If S is a closed surface enclosing a volume V and \(\hat{n}\) is the unit vector normal to the surface and \(\bar{r}\) is the position vector, then the value of the integral \(\iint_{S} \hat{n}\, d\,S\) is :
MH SET - 2020, SET Exam - 2020,
Answer: C
The integral is 0.
Apply the gradient form of Gauss's theorem, \(\oint_{S} \phi\, \hat{n}\, dS = \int_{V} \nabla\phi\, dV\), with the constant scalar \(\phi = 1\). Its gradient is zero, so the closed-surface integral of the unit normal vanishes:
\[ \oint_{S} \hat{n}\, dS = \int_{V} \nabla(1)\, dV = 0 \]
Geometrically, the vector areas of opposite parts of a closed surface cancel, as the six faces of a cube show. V and 3V are what one gets for the different integral \(\oint \bar{r}\cdot\hat{n}\, dS = 3V\) (or a third of it), which is a scalar, not this vector integral.
Please log in or register to report this question.