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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)

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If you just want a formula, Wikipedia has it. MathWorld has a derivation as well. – Rahul Jun 3 '12 at 22:15

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 $2\pi\cdot(1-1/2)=\pi$.

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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: $$A=r^2\int_0^{\pi}\int_0^{\pi}\sin ududv=2\pi r^2$$

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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

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:

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

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.

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

$$ = \pi $$

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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 $\pi$.

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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

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