1.1 A point \(P\) is on the line segment \(AB\) iff the vector \(\overrightarrow{AP}\) is colinear to vector \(\overrightarrow{AB}\), that is, \(\overrightarrow{AB}= t \, \overrightarrow{AB}\), with \(0\leq t\leq 1\). This gives the parametrization \[ (x-1,y-1,z) = t (-1,-1,1) \] or \[ \boxed{ \boldsymbol{\alpha}(t)= (1-t,1-t,t), \qquad 0\leq t\leq 1} \] With \(d\boldsymbol{\alpha}(t)= dt (-1,-1,1)\) and \(\boldsymbol{F}= (P,Q,R)=(2xy+y^2-1, 2xy+x^2, z)\) we find \[ \boldsymbol{F}\cdot d\boldsymbol{\alpha}(t)=(-P-Q+R) dt= (-4xy-y^2-x^2+z+1)dt= \big(-6(1-t)^2 + t+1\big) dt \] This gives \[ I = \int_{\cal L}\boldsymbol{F}\cdot d\boldsymbol{\alpha}= \int_0^1 \big(-6(1-t)^2 + t+1\big) dt= [2(1-t)^3+ \frac{1}{2}(t+1)^2]= \boxed{-\frac{1}{2}} \]
1.2 We have the equalities \(\frac{\partial R}{\partial x}= \frac{\partial P}{\partial z}=0\), \(\frac{\partial P}{\partial y}= \frac{\partial Q}{\partial x}=2x+2y\), and \(\frac{\partial Q}{\partial z}= \frac{\partial R}{\partial y}=0\). So it likely that a potential field \(\varphi(x,y,z)\) exists. We set \(\boldsymbol{F}=\boldsymbol{\nabla}\varphi\) and first integrate \(P= \frac{\partial \varphi}{\partial x}\): \[ \varphi(x,y,z)= x^2 y+y^2 x -x +\psi(y,z) \] Then we enforce \(Q= \frac{\partial \varphi}{\partial y}\): \[ 2xy +x^2 = 2xy+ x^2+\frac{\partial \psi}{\partial y} \Longrightarrow \psi(y,z)= f(z) \] Finally \(R= \frac{\partial \varphi}{\partial z}\) gives \[ z =\frac{df}{dz}\Longrightarrow f(z)= \frac{z^2}{2} + Cst \] Conclusion: \(\boldsymbol{F}\) is a gradient field with \(\varphi\) given by \[ \boxed{ \varphi(x,y,z)= x^2 y + xy^2-x+ \frac{z^2}{2} +Cst } \]
1.3 We note that \(\boldsymbol{\beta}(0)=(1,1,0)=A\) and \(\boldsymbol{\beta}(1)=(0,0,1)\). Hence, path \({\cal C}\) starts at \(A\) and ends at \(B\). Since \(\boldsymbol{F}\) is gradient field, its line integral should be independent of the path connecting \(A\) to \(B\): we should find the same answer as that found in 1.1. \[ \int_{\cal C}\boldsymbol{F}\cdot d\boldsymbol{\beta}= \int_{\cal C}d\varphi= \varphi(0,0,1)-\varphi(1,1,0)= \boxed{-\frac{1}{2}} \]
The partial derivatives of function \(g(u,v)=f(X(u,v),Y(u,v))\) w.r.t. \((u,v)\) can be found from those of \(f\) w.r.t. \((x,y)\) according to the chain rule: \[ \frac{\partial g}{\partial u} = \frac{\partial f}{\partial x} \frac{\partial X}{\partial u} + \frac{\partial f}{\partial y} \frac{\partial Y}{\partial u} = v \frac{\partial f}{\partial x} + 2u \frac{\partial f}{\partial y} \qquad(1) \] \[ \frac{\partial g}{\partial v} = \frac{\partial f}{\partial x} \frac{\partial X}{\partial v} + \frac{\partial f}{\partial y} \frac{\partial Y}{\partial v} = u \frac{\partial f}{\partial x} + 2v\frac{\partial f}{\partial y} \qquad(2) \] using \(X(u,v)=uv\) and \(Y(u,v)=u^2+v^2\).
