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)

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