Parameterization of Surface for Surface Integral

[KindofSlow]

I'm working on a problem in the chapter "Surface Integrals of Vector Fields".
Problem is to evaluate a vector function over the top half of a sphere.
In the solution, the Professor states: "In this case since the surface is a sphere we will need to use the parametric representation of the surface."
In the previous problem the surface was a paraboloid and the surface of the paraboloid did not need to be parameterized.

My question is: "Why does the fact that the surface is a sphere instead of a paraboloid require the surface to be parameterized?"

Thank you

 

[HallsofIvy]

I'm working on a problem in the chapter "Surface Integrals of Vector Fields".
Problem is to evaluate a vector function over the top half of a sphere.
In the solution, the Professor states: "In this case since the surface is a sphere we will need to use the parametric representation of the surface."
In the previous problem the surface was a paraboloid and the surface of the paraboloid did not need to be parameterized.

My question is: "Why does the fact that the surface is a sphere instead of a paraboloid require the surface to be parameterized?"

Thank you

I disagree with your statement "the paraboloid did not need to be parameterized". Any surface is a two dimensional set and must be written in terms of two parameters, not the three x, y, and z coordinates. What has happened in the case of the paraboloid is that you used x and y themselves as parameters. Using, say, \(\displaystyle z= 9- x^2- y^2\) is exactly the same as using the paremeterization x= u, y= v, \(\displaystyle z= 9- u^2- v^2\), with u and v as parameters instead of x and y. But either way is a "parameterization".

But, of course, that raises the question, probably what you really intended to ask, "why can we not use x and y as parameters in the case of the sphere?" The reason is that when we write "x= f(u, v), y= g(u,v), z= h(u,v)" we must have f, g, and h functions of u and v. \(\displaystyle z= 9- x^2- y^2\) is a function of x and y. \(\displaystyle x^2+ y^2+ z^2= 9\) does not give z as as function of x and y.

Recall the "vertical line test" for a function: if a vertical line crosses the graph only once then it is a function. If the vertical line crosses the graph more than once, it is not. Any vertical line (x and y constant, z changing), if it crosses through the sphere at all, crosses twice so z is NOT function of x and y. (And x is not a function of y and z and y is not a function of x and z.)

 

[KindofSlow]

That is a fantastic answer.
Thank you very much.