Find surface area of $ S=\{ (x,y,z)\in\mathbb{R}:x^2+y^2+z^2=4, (x-1)^2+y^2\leq1, z\ge0 \} $

Question

Define $$ S=\{ (x,y,z)\in\mathbb{R}:x^2+y^2+z^2=4, (x-1)^2+y^2\le1, z\ge0 \} $$ Find the surface area of $S$.

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The surface area of $z=f(x,y)$ where $(x,y)$ is a point from the region $D$ in the $xy$ plane is given by $$ S = \iint\limits_{D}{{\sqrt {\,{{\left[ {{f_x}} \right]}^2} + {{\left[ {{f_y}} \right]}^2} + 1} \,dA}} $$ I know how to solve it when $D$ given by $x^2+y^2\le 1$. It can be done by switching to polar coordinates and we end up with a neat integral.
Here $D$ is the set of points lying inside the circle centered at $(1,0)$ with radius $1$.
I tried shifting everything by $1$ unit. Finding surface area of $S$ is equivalent to finding the surface area of the sphere $(x+1)^2+y^2+z^2=4$ directly above $x^2+y^2\le 1$. But now I don't know how to calculate the integral.

$$z=\sqrt{4-(x+1)^2-y^2}$$ $$S=\int_{0}^{2\pi}\int_{0}^{1}\frac{8r}{4-(r\cos\theta+1)^2-(r\sin\theta)^2}~\mathbb{d}r~\mathbb{d}\theta$$

Answer 1

$ S = \{ (x, y, z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 4, \, (x-1)^2 + y^2 \leq 1, \, z \geq 0 \} $

This set $S$ is the portion of the sphere $x^2 + y^2 + z^2 = 4$ (radius $2$) that lies within the cylinder $(x-1)^2 + y^2 \leq 1$ (radius $1$, centered at $(1, 0)$) and above the plane $z = 0$

$1.$ The sphere $x^2 + y^2 + z^2 = 4$ is centered at the origin with radius $2$

$2.$ The cylinder $(x-1)^2 + y^2 \leq 1$ is centered at $(1, 0, 0)$ with radius $1$

$3.$ The intersection of the sphere and the cylinder is a curve on the sphere. The region $S$ is the portion of the sphere that lies inside the cylinder and above the $z = 0$ plane

We parameterize the sphere using spherical coordinates

$ x = 2 \sin \phi \cos \theta, \quad y = 2 \sin \phi \sin \theta, \quad z = 2 \cos \phi $

where $\phi \in [0, \pi/2]$ (since $z \geq 0$) and $\theta \in [0, 2\pi)$

The surface area element on the sphere is

$ dS = (2 \sin \phi) \cdot (2 \, d\phi \, d\theta) = 4 \sin \phi \, d\phi \, d\theta $

The cylinder $(x-1)^2 + y^2 \leq 1$ intersects the sphere. Substituting the spherical coordinates into the cylinder equation:

$ (x-1)^2 + y^2 = (2 \sin \phi \cos \theta - 1)^2 + (2 \sin \phi \sin \theta)^2 \leq 1 $

Expanding:

$ (2 \sin \phi \cos \theta - 1)^2 + (2 \sin \phi \sin \theta)^2 \leq 1 $

Simplify

$ 4 \sin^2 \phi \cos^2 \theta - 4 \sin \phi \cos \theta + 1 + 4 \sin^2 \phi \sin^2 \theta \leq 1 $

Combine terms

$ 4 \sin^2 \phi (\cos^2 \theta + \sin^2 \theta) - 4 \sin \phi \cos \theta + 1 \leq 1 $

Since $\cos^2 \theta + \sin^2 \theta = 1$, this simplifies to

$ 4 \sin^2 \phi - 4 \sin \phi \cos \theta \leq 0 $

Divide through by $4 \sin \phi$ (assuming $\sin \phi \neq 0$)

$ \sin \phi - \cos \theta \leq 0 $

Thus, the condition becomes

$ \sin \phi \leq \cos \theta $

The inequality $\sin \phi \leq \cos \theta$ defines the region of integration. To find the limits

$1.$ For a fixed $\theta$, $\phi$ ranges from $0$ to $\arcsin(\cos \theta)$.

$2.$ The angle $\theta$ ranges from $0$ to $2\pi$, but due to symmetry, we can integrate over $\theta \in [-\pi/2, \pi/2]$ and multiply by $2$.

The surface area $A$ is given by

$ A = \iint_S dS = \int_{\theta = -\pi/2}^{\pi/2} \int_{\phi = 0}^{\arcsin(\cos \theta)} 4 \sin \phi \, d\phi \, d\theta $

First, evaluate the inner integral

$ \int_{\phi = 0}^{\arcsin(\cos \theta)} 4 \sin \phi \, d\phi = 4 \left[ -\cos \phi \right]_0^{\arcsin(\cos \theta)} = 4 \left( -\cos(\arcsin(\cos \theta)) + \cos(0) \right) $

Simplify $\cos(\arcsin(\cos \theta))$

$ \cos(\arcsin(\cos \theta)) = \sqrt{1 - \cos^2 \theta} = |\sin \theta| $

Thus, the inner integral becomes

$ 4 \left( -|\sin \theta| + 1 \right) = 4(1 - |\sin \theta|) $

Now, integrate over $\theta$

$ A = \int_{\theta = -\pi/2}^{\pi/2} 4(1 - |\sin \theta|) \, d\theta $

Split the integral

$ A = 4 \int_{-\pi/2}^{\pi/2} 1 \, d\theta - 4 \int_{-\pi/2}^{\pi/2} |\sin \theta| \, d\theta $

Evaluate the integrals

$1.$ $\int_{-\pi/2}^{\pi/2} 1 \, d\theta = \pi$

$2.$ $\int_{-\pi/2}^{\pi/2} |\sin \theta| \, d\theta = 2$

Thus

$ A = 4\pi - 4\times 2 = 4\pi-8 $

The surface area of $S$ is

$$ \boxed{4\pi-8} $$

Answer 2

The given integral is not correct.

Using translated polar coordinates to parameterize $S$ with

$$\vec s(r,\theta) = \left\langle r\cos\theta+1, r\sin\theta, \sqrt{3-r^2-2r\cos\theta}\right\rangle$$

where $(r,\theta)\in[0,1]\times[0,2\pi]$, as in OP's integral, it follows that the normal to $S$ has magnitude

$$\left\|\vec n\right\| = \left\|\frac{\partial\vec s}{\partial r}\times\frac{\partial\vec s}{\partial\theta}\right\| = \frac{2r}{\sqrt{3-r^2-2r\cos\theta}}$$

Then the area of $S$ is $\displaystyle\int_0^{2\pi}\int_0^1\|\vec n\|\,dr\,d\theta$. Evaluation can be messy, but you might find some of the methods shown here to be productive.

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It would be easier to use standard polar coordinates, so that $x=r\cos\theta$ and $z=\sqrt{4-r^2}$. The domain is determined by the cylinder,

$$(x-1)^2+y^2=1 \implies r = 2\cos\theta \implies \begin{cases}\theta\in[0,\color{red}\pi] \\ r\in[0,\color{red}{2\cos\theta}]\end{cases}$$

Now, the normal has magnitude

$$\left\|\vec n\right\| = \frac{2r}{\sqrt{4-r^2}}$$

and so the area is

$$\begin{align*} & \int_0^\pi \int_0^{2\cos\theta} \frac{2r}{\sqrt{4-r^2}} \, dr \, d\theta \\ &= 4 \int_0^\pi (1-\sin\theta) \, d\theta \\ &= \boxed{4\pi-8} \end{align*}$$