Parametrize and find the surface area

In summary: We can see that the range for θ is [0,2π], but for the range of r we need to find where the two curves intersect.r = 2·sinθ = 2·r·sinθ → r = 0 or r=2.The two curves intersect at the origin and at r=2, so the range for r is [0,2].Therefore, the parametric equations for the upper half of the sphere contained within the cylinder are:x = r·cosθ y = r·sinθ z = \sqrt{4-r^2}\,and: θ = [0,2π], r = [0,2], where r=0 is
  • #1
vDrag0n
3
0

Homework Statement


a)parametrize the upper surface of that portion of the sphere x^2 + y^2 +z^2 = 4 contained within the cylinder, x^2 + y^2 = 2y
b)find the area of the surface


Homework Equations


how to find the range for phi. for paramaetrizing into spherical coordinates


The Attempt at a Solution



a) because its a sphere i tried parmetrizing into spherical coordinates
and thus
X (u,v) = (2sinucosv, 2sinusinv,2cosu)
u,v = [0,?]x[0,2pi]
I've tried substuting the eq of the cylinder into the eq of the sphere but i ended up with
z^2 = 4-2y => (4 - 4sinusinv)^0.5
which isn't very helpful in finding the ranges of U



b) without finding the ranges for u, I won't be able to use the formula doubleint ||Xu x Xv|| du dv

but i can try A(S) = double int (1 + ||gradF||^2)^0.5 dx dy
i can have F = (4-x^2 - y^2)^0.5 and D == x^2 + y^2 - 2y <=0
looks like polar coordinates would be best so i set
x = r cos u
y = r sin u
r,u = [0,2sinu] x [0, 2pi]
and thus end up with an integral of
int ( 2/ (4 - r^2)^0.5 dr du
but this will give me an integral of
int 2 *( [arcsin (r/2)] from 0 to rsin ) du which is not an integral i would like to solve.
 
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  • #3
vDrag0n said:

Homework Statement


a)parametrize the upper surface of that portion of the sphere x^2 + y^2 +z^2 = 4 contained within the cylinder, x^2 + y^2 = 2y
b)find the area of the surface
Hi vDrag0n.

In cylindrical coordinates, x2 + y2 = r2 and y=r·sinθ .

So the equation for the cylinder is: r2 = 2·r·sinθ → r = 2·sinθ.

The equation for the sphere is: r2 +z2 = 4, so the upper half is:

[tex]z=\sqrt{4-r^2}\,.[/tex]
 

Related to Parametrize and find the surface area

1. What is parametrization and how is it used to find the surface area?

Parametrization is a mathematical technique used to describe and represent a surface in terms of parameters. These parameters can be variables such as x, y, and z coordinates. The surface area can be found by using parametrization to create a function that represents the surface and then calculating the integral of that function over the desired region.

2. What are the benefits of using parametrization to find surface area?

Parametrization allows for the representation of complex surfaces in a simpler form, making it easier to calculate surface area. It also allows for the use of integration techniques, which can greatly simplify the calculation process.

3. Can any surface be parametrized to find its surface area?

Yes, any surface can be parametrized as long as it is continuous and has well-defined boundaries. However, some surfaces may require more complex parametrization techniques compared to others.

4. How does the parametrization process differ for different types of surfaces?

The parametrization process may differ depending on the type of surface. For example, parametrization of a plane or a sphere is relatively straightforward, while parametrization of a torus or a hyperboloid may require more complex equations.

5. Are there any limitations to using parametrization to find surface area?

While parametrization is a powerful tool for calculating surface area, it may not be suitable for all types of surfaces. Some surfaces may be too complex to be accurately represented by parametrization, and in those cases, other methods may need to be used.

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