Area of Surface: x^2+y^2+z^2=R^2, z>=h, 0<=h<=R

[Incand]

Homework Statement

Calculate the area of the surface $x^2+y^2+z^2 = R^2 , z \ge h , 0 \le h \le R$

Homework Equations

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

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 I am 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!

 

[LCKurtz]

Homework Statement

Calculate the area of the surface $x^2+y^2+z^2 = R^2 , z \ge h , 0 \le h \le R$

Homework Equations

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

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 I am 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!

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 $\theta$.

 

[HallsofIvy]

Homework Statement

Calculate the area of the surface $x^2+y^2+z^2 = R^2 , z \ge h , 0 \le h \le R$

Homework Equations

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

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 I am 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 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).

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!

 

[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 $0 \le \theta \le \arccos (\frac{h}{R} )$

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