1. **Problem Statement:** Verify Green's theorem for the given line integrals and curves, and explain its application in mesh processing. --- ### Part i 2. **Green's Theorem Statement:** Green's theorem relates a line integral around a simple closed curve $C$ to a double integral over the region $D$ enclosed by $C$: $$\oint_C P\,dx + Q\,dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA$$ where $P=2x - y$ and $Q=x + 3y$. 3. **Calculate partial derivatives:** $$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x + 3y) = 1$$ $$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(2x - y) = -1$$ 4. **Compute the double integral:** $$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (-1) = 2$$ The region $D$ is the triangle with vertices $(0,0), (1,0), (0,1)$. Set up the integral: $$\iint_D 2 \, dA = 2 \times \text{Area}(D)$$ Area of the triangle: $$\frac{1}{2} \times 1 \times 1 = \frac{1}{2}$$ So, $$\iint_D 2 \, dA = 2 \times \frac{1}{2} = 1$$ 5. **Calculate the line integral directly:** Parametrize each side of the triangle and compute $\oint_C P\,dx + Q\,dy$. - Side 1: $(0,0)$ to $(1,0)$, $x=t$, $y=0$, $t \in [0,1]$ $$dx = dt, dy=0$$ Integral: $$\int_0^1 (2t - 0) dt = \int_0^1 2t dt = [t^2]_0^1 = 1$$ - Side 2: $(1,0)$ to $(0,1)$, parametrize $x=1 - t$, $y=t$, $t \in [0,1]$ $$dx = -dt, dy = dt$$ Integral: $$\int_0^1 \left[(2(1 - t) - t)(-1) + ((1 - t) + 3t)(1)\right] dt$$ Simplify inside: $$ (2 - 2t - t)(-1) + (1 - t + 3t) = (2 - 3t)(-1) + (1 + 2t) = -2 + 3t + 1 + 2t = -1 + 5t$$ Integral: $$\int_0^1 (-1 + 5t) dt = [-t + \frac{5t^2}{2}]_0^1 = -1 + \frac{5}{2} = \frac{3}{2}$$ - Side 3: $(0,1)$ to $(0,0)$, $x=0$, $y=1 - t$, $t \in [0,1]$ $$dx=0, dy = -dt$$ Integral: $$\int_0^1 \left[(2\cdot0 - (1 - t))\cdot 0 + (0 + 3(1 - t))(-1)\right] dt = \int_0^1 -3(1 - t) dt = \int_0^1 (-3 + 3t) dt$$ Calculate: $$[-3t + \frac{3t^2}{2}]_0^1 = -3 + \frac{3}{2} = -\frac{3}{2}$$ 6. **Sum all integrals:** $$1 + \frac{3}{2} - \frac{3}{2} = 1$$ This matches the double integral result, verifying Green's theorem. --- ### Part ii 7. **Given:** $$P = y \sin x, \quad Q = x \cos y$$ Curve $C$ is the square with vertices $(0,0), (a,0), (a,a), (0,a)$. 8. **Calculate partial derivatives:** $$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x \cos y) = \cos y$$ $$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(y \sin x) = \sin x$$ 9. **Compute the double integral:** $$\iint_D (\cos y - \sin x) dA$$ Region $D$ is the square $0 \leq x \leq a$, $0 \leq y \leq a$. Set up the integral: $$\int_0^a \int_0^a (\cos y - \sin x) dx dy = \int_0^a \left[ \int_0^a \cos y dx - \int_0^a \sin x dx \right] dy$$ Since $\cos y$ is constant w.r.t. $x$: $$= \int_0^a (a \cos y) dy - \int_0^a \left[-\cos x \right]_0^a dy = a \int_0^a \cos y dy - \int_0^a (-\cos a + 1) dy$$ Calculate each: $$a [\sin y]_0^a - \int_0^a (1 - \cos a) dy = a (\sin a - 0) - (1 - \cos a) a = a \sin a - a + a \cos a$$ So, $$\iint_D (\cos y - \sin x) dA = a(\sin a + \cos a - 1)$$ 10. **Calculate the line integral:** Parametrize each side and compute $\oint_C P dx + Q dy$. - Side 1: $(0,0)$ to $(a,0)$, $x=t$, $y=0$, $t \in [0,a]$ $$dx = dt, dy=0$$ Integral: $$\int_0^a y \sin x dx = \int_0^a 0 dt = 0$$ - Side 2: $(a,0)$ to $(a,a)$, $x=a$, $y=t$, $t \in [0,a]$ $$dx=0, dy=dt$$ Integral: $$\int_0^a Q dy = \int_0^a a \cos t dt = a [\sin t]_0^a = a \sin a$$ - Side 3: $(a,a)$ to $(0,a)$, $x=t$, $y=a$, $t \in [a,0]$ $$dx=dt, dy=0$$ Integral: $$\int_a^0 y \sin x dx = \int_a^0 a \sin t dt = - \int_0^a a \sin t dt = -a [-\cos t]_0^a = -a (-\cos a + 1) = a (\cos a - 1)$$ - Side 4: $(0,a)$ to $(0,0)$, $x=0$, $y=t$, $t \in [a,0]$ $$dx=0, dy=dt$$ Integral: $$\int_a^0 Q dy = \int_a^0 0 \cdot \cos y dy = 0$$ 11. **Sum all integrals:** $$0 + a \sin a + a (\cos a - 1) + 0 = a (\sin a + \cos a - 1)$$ This matches the double integral, verifying Green's theorem. --- ### Part iii 12. **Application of Green's Theorem in Mesh Processing:** Green's theorem converts line integrals around a closed curve into double integrals over the enclosed area. In 2D mesh processing, this allows efficient computation of surface integrals by evaluating boundary integrals instead of integrating over the entire surface. This is useful in graphics and computer vision for calculating properties like area, centroid, and moments of 2D shapes represented as polygonal meshes. It simplifies computations and improves numerical stability by working with mesh edges rather than interior points. --- **Final answers:** - Part i: Verified, line integral and double integral both equal 1. - Part ii: Verified, line integral and double integral both equal $a(\sin a + \cos a - 1)$. - Part iii: Explained the use of Green's theorem in mesh processing for surface integrals.