2.5: Practice Set

Exercise \(\PageIndex{1a}\)

Determine whether \( \vecs{F}(x,y)=\langle 2x,4y \rangle \) is conservative or not. If it is conservative, find a function \(f\) such that \(\vecs{F}=\nabla f\).

Answer

We check cross partials to determine if \(\vecs{F}=\langle P(x,y),Q(x,y)\rangle \) is conservative:

\(P_y=\dfrac{\partial }{\partial y}\left( 2x \right)=0\)

\(Q_x=\dfrac{\partial }{\partial x}\left( 4y \right)=0\)

Since these match, this vector field is conservative.

To find the potential function we solve each integral separately:

\(\displaystyle \int P dx = \int 2x \, dx =x^2+h_1(y) \)

\(\displaystyle \int Q dy = \int 4y \, dy =2y^2+h_2(x) \)

Letting \(h_1(y)=2y^2\) and \(h_2(x)=x^2\) we see that these are equal, giving us the potential function:

\(f(x,y)=x^2+2y^2\) satisfying \(\nabla f=\vecs{F}\)

Exercise \(\PageIndex{1b}\)

Determine whether \( \vecs{F}=\langle 4y,2x\rangle \) is conservative or not. If it is conservative, find a function \(f\) such that \(\vecs{F}=\nabla f\).

Answer

We check cross partials to determine if \(\vecs{F}=\langle P(x,y),Q(x,y)\rangle \) is conservative:

\(P_y=\dfrac{\partial }{\partial y}\left( 4y \right)=4\)

\(Q_x=\dfrac{\partial }{\partial x}\left( 2x \right)=2\)

Since these do not match, this vector field is not conservative.

Exercise \(\PageIndex{1c}\)

Determine whether \( \vecs{F}(x,y)=\langle y\cos(xy),x\cos(xy)\rangle \) is conservative or not. If it is conservative, find a function \(f\) such that \(\vecs{F}=\nabla f\).

Answer

We check cross partials to determine if \(\vecs{F}=\langle P(x,y),Q(x,y)\rangle \) is conservative:

\(P_y=\dfrac{\partial }{\partial y}\left( y\cos(xy) \right)=\cos(xy)-xy\sin(xy)\)

\(Q_x=\dfrac{\partial }{\partial x}\left( x\cos(xy) \right)=\cos(xy)-xy\sin(xy)\)

Since these match, this vector field is conservative.

To find the potential function we solve each integral separately:

\(\displaystyle \int P dx = \int y\cos(xy) \, dx =\sin(xy)+h_1(y) \)

\(\displaystyle \int Q dy = \int x\cos(xy) \, dy =\sin(xy)+h_2(x) \)

Letting \(h_1(y)=0\) and \(h_2(x)=0\) we see that these are equal, giving us the potential function:

\(f(x,y)=\sin(xy)\) satisfying \(\nabla f=\vecs{F}\)

Exercise \(\PageIndex{1d}\)

Determine whether \( \vecs{F}(x,y)=\left\langle \dfrac{1}{x-y}+\cos(x),\dfrac{1}{y-x}+\sin(y)\right\rangle \) is conservative or not. If it is conservative, find a function \(f\) such that \(\vecs{F}=\nabla f\).

Answer

We check cross partials to determine if \(\vecs{F}=\langle P(x,y),Q(x,y)\rangle \) is conservative:

\(P_y=\dfrac{\partial }{\partial y}\left( \dfrac{1}{x-y}+\cos(x)\right)=\dfrac{1}{(x-y)^2}\)

\(Q_x=\dfrac{\partial }{\partial x}\left( \dfrac{1}{y-x} +\sin(y)\right)=\dfrac{1}{(y-x)^2}\)

Since these match, this vector field is conservative.

To find the potential function we solve each integral separately:

\(\displaystyle \int P dx = \int \dfrac{1}{x-y} +\cos(x) \, dx =\ln|x-y|+\sin(x)+h_1(y) \)

\(\displaystyle \int Q dy = \int \dfrac{1}{y-x} +\sin(y) \, dy =\ln|y-x|-\cos(y)+h_2(x) \)

Letting \(h_1(y)=-\cos(y)\) and \(h_2(x)=\sin(x)\) we see that these are equal, giving us the potential function:

\(f(x,y)=\ln|x-y|+\sin(x)-\cos(y) \) satisfying \(\nabla f=\vecs{F}\)

Exercise \(\PageIndex{1e}\)

Determine whether \( \vecs{F}(x,y,z)=\langle x^2,z^2,y^2\rangle \) is conservative or not. If it is conservative, find a function \(f\) such that \(\vecs{F}=\nabla f\).

Answer

We check the curl to determine if \(\vecs{F}=\langle P(x,y,z),Q(x,y,z), R(x,y,z)\rangle \) is conservative:

\(\text{curl}(\vecs{F}) = \left| \begin{array} {ccc} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ x^2 & z^2 & y^2 \end{array} \right| = \left\langle 2y-2z, 0,0\right\rangle\)

Since the curl is not the zero vector, this vector field is not conservative.

