Question

(a) Show that the vector field F = sin y cos x i + cos y sin x j + z k is conservative, or equivalently satisfies the equations of the (curl) component test. (b) Find a scalar potential function f(x, y, z) for the vector field F(x, y, z), and show all steps of your construction of f. (c) Evaluate the line integral ∫cF *dr where C is a path from the point A(0, 1, 0) to B(π/2, π/2, 2).

Answer 1

a. If a vector field $\vec F$ is conservative, then its curl is 0:

$\mathrm{curl}\vec F=\left(\dfrac{\partial(\cos y\sin x)}{\partial z}-\dfrac{\partial(z)}{\partial y}\right)\,\vec\imath-\left(\dfrac{\partial(z)}{\partial x}-\dfrac{\partial(\sin y\cos x)}{\partial z}\right)\,\vec\jmath+\left(\dfrac{\partial(\cos y\sin x)}{\partial x}-\dfrac{\partial(\sin y\cos x)}{\partial y}\right)\,\vec k=\boxed{\vec 0}$

(Note that, generally, if the curl is 0, the field need not be conservative!)

b. If there is a scalar function $f$ for which $\nabla f=\vec F$, then

$\dfrac{\partial f}{\partial x}=\sin y\cos x$

$\dfrac{\partial f}{\partial y}=\cos y\sin x$

$\dfrac{\partial f}{\partial z}=z$

From these equations we find (first by integrating both sides of the first equation with respect to $x$)

$f(x,y,z)=\sin y\sin x+g(y,z)$

$\implies\dfrac{\partial f}{\partial y}=\cos y\sin x=\cos y\sin x+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

$\implies\dfrac{\partial f}{\partial z}=z=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac{z^2}2+C$

So we have

$\boxed{f(x,y,z)=\sin y\sin x+\dfrac{z^2}2+C}$

c. By the gradient theorem (i.e. fundamental theorem of calculus), we have for any path $C$ beginning at $(x_1,y_1,z_1)$ and terminating at $(x_2,y_2,z_2)$

$\displaystyle\int_C\nabla f\cdot\mathrm d\vec r=f(x_2,y_2,z_2)-f(x_1,y_1,z_1)$

so that

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=f\left(\frac\pi2,\frac\pi2,2\right)-f(0,1,0)=\boxed3$