Question 27.19: A cube of side 2 units is constructed with five solid faces ......
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
v = (x+y)i+(y+z)j+(x+z)k
(a) Evaluate \oint_{C}\mathbf{V}.dr.
(b) Evaluate curl v.
(c) Evaluate \int_{s} curl v dS, 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} v · dr must be performed in the sense shown. In this plane z = 2 and dz = 0, and hence
v· dr = (x+y)dx+ (y+2)dy
We 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}}+2x\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}}+2y\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) \mathrm{curlv}=\left|\begin{array}{c c c}{{\mathrm{i}}}&{{\mathrm{j}}}&{{\mathrm{k}}}\\ {{\frac{\partial}{\partial x}}}&{{\frac{\partial}{\partial y}}}&{{\frac{\partial}{\partial z}}}\\ \mathrm{x}+y & y+z & x+z\end{array}\right|=-\mathrm{i} – \mathrm{j}-\mathrm{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
\int_{A}\mathrm{curl} \mathbf{v}\cdot\mathrm{d}\mathbf{S}=\int_{z=0}^{z=2}\int_{y=0}^{y=2}-\mathrm{d}\mathbf{y}\mathrm{d}z\\ =\int_{z=0}^{z=2}{[-y]^2_0} dz\\ =\int_{z=0}^{z=2}{-2} dz\\ =[-2z]^{2}_0\\= −4
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}{\mathrm{curl}}\,{\mathbf{v}}\cdot{\mathrm{d}}\mathbf{S}=-4+4-4+4+4=4Note from part (a) that this equals \oint{v.dr} and so we have verified Stokes’ theorem.