Interesting problem about latitude

[ArcainineFalls531]

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 ) 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.

 

[daveb]

You could try a surface integral with appropriate limits of integration.

 

[HallsofIvy]

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:
\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}.
Of course,
z= \sqrt{R^2- x^2- y^2}
so the integral for surface area, in polar coordinates would be
R\int \frac{rdrd\theta}{\sqrt{R^2- r^2}}
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)= \frac{R\sqrt{3}}{2} and thus r²+ z²= r²+ 3R²/4= R² which reduces to r= R/2. The limits of integration are 0\le \theta \le 2\pi and 0\le r \le R/2The 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:
\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}.
Of course,
z= \sqrt{R^2- x^2- y^2}
so the integral for surface area, in polar coordinates would be
R\int \frac{rdrd\theta}{\sqrt{R^2- r^2}}
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)= \frac{R\sqrt{3}}{2} and thus r²+ z²= r²+ 3R²/4= R² which reduces to r= R/2. The limits of integration are 0\le \theta \le 2\pi and 0\le r \le R/2The 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:
\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}.
Of course,
z= \sqrt{R^2- x^2- y^2}
so the integral for surface area, in polar coordinates would be
R\int \frac{rdrd\theta}{\sqrt{R^2- r^2}}
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)= \frac{R\sqrt{3}}{2} and thus r²+ z²= r²+ 3R²/4= R² which reduces to r= R/2. The limits of integration are 0\le \theta \le 2\pi and 0\le r \le R/2The 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:
\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}.
Of course,
z= \sqrt{R^2- x^2- y^2}
so the integral for surface area, in polar coordinates would be
R\int \frac{rdrd\theta}{\sqrt{R^2- r^2}}
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)= \frac{R\sqrt{3}}{2} and thus r²+ z²= r²+ 3R²/4= R² which reduces to r= R/2. The limits of integration are 0\le \theta \le 2\pi and 0\le r \le R/2

 

[ArcainineFalls531]

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.

 

[SpongeBobRhombusHat]

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