Interesting problem about latitude

In summary, there are multiple ways to solve the problem of finding the area of the surface between the equator and latitude 60 degrees north on a planet of radius a. One way is to use a surface integral with appropriate limits of integration. Another way is to project the sphere onto a cylinder and use equal area properties. This was also discovered by Archimedes and can be proven using basic differential geometry.
  • #1
ArcainineFalls531
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For a planet of radius a, find the area of the surface between the equator and latitude 60 degress north.

This problem was posed to me way back in my calc II class. Instructor (somehow :wink: ) used series to solve it, then asked us if we could solve it differently. I only vaguely remember his solution, otherwise I'd post it as well. Anyone have any interesting ways to solve this? Thought it might be fun.
 
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  • #2
You could try a surface integral with appropriate limits of integration.
 
  • #3
The equation of the sphere of radius r, x2+ y2+ z2= R2 can be thought of as a "level surface" of the function f(x,y,z)= x2+ y2+ z2. Its gradient, grad f= 2xi+ 2yj+ 2zk is perpendicular to that surface. "Normalizing" to integrate on the xy-axis by dividing through by z2, the differential of surface area is the length of the vector (x/z)i+ (y/z)j+ k:
[tex]\sqrt{\frac{x^2}{z^2}+ \frac{y^2}{z^2}+ 1}dxdy= \frac{\sqrt{x^2+ y^2+ z^2}}{z}dxdy= \frac{Rdxdy}{z}[/tex].
Of course,
[tex]z= \sqrt{R^2- x^2- y^2}[/tex]
so the integral for surface area, in polar coordinates would be
[tex]R\int \frac{rdrd\theta}{\sqrt{R^2- r^2}}[/tex]
The lower boundary is z= 0 which corresponds to the circle of radius R.
The upper limit is at "60 degrees north" where z= Rsin(60)= [itex]\frac{R\sqrt{3}}{2}[/itex] and thus r2+ z2= r2+ 3R2/4= R2 which reduces to r= R/2. The limits of integration are [itex]0\le \theta \le 2\pi[/itex] and [itex]0\le r \le R/2[/itex]The equation of the sphere of radius r, x2+ y2+ z2= R2 can be thought of as a "level surface" of the function f(x,y,z)= x2+ y2+ z2. Its gradient, grad f= 2xi+ 2yj+ 2zk is perpendicular to that surface. "Normalizing" to integrate on the xy-axis by dividing through by z2, the differential of surface area is the length of the vector (x/z)i+ (y/z)j+ k:
[tex]\sqrt{\frac{x^2}{z^2}+ \frac{y^2}{z^2}+ 1}dxdy= \frac{\sqrt{x^2+ y^2+ z^2}}{z}dxdy= \frac{Rdxdy}{z}[/tex].
Of course,
[tex]z= \sqrt{R^2- x^2- y^2}[/tex]
so the integral for surface area, in polar coordinates would be
[tex]R\int \frac{rdrd\theta}{\sqrt{R^2- r^2}}[/tex]
The lower boundary is z= 0 which corresponds to the circle of radius R.
The upper limit is at "60 degrees north" where z= Rsin(60)= [itex]\frac{R\sqrt{3}}{2}[/itex] and thus r2+ z2= r2+ 3R2/4= R2 which reduces to r= R/2. The limits of integration are [itex]0\le \theta \le 2\pi[/itex] and [itex]0\le r \le R/2[/itex]The equation of the sphere of radius r, x2+ y2+ z2= R2 can be thought of as a "level surface" of the function f(x,y,z)= x2+ y2+ z2. Its gradient, grad f= 2xi+ 2yj+ 2zk is perpendicular to that surface. "Normalizing" to integrate on the xy-axis by dividing through by z2, the differential of surface area is the length of the vector (x/z)i+ (y/z)j+ k:
[tex]\sqrt{\frac{x^2}{z^2}+ \frac{y^2}{z^2}+ 1}dxdy= \frac{\sqrt{x^2+ y^2+ z^2}}{z}dxdy= \frac{Rdxdy}{z}[/tex].
Of course,
[tex]z= \sqrt{R^2- x^2- y^2}[/tex]
so the integral for surface area, in polar coordinates would be
[tex]R\int \frac{rdrd\theta}{\sqrt{R^2- r^2}}[/tex]
The lower boundary is z= 0 which corresponds to the circle of radius R.
The upper limit is at "60 degrees north" where z= Rsin(60)= [itex]\frac{R\sqrt{3}}{2}[/itex] and thus r2+ z2= r2+ 3R2/4= R2 which reduces to r= R/2. The limits of integration are [itex]0\le \theta \le 2\pi[/itex] and [itex]0\le r \le R/2[/itex]The equation of the sphere of radius r, x2+ y2+ z2= R2 can be thought of as a "level surface" of the function f(x,y,z)= x2+ y2+ z2. Its gradient, grad f= 2xi+ 2yj+ 2zk is perpendicular to that surface. "Normalizing" to integrate on the xy-axis by dividing through by z2, the differential of surface area is the length of the vector (x/z)i+ (y/z)j+ k:
[tex]\sqrt{\frac{x^2}{z^2}+ \frac{y^2}{z^2}+ 1}dxdy= \frac{\sqrt{x^2+ y^2+ z^2}}{z}dxdy= \frac{Rdxdy}{z}[/tex].
Of course,
[tex]z= \sqrt{R^2- x^2- y^2}[/tex]
so the integral for surface area, in polar coordinates would be
[tex]R\int \frac{rdrd\theta}{\sqrt{R^2- r^2}}[/tex]
The lower boundary is z= 0 which corresponds to the circle of radius R.
The upper limit is at "60 degrees north" where z= Rsin(60)= [itex]\frac{R\sqrt{3}}{2}[/itex] and thus r2+ z2= r2+ 3R2/4= R2 which reduces to r= R/2. The limits of integration are [itex]0\le \theta \le 2\pi[/itex] and [itex]0\le r \le R/2[/itex]
 
