Surface integral of a vector field w/o div. theorem

In summary, the problem at hand involves calculating the flux of a vector field over a region with specified parameters. The solution can be found using the divergence theorem or by parametrizing the volume and integrating over its surface. However, there may be a mistake in the given volume and further clarification may be needed to solve the problem accurately.
  • #1
wildefire
2
0

Homework Statement


First a thanks for the existence of this site, i find it quite useful but had no need to actually post till now.

I am stuck on the following problem in "introduction to physics"
We should calculate the [itex]\oint[/itex] [itex]\vec{v}[/itex].d[itex]\vec{A}[/itex]

of a object with the following parameters Vol = x^2+y+2<4, 0<z<4-x^2-y^2
with [itex]\vec{v}[/itex] being x+y,y+z,x+z)
with and without the div. theorem,

Homework Equations


Divergence theorem,div = 3 - simple cylinder integral = possible solution =32pi
Parametrization of volume as
2cos(u)sin(v)
2sin(u)sin(v)
4cos(v)
u = (0,2pi) v= (0,pi/2)
Surface integration using [itex]\vec{v}[/itex](u,v).ru x rv

The Attempt at a Solution


I have 32Pi as a solution using the div theorem,
and i am in over 2 pages of writing now without using it,
Am i missing something or is this the kind of problem that only exist "to show you how useful something is" and i really should come to some horrible solution that is taking a double integral of something over 50 symbols long ?

Thanks in advance.
 
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  • #2
Yep, it's one of those - "To see how nice the theorem is, do it the hard way too", after one more page calculus and wolfram alpha both results match (dunno if i can delete the topic or close it first post here (technically second :) )
 
  • #3
wildefire said:

Homework Statement


First a thanks for the existence of this site, i find it quite useful but had no need to actually post till now.

I am stuck on the following problem in "introduction to physics"
We should calculate the [itex]\oint[/itex] [itex]\vec{v}[/itex].d[itex]\vec{A}[/itex]

of a object with the following parameters Vol = x^2+y+2<4, 0<z<4-x^2-y^2
Is that first inequality supposed to be ##x^2+y^2<4##?
with [itex]\vec{v}[/itex] being x+y,y+z,x+z)
with and without the div. theorem,

Homework Equations


Divergence theorem,div = 3 - simple cylinder integral = possible solution =32pi
Parametrization of volume as
2cos(u)sin(v)
2sin(u)sin(v)
4cos(v)
u = (0,2pi) v= (0,pi/2)
That is not a correct parameterization. You don't have a sphere, you have an upside down paraboloid.
Surface integration using [itex]\vec{v}[/itex](u,v).ru x rv

The Attempt at a Solution


I have 32Pi as a solution using the div theorem,
and i am in over 2 pages of writing now without using it,
Am i missing something or is this the kind of problem that only exist "to show you how useful something is" and i really should come to some horrible solution that is taking a double integral of something over 50 symbols long ?

Thanks in advance.

I don't think your volume is expressed correctly even with my assumption above. You have apparently worked the volume integral for the region under the paraboloid and above the xy plane. But what does the cylinder have to do with it? Something is wrong with your statement of the problem.
 

FAQ: Surface integral of a vector field w/o div. theorem

What is a surface integral of a vector field without using the divergence theorem?

A surface integral of a vector field without using the divergence theorem is a mathematical technique used to calculate the flux, or flow, of a vector field through a surface. It involves evaluating the dot product of the vector field and the surface's normal vector over the surface.

How is a surface integral of a vector field different from a line integral?

A surface integral of a vector field involves integrating over a two-dimensional surface, while a line integral involves integrating over a one-dimensional curve. Additionally, a surface integral takes into account the direction of the vector field at each point on the surface, while a line integral does not.

Can the surface integral of a vector field be negative?

Yes, the surface integral of a vector field can be negative. This can occur when the direction of the vector field at a certain point on the surface is opposite to the direction of the surface's normal vector. In this case, the dot product will be negative, resulting in a negative contribution to the overall integral.

What is the physical interpretation of a surface integral of a vector field?

The surface integral of a vector field has a physical interpretation of calculating the amount of fluid or energy passing through a surface. It is commonly used in fluid mechanics, electromagnetism, and other fields to calculate flux and work.

How is the surface integral of a vector field related to the divergence theorem?

The surface integral of a vector field and the divergence theorem are closely related. The divergence theorem states that the flux of a vector field through a closed surface is equal to the volume integral of the vector field's divergence over the enclosed volume. In other words, it provides a way to calculate a surface integral using a volume integral. This relationship is useful in simplifying calculations and solving certain problems.

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