(b)Since \(\mathbf{F}=\mathbf{F}(x,y)=(2xy+24x)\mathbf{i}+(x^{2}+16)\mathbf{j}\) is a conservative vector field, there is a potential function \(f\) for \(\mathbf{F}\) for which \[ \nabla\! \ f=\mathbf{F}=(2xy+24x)\mathbf{i}+(x^{2}+16)\mathbf{j} \]
Since \(\nabla\! \ f=\dfrac{\partial f}{\partial x}\mathbf{i}+\dfrac{\partial f}{\partial y}\mathbf{j}\), then \[\begin{equation*} \dfrac{\partial f}{\partial x}=2xy+24x\qquad \hbox{and}\qquad \dfrac{\partial f}{\partial y}=x^{2}+16 \end{equation*}\]
We integrate the first of these partial derivatives partially with respect to \(x\). \[\begin{equation*} f(x,y)=\int (2xy+24x)\,dx=x^{2}y+12x^{2}+h(y) \end{equation*}\]
where the “constant of integration,” \(h(y)\), is a function of \(y\). Now we differentiate \(f\) with respect to \(y\), obtaining \[\begin{equation*} \dfrac{\partial f}{\partial y}=\dfrac{\partial }{\partial y}[x^{2}y+12x^{2}+h(y)] =x^{2}+h' (y) \end{equation*}\]
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But \(\dfrac{\partial f}{\partial y}=x^{2}+16\), so \(h' (y)=16\). Now we integrate \(h'\) with respect to \(y\); then \[ h(y)=16y+K \]
where \(K\) is a constant. So, \[ f(x,y)=x^{2}y+12x^{2}+16y+K \]
is a potential function for \(\mathbf{F}.\)
(c) Since \(\mathbf{F}\) is independent of the path, we can use the Fundamental Theorem of Line Integrals. Since \(f(x,y)=x^{2}y+12x^{2}+16y + K\) and \(\mathbf{F}=\nabla\! \ f,\) \[\begin{equation*} \int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\int_{C}\nabla\! \ f\,{\cdot}\, d\mathbf{r}=\big[ f(x,y)\big] _{(0,1)}^{(1,2)}=f(1,2)-f(0,1)=46-16=30 \end{equation*}\]
for any piecewise-smooth curve \(C\) connecting \(( 0,1)\) to \(( 1,2)\).