  • #4
Wow, that's pretty interesting, certainly. Just reviewing the concepts of surface integrals. Thanks for all the input, these problems never cease to amaze me.
 
  • #5
Another way to do this was discovered by Archimedes, but is easy to prove using basic differential geometry.

Place the sphere inside a cylinder of radius a and height 2a so that the center of the sphere lies on the center axis of the cylinder. Define a map by projecting each point on the sphere radially onto the cylinder. Another way of saying this is that the image of a point x on the sphere is given by where the perpendicular to the axis which goes through x intersects the cylinder.

This projection is "equal area," meaning that if you want to measure the area of a region on the sphere, you will get the same answer looking at the area of its projection.

I'll leave the details to you. They aren't too difficult.

-SBRH
 

1. What is the significance of latitude in relation to climate?

Latitude is a measure of how far north or south a location is from the equator. It plays a crucial role in determining a region's climate, as areas closer to the equator receive more direct sunlight and therefore tend to be warmer than those closer to the poles.

2. How does latitude affect the length of daylight?

The Earth's tilt and rotation around the sun results in varying lengths of daylight at different latitudes throughout the year. Areas closer to the equator experience more consistent day lengths, while those closer to the poles may have long periods of daylight in the summer and long periods of darkness in the winter.

3. Is there a connection between latitude and biodiversity?

Latitude can have a significant impact on biodiversity. Areas near the equator tend to have higher levels of biodiversity due to the favorable climate and more direct sunlight, while areas closer to the poles may have less diverse ecosystems due to harsher conditions.

4. How does latitude impact agriculture and crop production?

The climate and length of growing season can vary greatly depending on latitude, which can have a significant impact on agriculture and crop production. Areas closer to the equator may have longer growing seasons and more favorable conditions for certain crops, while more extreme latitudes may have shorter growing seasons and more limited crop options.

5. Can latitude affect human health and well-being?

Latitude can have a direct impact on human health and well-being. Areas closer to the equator tend to have warmer climates, which can increase the risk of certain health issues such as heat stroke. In contrast, areas closer to the poles may have colder climates, which can increase the risk of health issues related to cold temperatures such as hypothermia.

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