Surface area of a solid

Question

So I need to find the surface area of this solid defined as: $x^2+y^2+z^2=1$ where $z\geq \frac{1}{2}$

I tried to do the following:

$x^2+y^2+z^2 \geq \frac{1}{2}$ into $z=\sqrt{1-x^2-y^2}$

$Fx(x,y)= -\frac{x}{\sqrt{1-x^2-y^2}}$ by using u-substitution to make $\frac{1}{2\sqrt{1-x^2-y^2}}(-2x)$

$Fy(x,y)= -\frac{y}{\sqrt{1-x^2-y^2}}$ by using u-substitution to make $\frac{1}{2\sqrt{1-x^2-y^2}}(-2y)$

Surface area formula for a sphere: $S=\int\int\sqrt{(Fx)^2+(Fy)^2+1}dA$

$S=\int\int\sqrt{(\frac{-x}{\sqrt{1-x^2-y^2}})^2+(\frac{-y}{\sqrt{1-x^2-y^2}})^2} = \int\int\sqrt{\frac{x^2}{1-x^2-y^2}+\frac{y^2}{1-x^2-y^2}+1} dA$

$\int\int\sqrt{\frac{1}{1-x^2-y^2}}dA$ from my book formula: $\int\int\sqrt{\frac{x^2}{a^2-x^2-y^2}+\frac{y^2}{a^2-x^2-y^2}+1}dA$ = $\int\int\sqrt{\frac{a^2}{a^2-x^2-y^2}}dA$

=$\frac{1}{\sqrt{1-x^2-y^2}}dA = \frac{1}{\sqrt{1-r^2}}rdrd\theta = \int\int(1-r^2)^\frac{-1}{2}rdrd\theta$ <--using $r^2=x^2+y^2$ as a polar coordinate

From this point onward I just need to figure out the integral boundaries and evaluate, but I want to make sure I got to this step correctly first. I'm not even sure I used the right formula, so I'd like some feedback on if this is the right approach or I messed something up along the way

Answer 1

The solid is the upper half of the sphere with radius 1. I assume you want to add the bottom of it too. So, that gives $2\pi+\pi=3\pi$. The $4\pi$ comes from the radial part of the surface integral wich you half. Do you want to calculate it explicitly?

EDIT: As Raffaele pointed out, I was wrong with the answer because the solid is only a quarter of its height not half. The correct answer is the surface equals $\pi+\frac{3}{4}\pi$. Where the first term is the surface of the spherical cap and the second term is the surface of the base circle.

To show this, here we shall use spherical coordinates. I choose $$x=r\sin\theta\cos\varphi\\y=r\sin\theta\sin\varphi\\z=r\cos\theta.$$ We will use the spherical surface element in this coordinates and integrate the part of the sphere we need. Here is a derivation of the surface element. It also includes a sketch of what $r,\theta,\varphi$ represent.

The surface element is $r^2\sin\theta d\theta d\varphi$. For us $r=1$ and $\varphi\in [0,2\pi)$. The angle $\theta$ determines how much of the sphere we integrate from top to bottom. We want only $\frac{1}{4}$. If we choose $\theta\in[0,\pi)$ and integrate over the surface we will get the surface of a sphere. We have to restrict $\theta$ to $[0,\pi-\alpha)$. We need to find what $\alpha$ is. I have made a sketch where the angle is $\alpha$.

In this picture $b=\frac{1}{2}$, i.e. the minimal $z$ value. This implies that $a=\sqrt{3}/2$. We can solve for the left angle in the triangle, lets call it $\alpha'=\arcsin(1/2)$. To get $\alpha$ we must the $\pi/2$ below to it. So, $$\alpha=\alpha'+\pi/2=\pi/3.$$ Finally this means $\theta\in(0,\pi/3]$. Let's integrate $$ S=\int_{\theta=0}^{\pi/3}\int_{\varphi=0}^{2\pi}\sin\theta d\varphi d\theta=\pi. $$ The base is $$S'=a^2\pi=\frac{3}{4}\pi.$$