# Area of surface

1. Feb 24, 2015

### Incand

1. The problem statement, all variables and given/known data
Calculate the area of the surface $x^2+y^2+z^2 = R^2 , z \ge h , 0 \le h \le R$

2. Relevant equations
$A(S_D) = \iint_D |\mathbf r'_s \times \mathbf r'_t|dsdt$
where $S_D$ is the surface over $D$.

3. The attempt at a solution
We write the surface in parametric form using spherical coordinates
$\mathbf r(\theta ,\phi) = R(\sin \theta \cos \phi , \sin \theta \sin \phi , \cos \theta)$
which gives us
$\mathbf r'_\theta = R(\cos \theta \cos \phi , \cos \theta \sin \phi , -\sin \theta)$
and
$\mathbf r'_\phi = R(-\sin \theta \sin \phi , \sin \theta \cos \phi , 0 )$
so we end up with
$|\mathbf r'_\theta \times \mathbf r'_\phi | = R^2|\sin \theta | |(\sin \theta \cos \phi , \sin \theta , \sin \theta \sin \phi , \cos \theta ) | = R^2|\sin \theta |$
and the area
$A(S_D) = \iint _D R^2|\sin \theta | = \int _?^? \int _0^{2\pi } R^2|\sin \theta | d\phi d? = 2\pi \int _?^? R^2|\sin \theta | d?$
So my problem is i don't know what im supposed to integrate over. I suppose it should be $\theta$ but I'm not sure between what limits. Or even if the switch to spherical coordinates was a good idea at all.

The answer according to the book should be $A(S_D) = 2\pi R(R-h)$
which makes me think i made another error somewhere else as well since I got an excess of $R$.

Appreciate any help. Cheers!

2. Feb 24, 2015

### LCKurtz

OK, you are using $\theta$ as the angle from the $z$ axis, and yes, spherical coordinates is what you want. Draw a cross section of your sphere of radius $R$ with a horizontal line at $z=h$. The triangle that forms should give you the limits for $\theta$.

3. Feb 25, 2015

### HallsofIvy

Staff Emeritus
The only two variables you have are "$\phi$" and "$\theta$" and you have already integrated with respect to $\phi$ so the only possible remaining variable is $\theta$. $\theta$ is the angle a line from (0, 0, R) to (x, y, z) makes with the z-axis. When (x, y, z)= (0, 0, R), $\theta= 0$. When $z= R cos(\theta)= h$, $cos(\theta)= \frac{h}{R}$ so $\theta= cos^{-1}\left(\frac{h}{R}\right)$.

4. Feb 25, 2015

### Incand

Thanks for the help both of you! I got it from Kurtz advice earlier but couldn't respond before now. The confusion with if i should integrate over $\theta$ was more that i couldn't find the limits and thought that perhaps spherical was a bad idea and i should use the radius somehow.

Gonna add the rest of the solution in case anyone else come upon the thread.
From the attachment pdf we can see that $cos(\theta)= \frac{h}{R}$ so therefore $0 \le \theta \le \arccos (\frac{h}{R} )$

$A = 2R^2\pi \int_0^{\arccos \frac{h}{R}} \sin \theta d\theta = 2R^2\pi ( -\frac{h}{R} + 1) = 2R\pi(R-h)$

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