I Integral equation for large surfaces

[Apashanka]

We often neglect the terms of a surface integral $\int_v(\nabla•A)dv=\int_s(A•ds)$ for $s$ to be very large or $v$ to be very large,
What is actually the reason behind this to neglect??

 

[Orodruin]

The field $\vec A$ is assumed to go to zero at infinity at a rate faster than the rate at which the surface grows.

 

[Apashanka]

The field $\vec A$ is assumed to go to zero at infinity at a rate faster than the rate at which the surface grows.

Okk got it ...thanks @Orodruin

 

[Apashanka]

The field $\vec A$ is assumed to go to zero at infinity at a rate faster than the rate at which the surface grows.

But if the rate at which the vector field $A$ goes to zero at infinity is slower than the rate at which the surface grows ,then also it would be zero ,isn't it??

 

[Orodruin]

No. You need the assumption of the field going to zero fast enough. Take the example of the field of a single point charge.

 

[Apashanka]

No. You need the assumption of the field going to zero fast enough. Take the example of the field of a single point charge.

Ok the field decreses as $\frac{1}{r^2}$ and the spherical surface element centered at the point charge grows as $r^2sin\theta d\theta d\phi$

 

[Orodruin]

Ok the field decreses as $\frac{1}{r^2}$ and the spherical surface element centered at the point charge grows as $r^2sin\theta d\theta d\phi$

And so the integral is not zero.