(a) Find a function \(f\) such that \({\bf{F}} = \nabla f\) and (b) use part (a) to evaluate \(\int_C {\bf{F}} \cdot d{\bf{r}}\) along the given curve \(C\).

\({\bf{F}}(x,y,z) = \sin y{\bf{i}} + (x\cos y + \cos z){\bf{j}} - y\sin z{\bf{k}}\),

Vector field is \({\bf{F}}(x,y,z) = \sin y{\bf{i}} + (x\cos y + \cos z){\bf{j}} - y\sin z{\bf{k}}\) and curve \({\bf{r}}(t) = \sin t{\bf{i}} + t{\bf{j}} + 2t{\bf{k}},0 \le t \le \frac{\pi }{2}.\)

Relation between the potential function \(f\) and vector field \({\bf{F}}\).

\(\nabla f = {\bf{F}}\)

Consider vector function \({\bf{r}}(t),a \le t \le b\) with a smooth curve \(C\). Consider \(f\) is a differentiable function two or three variables of gradient function \(\nabla f\) and is continuous on curve \(C\). Then,

\(\int_C \nabla f \cdot d{\bf{r}} = f({\bf{r}}(b)) - f({\bf{r}}(a))\)

a)

Vector field is \({\bf{F}}(x,y,z) = \sin y{\bf{i}} + (x\cos y + \cos z){\bf{j}} - y\sin z{\bf{k}}\).

Consider \(\nabla f = {f_x}(x,y,z){\bf{i}} + {f_y}(x,y,z){\bf{j}} + {f_z}(x,y,z){\bf{k}}\).

Write the relation between the potential function \(f\) and vector field \({\bf{F}}\).

\(\nabla f = {\bf{F}}\)

Substitute \({f_x}(x,y,z){\bf{i}} + {f_y}(x,y,z){\bf{j}} + {f_z}(x,y,z){\bf{k}}\) for \(\nabla f\),

\({\bf{F}} = {f_x}(x,y,z){\bf{i}} + {f_y}(x,y,z){\bf{j}} + {f_z}(x,y,z){\bf{k}}\)

Compare the equation \({\bf{F}} = {f_x}(x,y,z){\bf{i}} + {f_y}(x,y,z){\bf{j}} + {f_z}(x,y,z){\bf{k}}\) with

\({\bf{F}}(x,y,z) = \sin y{\bf{i}} + (x\cos y + \cos z){\bf{j}} - y\sin z{\bf{k}}\).

\(\begin{aligned}{l}{f_x}(x,y,z) & = \sin y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\{f_y}(x,y,z) & = x\cos y + \cos z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\\{f_z}(x,y,z) & = - y\sin z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)\end{aligned}\)

Integrate equation (1) with respect to x.

\(\begin{align} f(x,y,z) & =\int{(\sin y)}dx \\ & =\sin y\int{d}x \\ & =\sin y(x)+g(y,z)\left\{ \because \int{d}t=t \right\} \\ f(x,y,z) & =x\sin y+g(y,z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4) \end{align}\)

Apply partial differentiation with respect to \(y\) on both sides of equation (4).

\(\begin{align} {{f}_{y}}(x,y,z) & =\frac{\partial }{\partial y}(x\sin y+g(y,z)) \\ & =\frac{\partial }{\partial y}(x\sin y)+\frac{\partial }{\partial y}(g(y,z)) \\ & =x\frac{\partial }{\partial y}(\sin y)+{{g}_{y}}(y,z) \\ & =x(\cos y)+{{g}_{y}}(y,z)\left\{ \because \frac{\partial }{\partial t}(\sin t)=1 \right\}\end{align}\)

\({f_y}(x,y,z) = x\cos y + {g_y}(y,z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(5)\)

Compare the equations (2) and (5).

\({g_y}(y,z) = \cos z\)

Apply integration on both sides of equation.

\(\begin{align} \int{{{g}_{y}}}(y,z)dy & =\int{\cos }zdy \\ g(y,z) & =y\cos z+h(z)\quad \left\{ \because \int{d}t=t \right\} \end{align}\)

Here,

\(h(z)\) is function of z.

