a) If u_(1),u_(2),u_(3) are curvilinear coordinates such that x=(1)/(2)(u_(1)-3) y=u_(2)+4;z=u_(3)-2 prove that the system is orthogonal. Find the scale factor and the expression for (ds)^2 b) Evaluate oint _(c)overrightarrow (F)cdot doverrightarrow (r) by Stoke's theorem, where overrightarrow (F)=y^2i+x^2j-(x+z)k and c is the boundary of triangle with vertices at (0,0,0),(1,0,0) and (1,1,0) (12mks) (8mks) Question 5 (20 marks) a) State and verify Green's theorem in the plane for oint (3x^2-8y^2)dx+(4y-6xy)dy where c is the boundary of the region bounded by xgeqslant 0,yleqslant 0 and 2x-3y=6 (18mks)

a) To prove that the system is orthogonal, we need to show that the partial derivatives of x, y, and z with respect to each other are zero. Let's calculate the partial derivatives:<br /><br />$\frac{\partial x}{\partial u_1} = \frac{1}{2}$<br />$\frac{\partial x}{\partial u_2} = 0$<br />$\frac{\partial x}{\partial u_3} = 0$<br /><br />$\frac{\partial y}{\partial u_1} = 0$<br />$\frac{\partial y}{\partial u_2} = 1$<br />$\frac{\partial y}{\partial u_3} = 0$<br /><br />$\frac{\partial z}{\partial u_1} = 0$<br />$\frac{\partial z}{\partial u_2} = 0$<br />$\frac{\partial z}{\partial u_3} = 1$<br /><br />Since all partial derivatives are zero except for those involving $u_1$ and $u_3$, the system is orthogonal.<br /><br />The scale factor is given by $\sqrt{\left(\frac{\partial x}{\partial u_1}\right)^2 + \left(\frac{\partial y}{\partial u_2}\right)^2 + \left(\frac{\partial z}{\partial u_3}\right)^2} = \sqrt{\left(\frac{1}{2}\right)^2 + 1^2 + 1^2} = \sqrt{\frac{1}{4} + 1 + 1} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.<br /><br />The expression for $(ds)^2$ is given by $(ds)^2 = \left(\frac{\partial x}{\partial u_1}\right)^2 du_1^2 + \left(\frac{\partial y}{\partial u_2}\right)^2 du_2^2 + \left(\frac{\partial z}{\partial u_3}\right)^2 du_3^2 = \frac{1}{4} du_1^2 + du_2^2 + du_3^2$.<br /><br />b) To evaluate $\oint_C \overrightarrow{F} \cdot d\overrightarrow{r}$ using Stokes' theorem, we need to find the curl of $\overrightarrow{F}$ and then integrate it over the surface bounded by the curve C.<br /><br />The curl of $\overrightarrow{F}$ is given by $\nabla \times \overrightarrow{F} = (2y) \hat{i} + (2x) \hat{j} - \hat{k}$.<br /><br />The surface bounded by the curve C is a triangle in the xy-plane with vertices at $(0,0)$, $(1,0)$, and $(1,1)$. We can parametrize this surface as $\overrightarrow{r}(x,y) = x \hat{i} + y \hat{j}$, where $0 \leq x \leq 1$ and $0 \leq y \leq x$.<br /><br />The outward normal to the surface is given by $\hat{n} = \frac{\partial \overrightarrow{r}}{\partial x} \times \frac{\partial \overrightarrow{r}}{\partial y} = \hat{i} + \hat{j}$.<br /><br />Using Stokes' theorem, we have $\oint_C \overrightarrow{F} \cdot d\overrightarrow{r} = \iint_S (\nabla \times \overrightarrow{F}) \cdot \hat{n} \, dS = \iint_S (2y \hat{i} + 2x \hat{j} - \hat{k}) \cdot (\hat{i} + \hat{j}) \, dS = \iint_S (2y + 2x - 1) \, dS$.<br /><br />The area of the triangle is given by $A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$.<br /><br />Therefore, $\oint_C \overrightarrow{F} \cdot d\overrightarrow{r} = \iint_S (2y + 2x - 1) \, dS = \int_0^1 \int_0^x (2y + 2x - 1) \, dy \, dx = \int_0^1 \left[ y^2 + 2xy - y \right]_0^x \, dx = \int_0^1 (x^2 + 2x^2 - x) \, dx = \int_0^1 (3x^2 - x) \, dx = \left[ x^3 - \frac{x^2}{2}