Following the same process, we find \(\frac{\partial}{\partial u} \frac{\partial f}{\partial x}\): we apply formula (1), replacing function \(f\) by \({\partial f}/{\partial x}\) \[ \frac{\partial}{\partial u} \frac{\partial f}{\partial x} (u+v,uv^2) = v\frac{\partial^2 f}{\partial x^2} (u+v,uv^2) + 2u \frac{\partial^2 f}{\partial y\partial x} (u+v,uv^2) \] Likewise, \[ \frac{\partial}{\partial u} \frac{\partial f}{\partial y} (u+v,uv^2) = v\frac{\partial^2 f}{\partial x\partial y} (u+v,uv^2) + 2u \frac{\partial^2 f}{\partial y^2} (u+v,uv^2) \] Now we can find \(\frac{\partial^2 g}{\partial u\partial v}\) \[ \frac{\partial^2 g}{\partial u\partial v} (u,v)= uv \frac{\partial^2 f}{\partial x^2} +2u^2 \frac{\partial^2 f}{\partial y \partial x} +2v^2 \frac{\partial^2 f}{\partial x \partial y} +4uv \frac{\partial^2 f}{\partial y^2} + \frac{\partial f}{\partial x} \] Since \(f\) is of class \(C^2\), we have equality of the mixed partial derivatives: \[ \boxed{ \frac{\partial^2 g}{\partial u\partial v} (u,v)= uv \frac{\partial^2 f}{\partial x^2} +2(u^2+v^2) \frac{\partial^2 f}{\partial x \partial y} +4uv \frac{\partial^2 f}{\partial y^2} + \frac{\partial f}{\partial x} = \frac{\partial^2 g}{\partial v\partial u} (u,v) } \] With the numerical values \[ \frac{\partial f}{\partial x}=1, \quad \frac{\partial f}{\partial y}=-1, \quad \frac{\partial^2 f}{\partial x^2}= \frac{\partial^2 f}{\partial y^2}=\frac{\partial^2 f}{\partial x\partial y}= \frac{\partial^2 f}{\partial y\partial x}=1 \] and \((u,v)=(1,2)\) we find \[ \boxed{ \frac{\partial^2 g}{\partial u\partial v} (1,2)= 2+10+9=21 = \frac{\partial^2 g}{\partial v\partial u} (1,2)} \]
3.1. With \(\boldsymbol{F}= (x,y,z)\) we find \(\mathop{\mathrm{div}}\boldsymbol{F}=3\). Then the integral of \(\mathop{\mathrm{div}}\boldsymbol{F}\) over \({\cal R}\) can simply be found from the knowledge of the volume of 1/8th of the unit sphere: \[ \iiint_{\cal R}\mathop{\mathrm{div}}\boldsymbol{F}\, dV = 3 \text{Vol}({\cal R})= \boxed{\frac{\pi}{2}} \]
3.2 Surface \(S_0\) can be parametrized as follows (spherical coordinates) \[ (u,v)\in\Omega=[0,\frac{\pi}{2}]\times [0,\frac{\pi}{2}] \to \boldsymbol{r}= (\sin u \cos v, \sin u \sin v, \cos u) \] Then we can find normal \(\boldsymbol{N}\): \[ \boldsymbol{N}= \frac{\partial \boldsymbol{r}}{\partial u} \times \frac{\partial \boldsymbol{r}}{\partial v} = \begin{pmatrix} \cos u \cos v \\ \cos u \sin v \\ -\sin u\end{pmatrix} \times \begin{pmatrix} -\sin u \sin v \\ \sin u \cos v \\ 0\end{pmatrix} \] This gives \[ \boldsymbol{N}= \begin{pmatrix}\sin^2 u \cos v \\ \sin^2 u \sin v \\ \cos u \sin u (\cos^2 v+\sin^2v) \end{pmatrix} = \sin u \begin{pmatrix}\sin u \cos v \\ \sin u \sin v \\ \cos u \end{pmatrix} = \sin u \begin{pmatrix}x \\ y \\ z \end{pmatrix} \] Now we can find the flux of \(\boldsymbol{F}\) through surface \(S_0\): \[ \iint_{S_0}\boldsymbol{F}\cdot \boldsymbol{N}dA = \iint_\Omega (x^2 +y^2 +z^2) \sin u \, du dv= \iint_\Omega \sin u \, du dv= \frac{\pi}{2}\int_0^{\pi/2} \sin u du = \boxed{\frac{\pi}{2}} \] where we have use \(x^2+y^2+z^2=1\) on \(S_0\).
3.3 On surface \(S_1\) we have \(x=0\) and \(\boldsymbol{\hat{n}}= (-1,0,0)\): this gives \(\boldsymbol{F}\cdot \boldsymbol{\hat{n}}=0\) at all points of \(S_1\): \[ \iint_{S_1}\boldsymbol{F}\cdot\boldsymbol{\hat{n}}dA = \boxed{0} \] The same thing happens for the flux of \(\boldsymbol{F}\) through surface \(S_2\) (\(y=0\) and \(\boldsymbol{\hat{n}}=(0,-1,0)\)) and through surface \(S_3\) (\(z=0\) and \(\boldsymbol{\hat{n}}= (0,0,-1)\)): \[ \iint_{S_2}\boldsymbol{F}\cdot\boldsymbol{\hat{n}}dA = \iint_{S_3}\boldsymbol{F}\cdot\boldsymbol{\hat{n}}dA = \boxed{0} \] We verify that the divergence theorem is satisfied: \[ \iiint_{\cal R}\mathop{\mathrm{div}}\boldsymbol{F}dV = \iint_{S_0}\boldsymbol{F}\cdot\boldsymbol{\hat{n}}dA + \iint_{S_1}\boldsymbol{F}\cdot\boldsymbol{\hat{n}}dA + \iint_{S_2}\boldsymbol{F}\cdot\boldsymbol{\hat{n}}dA + \iint_{S_3}\boldsymbol{F}\cdot\boldsymbol{\hat{n}}dA = \frac{1}{2} \]
3.4 According to Stokes’ Theorem, we have \[ \oint_{\cal C}\boldsymbol{F}\cdot d\boldsymbol{\alpha}= \iint_{S_0}\mathop{\mathrm{curl}}\boldsymbol{F}\cdot \boldsymbol{\hat{n}}\, dA \] It is straightforward to verify that \(\mathop{\mathrm{curl}}\boldsymbol{F}=\boldsymbol{0}\) everywhere. We conclude that \[ \boxed{\oint_{\cal C}\boldsymbol{F}\cdot d\boldsymbol{\alpha}= 0} \]