What is the Fundamental Vector Product for Calculating Surface Area on a Sphere?

[rohanprabhu]

I was trying my hand at integrals with 2 variables... So.. my first excercise was to find out the surface area of a sphere. So, the co-ordinate system is something like this:

i] The whole co-ordinate system is mapped on the surface of a sphere.
ii] The x-axis of the sphere is like the equator of the earth.
iii] The y-axis of the sphere is like the prime meredian of the earth.

So, for the area, I used:

<br /> A = \int^{2\pi r}_{0}dx\int^{2\pi r}_{0}dy<br />

giving me the area:

<br /> A = 4 \pi^2 r^2<br />

Which is ofcourse wrong.. because if i look from the first equation, it is more like I'm calculating the area of a square lamina having lengths 2 \pi r each.

 

[Hootenanny]

Which is ofcourse wrong.. because if i look from the first equation, it is more like I'm calculating the area of a square lamina having lengths 2 \pi r each.

And that is exactly what you are doing. What you need to do, is figure out the equation of the projection of a sphere on the xy plane (imagine slicing a ball in half) and evaluate the double integral over that region (with correct limits).

 

[mathman]

I was trying my hand at integrals with 2 variables... So.. my first excercise was to find out the surface area of a sphere. So, the co-ordinate system is something like this:

i] The whole co-ordinate system is mapped on the surface of a sphere.
ii] The x-axis of the sphere is like the equator of the earth.
iii] The y-axis of the sphere is like the prime meredian of the earth.

So, for the area, I used:

<br /> A = \int^{2\pi r}_{0}dx\int^{2\pi r}_{0}dy<br />

giving me the area:

<br /> A = 4 \pi^2 r^2<br />

Which is ofcourse wrong.. because if i look from the first equation, it is more like I'm calculating the area of a square lamina having lengths 2 \pi r each.

The angle around the equator goes from 0 to 2pi. However, the polar angle goes from -pi/2 to pi/2 and even more important its differential is cosydy, not dy.

 

[HallsofIvy]

In general, the integral
\int_a^b\int_c^d du dv
is the area of a rectangle in uv-space, not the surface of a sphere. For a problem like that, it is not enough to know how to do multiple integrals. You also have to know how to find the "differential of area" for a curved surface. The differential of area is "du dv" only for u and v being orthogonal coordinates in a plane.

 

[rohanprabhu]

The angle around the equator goes from 0 to 2pi. However, the polar angle goes from -pi/2 to pi/2 and even more important its differential is cosydy, not dy.

Thanks for replying. Can you please ellaborate on how the differential is cosy dy?

 

[Winzer]

Here's a way to do it.
Lets take this surface and break it down.
First we require a parametric equation for a sphere--> a^2=x^2+z^2+y^2
a is the radius of the sphere.
We can then turn this into a vector equation representing the parametric surface
\vec{r}=&lt;x,y,(+/-)\sqrt(x^2+y^2-a^2)&gt;
We can then take partial derivatives with respect to x and y.
Then take the cross products of x with y, then the magnitude.
Basically:

\int\int|\vec{r}_x X\vec{r}_y|dxdy
Change to polar for convenience.

 

[HallsofIvy]

Or, since this is a sphere, start from spherical coordinates:
x= \rho cos(\theta) sin(\phi)
y= \rho sin(\theta) cos(\phi)
z= \rho cos(\phi)[/itex]<br /> <br /> On the surface of the sphere of radius R, \rho= R so we have<br /> x= R cos(\theta) sin(\phi)<br /> y= R sin(\theta) cos(\phi)<br /> z= R cos(\phi)[/itex]&lt;br /&gt; in terms of the two parameters \theta and \phi.&lt;br /&gt; &lt;br /&gt; Now, \vec{r}= R cos(\theta)sin(\phi)\vec{i}+ Rsin(\theta)\sin(\phi)\vec{j}+ R cos(\phi)\vec{k}&lt;br /&gt; &lt;br /&gt; The cross product \left|\vec{r_\theta} \times \vec{r_\phi}\right| is called the &amp;quot;fundamental vector product&amp;quot; of the surface and the &amp;quot;differential of surface area&amp;quot; is &lt;br /&gt; \left|\vec{r_\theta} \times \vec{r_\phi}\right|d\theta d\phi