- #1

LCSphysicist

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- 153

- Homework Statement:
- I am arguing with myself if the way i solved this problem is right

- Relevant Equations:
- .

It is more or less a generic problem of stokes theorem:

##\int_{\gamma} F dr##, where ##F = (-y/(x²+y²) + z,x/(x²+y²),ln(2+z^10))## and gamma is the intersection of ##z=y^2, x^2 + y^2 = 9## oriented in such way that its projection in xy is traveled clockwise.

So, i decided to apply stokes theorem to the cone surface, considering its boundary to be the one with z=0 and the other being the intersection said above. Call the answer A

##A + \int_{\alpha} (-y/(x^2+y^2), x/(x^2+y^2), ln(2)) * dC = \int \int grad F * ds##

Alpha is the circle at z=0, x²+y²=9, and the right side of the equation will give zero if you calculate it. Doing all the integrals it end as:

##A + 2\pi = 0 => A = -2\pi##

In fact the answer is right. But that is not the problem, i want to know about the singularities!

I mean i just ignored it, following the reasoning that

1: the grad will be calculate on the cylinder surface, which does not intersect the origin

2: the line integral of the circle at z=0 will also not intersect the origin

So, taking in account that the integral line and integral surface "does not know" about the singularity at the origin, i just ignored it. Is it right? (If it were necessary to calculate div F over the volume, in this case, i would need to pay attention at the singularity since the volume spread over there)

##\int_{\gamma} F dr##, where ##F = (-y/(x²+y²) + z,x/(x²+y²),ln(2+z^10))## and gamma is the intersection of ##z=y^2, x^2 + y^2 = 9## oriented in such way that its projection in xy is traveled clockwise.

So, i decided to apply stokes theorem to the cone surface, considering its boundary to be the one with z=0 and the other being the intersection said above. Call the answer A

##A + \int_{\alpha} (-y/(x^2+y^2), x/(x^2+y^2), ln(2)) * dC = \int \int grad F * ds##

Alpha is the circle at z=0, x²+y²=9, and the right side of the equation will give zero if you calculate it. Doing all the integrals it end as:

##A + 2\pi = 0 => A = -2\pi##

In fact the answer is right. But that is not the problem, i want to know about the singularities!

I mean i just ignored it, following the reasoning that

1: the grad will be calculate on the cylinder surface, which does not intersect the origin

2: the line integral of the circle at z=0 will also not intersect the origin

So, taking in account that the integral line and integral surface "does not know" about the singularity at the origin, i just ignored it. Is it right? (If it were necessary to calculate div F over the volume, in this case, i would need to pay attention at the singularity since the volume spread over there)

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