# surface area

• Aug 11th 2007, 07:31 PM
kittycat
surface area
Find the surface area of the portion of the sphere of radius 4 that lies inside the cylinder x^2 +y^2 =12 and above the xy-plane.

Could you please show me the steps of solving this question? Thank you very much.
• Aug 11th 2007, 07:37 PM
ThePerfectHacker
Quote:

Originally Posted by kittycat
Find the surface area of the portion of the sphere of radius 4 that lies inside the cylinder x^2 +y^2 =12 and above the xy-plane.

Could you please show me the steps of solving this question? Thank you very much.

$\displaystyle \int_0^{2\pi} \int_0^{\pi/4} \int_0^4 \rho^2 \sin \phi d\rho \ d\phi \ d\theta$

Because the cone creates a 45 degree opening angle with the z axis.
• Aug 11th 2007, 10:44 PM
CaptainBlack
Quote:

Originally Posted by kittycat
Find the surface area of the portion of the sphere of radius 4 that lies inside the cylinder x^2 +y^2 =12 and above the xy-plane.

Could you please show me the steps of solving this question? Thank you very much.

In spherical polars the surface area element of a sphere is $\displaystyle r^2 \sin(\phi) ~d\phi d \theta$, where $\displaystyle \phi$ is the zenithal angle
of the patch and $\displaystyle \theta$ the azimuthal angle.

The part of the surface of the sphere that is inside the cylinder is that part for which $\displaystyle \phi \in [\arcsin(\sqrt{12}/4), \pi/2]=[\pi/6, \pi/2]$

So the required area is:

$\displaystyle A=\int_{\phi=\pi /6}^{\pi}\int_{\theta=0}^{2\pi} 16 \sin(\phi) ~d\theta d \phi$

RonL
• Aug 12th 2007, 03:02 AM
Soroban
Hello, kittycat!

The problem has be solved "head on".

Quote:

Find the surface area of the portion of the sphere of radius 4
that lies inside the cylinder $\displaystyle x^2 +y^2 \:=\:12$ and above the xy-plane.

Turn the diagram 90° "clockwise".
Code:

                |               * * *           *    |    *         *-------*-------*A       *      B|    * :*                 |  *  :       *        | *    : *   - - * - - - - + - - - + * - -       *        |O      C *                 |       *        |        *         *-------+-------*           *    |    *               * * *                 |

In right triangle $\displaystyle ABO\!:\;\;OA = 4,\:OB = \sqrt{12}$

. . Hence: .$\displaystyle BA\,=\,OC\,=\,2$ .**

We want the area of the "polar cap" formed by the segment at the far right.

The formula for the area of a surface of revolution is: .$\displaystyle S \;=\;2\pi\int^b_a y\,\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$

The equation of the circle is: .$\displaystyle x^2+y^2\:=\:16\quad\Rightarrow\quad y \:=\:\sqrt{16-x^2}$

Then: .$\displaystyle \frac{dy}{dx}\:=\:\frac{-x}{\sqrt{16-x^2}}$

. . . . $\displaystyle 1 + \left(\frac{dy}{dx}\right)^2\;=\;1 + \frac{x^2}{16-x^2} \;=\;\frac{16}{16-x^2}$

. . . .$\displaystyle \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\frac{4}{\sqrt{16-x^2}}$

Hence, we have: .$\displaystyle S \;=\;2\pi\int^4_2\sqrt{16-x^2}\cdot\frac{4}{\sqrt{16-x^2}}\,dx \;=\;8\pi\int^4_2\!\!dx \;=\;8\pi x\bigg]^4_2$

. . . . $\displaystyle = \;8\pi(4 - 2) \;=\;\boxed{16\pi}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**
Note that: .$\displaystyle \angle AOC \,=\,60^o$

Fascinating bit of trivia:

A central cone with vertex angle 120° subtends ¼ the area of the sphere.

With such a cone at the north pole and the south pole,
. . the "equatorial band" has half the area of the sphere.