multivariable-calculus
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)
The intersection of a cylinder with a plane is an ellipse. Find the semiaxes of the ellipse and you get $$S=\pi ab$$
The minor semiaxis is always the same as the radius of the cylinder, in this case $b=r=2$.
The major semiaxis can be calculated from the angle between the plane and the cylinder axis. The angle the plane makes with the $z$ axis can be extracted from the plane normal, as the dot product gives us a cosine between two vectors, $\cos\alpha=(n_x,n_y,n_z)\cdot(0,0,1)=n_z$.
The normal of your plane can be read directly from the coefficient of the equation. An equation for a plane can be written as a dot product $\vec{n}\cdot\vec{r}=\rm const$, in your case $(1,2,1)\cdot(x,y,z)=4$. Rescale the normal to unit size and you get: $$\vec{n}=\frac{(1,2,1)}{\sqrt 6}$$ and as we demonstrated above, also by using the dot product, the $z$ component of the normal equals the cosine of the angle with the $z$ axis: $$\cos\alpha=\frac{1}{\sqrt 6}$$
If you draw the vertical cross section (the figure on the right), you can see a right triangle that relates the radius of the cylinder with the hypotenuse (the semiaxis):
$$a=\frac{r}{\cos\alpha}=2\sqrt 6$$ leading to the solution $$S=4\pi \sqrt 6$$
EDIT:
"Stokes' Theorem" is a name given to a much broader theorem than you appear to be refering to here. I am assuming you are referring to the Curl Theorem (one special case of Stokes' Theorem): $$\int_S \nabla \times \mathbf{F}\cdot d\mathbf{a} = \int_{\partial S}\mathbf{F}·d\mathbf{s}$$ In this case it is $d\mathbf{a}$ that represents the area element on the surface. $d\mathbf{s}$ is the element of arclength along the curve $\partial S$ which forms the boundary of the surface $S$.
Best Answer
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 $2\pi\cdot(1-1/2)=\pi$.