Question 27.17: Evaluate ∮S v· dS where v = x²i + 1/2y²j + 1/2 z²k and where......
Evaluate \oint _S{v· dS} where v = x^{2}i + \frac{1}{2}y^{2}j + \frac{1}{2}z^{2}k and where S is the surface of the unit cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1. The vector dS should be drawn on each of the six faces in an outward sense.
The cube is shown in Figure 27.22.We evaluate the surface integral over each of the six faces separately and then add the results.
On surface A, x = 1 and 0 ≤ y ≤ 1, 0 ≤ z ≤ 1. dS is a vector normal to this surface, drawn in an outward sense, and so we can write it as dy dz i. Then
=dy dz since on A, x = 1
The required surface integral over A is then
\int_{{z}=0}^{{z}=1}\int_{y=0}^{y=1}\,1\,\mathrm{d}y\,\mathrm{d}z=\int_{z=0}^{z=1}[y]_{0}^{1}\,\mathrm{d}z=1
On surface B, x = 0 and 0 ≤ y ≤ 1, 0 ≤ z ≤ 1. dS is a vector normal to this surface and so we can write it as −dy dz i. Then v · dS becomes −x² dy dz = 0 since x = 0. Over this surface, the integral is zero.
You should verify in a similar manner that over each of C and E the integral is \frac{1}{2}, whilst integrals over D and F are zero. The total surface integral is then 2.
