Parameterizing and finding its boundaries (vector calculus)

In summary, the conversation discusses parameterizing a surface and finding its area, specifically for the equation x2 + y2 + z2 = 4 with the constraint z ≥ √2. The discussion involves determining the boundaries for the parameter phi and clarifying the reasoning behind reversing the inequality when taking the arc cos.
  • #1
pcjang
3
0

Homework Statement



I was studying for my finals, and then the only thing that I got stuck on was parameterizing a surface and finding the area of the surface.

and my problem states that x2 + y2 + z2 = 4
and z [tex]\geq[/tex][tex]\sqrt{2}[/tex]

Homework Equations



so when parameterizing a sphere, it comes out to be
x = r sin[tex]\phi[/tex]cos[tex]\theta[/tex]
y = r sin[tex]\phi[/tex]sin[tex]\theta[/tex]
z = r cos[tex]\phi[/tex]



The Attempt at a Solution



so I'm sure that [tex]\theta[/tex] goes from 0 to 2[tex]\pi[/tex]. But i was getting confused what boundaries should be for [tex]\phi[/tex].

What i tried is that
since z [tex]\geq[/tex][tex]\sqrt{2}[/tex] and should be less than 2 since 2 is the radius, i put [tex]\sqrt{2}[/tex] [tex]\leq[/tex] z [tex]\leq[/tex] 2.
which z = 2 cos[tex]\phi[/tex]
so when i solved it, it came out to be [tex]\pi[/tex]/4 [tex]\leq[/tex] [tex]\phi[/tex] [tex] leq[/tex] 0.

but it just doesn't make sense to me how the boundary can go backwards. can someone explain to me~??
 
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  • #2
oops, last one, i meant to say

phi is greater than or equal to pi/4
and
phi is less than or equal to 0
 
  • #3
When you have [tex]\cos \phi > \frac{1}{\sqrt{2}} [/tex], you should note that the graph of cos x decreases continuously from 1 to 0 in the interval 0 to pi/2. Which is why you have to reverse the inequality when taking the arc cos the interval for phi.
 

1. What is parameterization in vector calculus?

Parameterization in vector calculus refers to the process of representing a curve or surface in terms of one or more parameters. These parameters are typically variables such as time, distance, or angle, which allow us to describe the curve or surface in a more simplified and general form.

2. Why is parameterization important in vector calculus?

Parameterization allows us to express complicated curves and surfaces in a more concise and manageable way. It also allows us to easily manipulate and analyze these curves and surfaces using mathematical techniques such as differentiation and integration.

3. How do you parameterize a curve in vector calculus?

To parameterize a curve, we typically use a vector function that maps a parameter (such as time) to a point on the curve. This function is often written in the form of r(t) = , where x(t), y(t), and z(t) are the coordinates of the point on the curve at time t.

4. What are the boundaries of a parameterized curve?

The boundaries of a parameterized curve refer to the starting and ending points of the curve, typically represented by the minimum and maximum values of the parameter. These boundaries help define the range of the curve and can be used to calculate important properties such as arc length and curvature.

5. How do you find the boundaries of a parameterized surface?

Finding the boundaries of a parameterized surface involves determining the range of values for each parameter that defines the surface. This can be done by setting each parameter equal to its minimum and maximum values, and then solving for the corresponding coordinates. The resulting points represent the boundaries of the surface.

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