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

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SUMMARY

The forum discussion centers on calculating the surface integral \(\oint \vec{v} \cdot d\vec{A}\) for a specified volume defined by the inequalities \(x^2 + y + 2 < 4\) and \(0 < z < 4 - x^2 - y^2\). The vector field \(\vec{v}\) is given as \((x+y, y+z, x+z)\). Participants confirm that the divergence theorem yields a solution of \(32\pi\), while also noting the complexity of solving the integral without this theorem. The discussion highlights the importance of correctly parameterizing the volume, as the initial parameterization was deemed incorrect for the described geometry.

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wildefire
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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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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 :) )
 
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.
 

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