A cube of side 2 units is constructed with five solid faces and one open face. It is located in the region defined by 0 ≤ x ≤ 2, 0 ≤ y ≤ 2, 0 ≤ z ≤ 2 and its open face is its top face, bounded by the curve C, lying in the plane z = 2, as shown in Figure 27.26. Throughout this region a vector field is given by
\mathbf{v}=(x+y) \mathbf{i}+(y+z) \mathbf{j}+(x+z) \mathbf{k}(a) Evaluate \oint_C \mathbf{v} \cdot \mathrm{d} \mathbf{r}
(b) Evaluate curl v.
(c) Evaluate \int_S \text { curl } \mathbf{v} \mathrm{d} \mathbf{S} \text {, and verify Stokes' theorem. }
(a) The open face is highlighted in Figure 27.26. It is bounded by the curve C around which the line integral \oint_C \mathbf{v} \cdot \mathrm{d} \mathbf{r} must be performed in the sense shown. In this plane z = 2 and dz = 0, and hence
\mathbf{v} \cdot \mathrm{d} \mathbf{r}=(x+y) \mathrm{d} x+(y+2) \mathrm{d} yWe perform the line integral around C in four stages.
On I, x = 0, dx = 0 and hence v • dr = (y + 2) dy. Noting that y varies from 0 to 2, the contribution to the line integral is
On II, y = 2, dy = 0 and hence v • dr = (x + 2) dx. Noting that x varies from 0 to 2, the contribution to the line integral is
\int_0^2 x+2 \mathrm{~d} x=\left[\frac{x^2}{2}+2 x\right]_0^2=6On III, x = 2, dx = 0 and hence v • dr = (y + 2) dy. Here y varies from 2 to 0. This contribution to the line integral is
\int_2^0 y+2 \mathrm{~d} y=\left[\frac{y^2}{2}+2 y\right]_2^0=-6On IV, y = 0, dy = 0 and hence v • dr = x dx. Here x varies from 2 to 0. The contribution to the line integral is
\int_2^0 x \mathrm{~d} x=\left[\frac{x^2}{2}\right]_2^0=-2Putting all these results together we find
\oint_C \mathbf{v} \cdot \mathrm{d} \mathbf{r}=6+6-6-2=4(b) \operatorname{curl} \mathbf{v}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x+y & y+z & x+z \end{array}\right|=-\mathbf{i}-\mathbf{j}-\mathbf{k}
(c) Now we calculate the surface integral which must be performed over the five solid surfaces separately. Refer to Figure 27.26. On surface A, the front face lying in the plane x = 2, dS = dy dz i. Hence curl v • dS = -dy dz. Then
\begin{aligned} \int_A \operatorname{curl} \mathbf{v} \cdot \mathrm{d} \mathbf{S} & =\int_{z=0}^{z=2} \int_{y=0}^{y=2}-\mathrm{d} y \mathrm{~d} z \\ & =\int_{z=0}^{z=2}[-y]_0^2 \mathrm{~d} z \\ & =\int_{z=0}^{z=2}-2 \mathrm{~d} z \\ & =[-2 z]_0^2 \\ & =-4 \end{aligned}On B, the back face lying in the plane x = 0, dS = -dy dz i. It follows that curl v • dS = dy dz. The required integral over B is
\int_{z=0}^{z=2} \int_{y=0}^{y=2} \mathrm{~d} y \mathrm{~d} z=4On C, the right-hand face, dS = dx dz j. Hence curl v • dS = -dx dz. The required integral over C is
\int_{z=0}^{z=2} \int_{x=0}^{x=2}-\mathrm{d} x \mathrm{~d} z=-4Similarly, on D, the left-hand face, dS = -dx dz j. Hence curl v • dS = dx dz. The required integral over D is
\int_{z=0}^{z=2} \int_{x=0}^{x=2} \mathrm{~d} x \mathrm{~d} z=4On surface F, the base, z = 0 and dS = -dx dy k. Hence curl v • dS = dx dy. The required integral over F is
\int_{y=0}^{y=2} \int_{x=0}^{x=2} \mathrm{~d} x \mathrm{~d} y=4Recall that the top surface E is open, and so we have completed the surface integrals. Finally, putting these results together,
\int_S \operatorname{curl} \mathbf{v} \cdot \mathrm{d} \mathbf{S}=-4+4-4+4+4=4Note from part (a) that this equals \oint \mathbf{v} \cdot \mathrm{d} \mathbf{r} and so we have verified Stokes’ theorem.