Discussion
Given a function \(\varphi = \frac{1}{2}(x^{2} + y^{2} + z^{2})\) in three-dimensional Cartesian space, the value of the surface integral
\[ \oint\!\!\oint_{S}\hat{\mathbf{n}}\cdot\nabla\varphi\,dS , \]
where \(S\) is the surface of a sphere of unit radius and \(\hat{\mathbf{n}}\) is the outward unit normal vector on \(S\), is
Answer: A
The surface integral equals \(4\pi\).
The gradient of \(\varphi = \frac{1}{2}(x^{2} + y^{2} + z^{2})\) is the position vector \(\nabla\varphi = (x, y, z) = \vec{r}\). On the unit sphere the outward normal is \(\hat{\mathbf{n}} = \vec{r}\) itself, so \(\hat{\mathbf{n}} \cdot \nabla\varphi = |\vec{r}|^{2} = 1\) everywhere, and the integral is just the area of the sphere, \(4\pi\).
The divergence theorem confirms it: \(\nabla^{2}\varphi = 3\), and \(3 \times \frac{4}{3}\pi = 4\pi\).
\(4\pi/3\) is the volume of the sphere without the factor 3, \(3\pi\) has no basis, and 0 would require \(\varphi\) to be harmonic.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development