Calculating the surface area of sphere above a plane 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)
 A: 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 $2\pi\cdot(1-1/2)=\pi$.
A: 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.
$$
\int_{0}^{2\pi} \int_{0}^{\frac{\sqrt{3}}{2}} \frac{1}{\sqrt{1-r^2}} \ r \ dr \ d\theta
$$
$$
= \pi
$$
A: 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:
$$A=r^2\int_0^{\pi}\int_0^{\pi}\sin ududv=2\pi r^2$$
A: 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 $\pi$. 