Substitute \(y\cos z + h(z)\) for \(g(y,z)\) in equation (4),

\(f(x,y,z) = x\sin y + y\cos z + h(z){\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{(6) }}\)

Apply partial differentiation with respect to \(z\) on both sides of equation (6).

\(\begin{align} {{f}_{z}}(x,y,z) & =\frac{\partial }{\partial z}(x\sin y)+\frac{\partial }{\partial z}(y\cos z)+\frac{\partial }{\partial z}(h(z)) \\ & =x\sin y\frac{\partial }{\partial z}(1)+y\frac{\partial }{\partial z}(\cos z)+{{h}^{\prime }}(z) \\ & =x\sin y(0)+y(-\sin z)+{{h}^{\prime }}(z)\left\{ \because \frac{\partial }{\partial t}(k)=0,\frac{\partial }{\partial t}(\cos t)=-\sin t \right\} \\ {{f}_{z}}(x,y,z) & =-y\sin z+{{h}^{\prime }}(z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(7) \end{align}\)

Compare the equation (3) and (7).

\({h^\prime }(z) = 0\)

Integrate equation with respect to \(z\).

\(\begin{align} h(z) & =\int{0}dz \\ & =K\left\{ \because \int{0}dt=K \right\} \end{align}\)

Here,

\(K\) is constant.

Consider the value of \(K\) as 0 . Therefore, \(h(z) = 0\).

Substitute 0 for \(h(z)\) in equation (6),

\(\begin{aligned}f(x,y,z) & = x\sin y + y\cos z + 0\\ & = x\sin y + y\cos z\end{aligned}\)

Thus, The potential function \(f\) of vector field

\({\bf{F}}(x,y,z) = \sin y{\bf{i}} + (x\cos y + \cos z){\bf{j}} - y\sin z{\bf{k}}\)is \(f(x,y,z) = x\sin y + y\cos z.\)

b)

Consider vector function \({\bf{r}}(t),a \le t \le b\) with a smooth curve \(C\). Consider \(f\) is a differentiable function two or three variables of gradient function \(\nabla f\) and is continuous on curve \(C\). Then,

\(\int_C \nabla f \cdot d{\bf{r}} = f({\bf{r}}(b)) - f({\bf{r}}(a))\,\,\,\,\,\,\,\,\,\,\,(8)\)

The point on curve \({\bf{r}}(t) = \sin t{\bf{i}} + t{\bf{j}} + 2t{\bf{k}}\) is,

\((x,y,z) = (\sin t,t,2t)\,\,\,\,\,\,\,\,\,\,\,\,\,(9)\)

Substitute 0 for \(t\),

\(\begin{array}{c}(x,y,z) = (\sin (0),0,2(0))\\ = (0,0,0)\end{array}\)

Hence, the lower limit \((a)\) is \((0,0,0)\).

Substitute \(\frac{\pi }{2}\) for \(t\),

\(\begin{array}{c}(x,y,z) = \left( {\sin \left( {\frac{\pi }{2}} \right),\frac{\pi }{2},2\left( {\frac{\pi }{2}} \right)} \right)\\ = \left( {1,\frac{\pi }{2},\pi } \right)\end{array}\)

Hence, the upper limit \((b)\) is \(\left( {1,\frac{\pi }{2},\pi } \right)\).

Find the common value \(\int_C \nabla f \cdot d{\bf{r}}\) by the use of equation (8).

\(\begin{aligned} \int_{C}{\nabla }f\cdot d\mathbf{r} & =f\left( 1,\frac{\pi }{2},\pi \right)-f(0,0,0) \\ & =\left\{\begin{aligned}{*{35}{l}} \left. \left( (1)\sin \left( \frac{\pi }{2} \right)+\left( \frac{\pi }{2} \right)\cos (\pi ) \right)- \right\} \\ ((0)\sin (0)+(0)\cos (0)) \\ \end{aligned} \right\}\{\because f(x,y,z)=x\sin y+y\cos z\} \\ & =(1)(1)+\left( \frac{\pi }{2} \right)(-1)-0 \\ & =1-\frac{\pi }{2} \end{aligned}\)

Thus, the value of \(\int_C \nabla f \cdot d{\bf{r}}\) along the curve \(C\) is \(1 - \frac{\pi }{2}\).