Question 27.15: (a) Evaluate ∮C xy dx + x² dy around the sides of the square......
(a) Evaluate \oint_{C}{xy } dx + x^{2 dy} around the sides of the square with vertices D(0, 0), E(1, 0), F(1, 1) and G(0, 1).
(b) Convert the line integral to a double integral and verify Green’s theorem.
(a) The path of integration is shown in Figure 27.15. To apply Green’s theorem the path of integration should be followed in such a way that the region of integration is always to its left. We must therefore travel anticlockwise around C.
On DE, y = 0, dy = 0 and 0 ≤ x ≤ 1.
On EF, x = 1, dx = 0 and 0 ≤ y ≤ 1.
On FG, y = 1, dy = 0 and x decreases from 1 to 0.
On GD, x = 0, dx = 0 and y decreases from 1 to 0.
The integral around the curve C can then be written as
\oint_{C}=\int_{\mathrm{D}}^{\mathrm{E}}+\int_{\mathrm{E}}^{\mathrm{F}}+\int_{\mathrm{F}}^{\mathrm{G}}+\int_{G}^{\mathrm{D}}Therefore
\oint_{C}x y\,\mathrm{d}x+x^{2}\,\mathrm{d}y=0+\int_{0}^{1}\,1\,\mathrm{d}y+\int_{1}^{0}x\,\mathrm{d}x+0\\ =[y]_{0}^{1}+\left[\frac{x^{2}}{2}\right]_{1}^{0}\\ =1-{\frac{1}{2}}\\ {}={\frac{1}{\operatorname{2}}}(b) Applying Green’s theorem with P(x, y) = xy and Q(x, y) = x² we can convert the line integral into a double integral. Note that \frac{\partial Q}{\partial x}=2x and {\frac{\partial P}{\partial y}}=x. Clearly the region of integration is the square R. We find
\oint_{C}x y\,\mathrm{d}x+x^{2}\,\mathrm{d}y=\iint_{R}(2x-x)\,\mathrm{d}x\,\mathrm{d}y \\ =\int_{0}^{1}\int_{0}^{1}x\,\mathrm{d}x\,\mathrm{d}y \\ =\int_{0}^{1}\biggl[{\frac{x^{2}}{2}}\biggr]^{1}_0\mathrm{d}y \\ =\int_{0}^{1}{\frac{1}{2}}\,\mathrm{d}y \\ =\left[{\frac{1}{2}}y\right]_{0}^{1}\\ {}={\frac{1}{2}}We see that the same result as that in part (a) is obtained and so Green’s theorem has been verified.
