$$ Then by direct calculation: $$\left|X_{u}\times X_{v}\right|=\left|\begin{array}{ccc} i & j & k\\ r\cos u\cos v & r\cos u\sin v & -r\sin u\\ -r\sin u\sin v & r\sin u\cos v & 0 \end{array}\right|=\left|\left(r^{2}\sin^{2}u\cos v\right)i+\left(-r^{2}\sin^{2}u\sin v\right)j+\left(r^{2}\sin u\cos u\right)k\right|=r^{2}\sin u$$ The area of half a sphere is found as follows: $$A=r^2\int_0^{\pi}\int_0^{\pi}\sin ududv=2\pi r^2$$

share|improve this answer
so your approach is similar to ananda's below, but you parameterized everything into polar coordinates first right? –  Mike Jun 3 '12 at 22:30
also, what exactly (intuitively) is the cross product doing? and why are the limits of the integral just 0 to pi and not to 2pi? –  Mike Jun 3 '12 at 22:32
aha found that it represents the area of a parallelogram! –  Mike Jun 3 '12 at 22:39
yes, exactly. also if you need a slice above $z=1/2$, then $u$ ranges from 0 to $\arccos{\frac{1}{2}}$. I wrongly assumed you were looking for the surface of half a sphere –  Valentin Jun 3 '12 at 22:45
add comment

Surface area is given by

$$ \iint_R \left| \vec r_u \times \vec r_v \right| \ dA $$

where $\vec r(u,v)$ is the parametrization of the surface. We can rewrite this as (derivation shown here: http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx):

$$ \iint_D \sqrt{ \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 + 1} \ dA $$

for a function $z = f(x,y)$ where $D$ is the projection of the surface onto the xy-plane.

Since we are only concerned with the portion of the unit sphere above $z = 0$, we can write it as

$$ z = \sqrt{1-x^2-y^2} $$

Computing the partial derivatives with respect to $x$ and $y$,

$$ \frac{\partial z}{\partial x} = \frac{-x}{\sqrt{1-x^2-y^2}} \rightarrow \left(\frac{\partial z}{\partial x}\right)^2 = \frac{x^2}{1-x^2-y^2} $$

$$ \frac{\partial z}{\partial y} = \frac{-y}{\sqrt{1-x^2-y^2}} \rightarrow \left(\frac{\partial z}{\partial y}\right)^2 = \frac{y^2}{1-x^2-y^2} $$

Substituting these into our expression for surface area,

$$ \iint_D \sqrt{ \frac{x^2}{1-x^2-y^2} + \frac{y^2}{1-x^2-y^2} + 1} \ dA $$

which simplifies to (omitting a bit of algebra)

$$ \iint_D \frac{1}{\sqrt{1-x^2-y^2}} \ dA $$

Observe that $D$ (the projection of our surface into the xy-plane) is given by

$$ z = \sqrt{1-x^2-y^2} $$

$$ \frac{1}{2} = \sqrt{1-x^2-y^2} $$

$$ \frac{1}{4} = 1-x^2-y^2 $$

$$ x^2+y^2 = \frac{3}{4} $$

which is a circle of radius $\frac{\sqrt{3}}{2}$. The integral over $D$ is easiest done in polar coordinates. I'll assume you know how to do that and omit the computation.

$$ \int_{0}^{2\pi} \int_{0}^{\frac{\sqrt{3}}{2}} \frac{1}{\sqrt{1-r^2}} \ r \ dr \ d\theta $$

$$ = \pi $$

share|improve this answer
add comment

We will basically project the part of the unit sphere above $z=\frac1 2$ onto $xy$ plane. I will assume that $\int \int_s||\frac {\partial r } {\partial x }\times \frac {\partial r } {\partial y }|| dy dx$ Now $r= f(x,y,z) = f(x,y,z(x,y))$. So $\frac {\partial r } {\partial x }=f(1,0,\frac {\partial z } {\partial x}) $ and $\frac {\partial r } {\partial y }=f(0,1,\frac {\partial z } {\partial y}) $. so $||\frac {\partial r } {\partial x }\times \frac {\partial r } {\partial y }||=$ $({\frac {\partial z } {\partial x}}^2+{\frac {\partial z } {\partial y}}^2+1)^{1/2}$

so now you have just find the derivatives and plug in . and the limit of the integral will be around the circle $x^2+y^2=3/4$. you can use polar co-ordinates . let me know if u have doubts , i think the answer will be $3/2$ times $\pi$.

share|improve this answer
So, z=sqrt(1-x^2-y^2), right and I need to take the partial w.r.t. x and y, respectively? what do you mean by the limit of the integral though? –  Mike Jun 3 '12 at 22:12
yes , after that u have to evaluate double integral for which u need the limits . Can you see what the region $S$ looks like ? its just a circle . –  Theorem Jun 3 '12 at 22:15
ok I get a very messy formula for the cross product: sqrt($(x+y+1-x^2-y^2)/(1-x^2-y^2)$)... I take that this is right and I set x^2+y^2=r^2? the limits would then be r=0 to r=sqrt(3/4) and $\theta from 0 to 2pi? how did you "guess" the answer so quickly when this is so complicated thought? –  Mike Jun 3 '12 at 22:28
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.