Exercise \(\PageIndex{1f}\)

Determine whether \( \vecs{F}(x,y,z)=\left\langle \dfrac{2x}{(x^2-y^2)^2}-1, 3-\dfrac{2y}{(x^2-y^2)^2},4z\right\rangle \) is conservative or not. If it is conservative, find a function \(f\) such that \(\vecs{F}=\nabla f\).

Answer

We check the curl to determine if \(\vecs{F}=\langle P(x,y,z),Q(x,y,z), R(x,y,z)\rangle \) is conservative:

\(\text{curl}(\vecs{F}) = \left| \begin{array} {ccc} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ \dfrac{2x}{(x^2-y^2)^2}-1 & 3-\dfrac{2y}{(x^2-y^2)^2} & 4z \end{array} \right| = \left\langle 0, 0, 8xy(x^2-y^2)^{-3}-8xy(x^2-y^2)^{-3}\right\rangle=\vecs{0}\)

Since the curl is the zero vector, this vector field is conservative.

To find the potential function we solve each integral separately:

\(\displaystyle \int P dx = \int \dfrac{2x}{(x^2-y^2)^2}-1 \, dx = -(x^2-y^2)^{-1}-x+h_1(y,z) \)

\(\displaystyle \int Q dy = \int 3-\dfrac{2y}{(x^2-y^2)^2}\, dy =3y-(x^2-y^2)^{-1}+h_2(x,z) \)

\(\displaystyle \int R dz = \int 4z \, dz =2z^2+h_3(x,y) \)

Letting \(h_1(y,z)=3y+2z^2\), \(h_2(x,z)=-x+2z^2\), and \(h_3(x,y)=-x+3y\) we see that these are equal, giving us the potential function:

\(f(x,y,z)=-(x^2-y^2)^{-1}-x+3y+2z^2\) satisfying \(\nabla f=\vecs{F}\)

Exercise \(\PageIndex{1g}\)

Determine whether \( \vecs{F}(x,y,z)=\left\langle 2y+3z,x+4z,3x+4y+6z \right\rangle \) is conservative or not. If it is conservative, find a function \(f\) such that \(\vecs{F}=\nabla f\).

Answer

We check the curl to determine if \(\vecs{F}=\langle P(x,y,z),Q(x,y,z), R(x,y,z)\rangle \) is conservative:

\(\text{curl}(\vecs{F}) = \left| \begin{array} {ccc} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ 2y+3z & x+4z & 3x+4y+6z \end{array} \right| = \left\langle 4-4, 3-3, 1-2\right\rangle=\langle 0,0,-1\rangle \)

Since the curl is not the zero vector, this vector field is not conservative.

Exercise \(\PageIndex{1h}\)

Determine whether \( \vecs{F}(x,y,z)=\left\langle (1+xyz^2)e^{xyz^2}+e^x, x^2z^2e^{xyz^2}+2yz^2, 2x^2yze^{xyz^2}+2y^2z \right\rangle \) is conservative or not. If it is conservative, find a function \(f\) such that \(\vecs{F}=\nabla f\).

Answer

Checking the curl of this vector field is unpleasant, so we attempt the integrals and see if they match:

\(\displaystyle \int P dx = \int (1+xyz^2)e^{xyz^2}+e^x\, dx = xe^{xyz^2}+e^x+h_1(y,z) \)

\(\displaystyle \int Q dy = \int x^2z^2e^{xyz^2}+2yz^2 \, dy =xe^{xyz^2}+y^2z^2+h_2(x,z) \)

\(\displaystyle \int R dz = \int 2x^2yze^{xyz^2}+2y^2z\, dz =xe^{xyz^2}+y^2z^2+h_3(x,y) \)

Letting \(h_1(y,z)=y^2z^2\), \(h_2(x,z)=e^x\), and \(h_3(x,y)=e^x\) we see that these are equal, giving us the potential function:

\(f(x,y,z)=xe^{xyz^2}+e^x+y^2z^2\) satisfying \(\nabla f=\vecs{F}\)

Since this potential function exists, we know the vector field is conservative.

Exercise \(\PageIndex{2a}\)

Find the expression denoted in terms of divergence and curl (if possible) for \(\vecs{A}=\left\langle 2x,3y,4z \right\rangle \)

(a) \(\text{curl}(\vecs{A})\)

(b) \(\text{div}(\vecs{A})\)

(c) \(\text{curl}(\text{curl}(\vecs{A}))\)

(d) \(\text{curl}(\text{div}(\vecs{A}))\)

Answer

(a) \(\text{curl}(\vecs{A}) = \left| \begin{array} {ccc} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ 2x & 3y & 4z \end{array} \right| = \left\langle 0-0, 0-0, 0-0\right\rangle=\vecs{0} \)

(b) \(\text{div}(\vecs{A})=\dfrac{\partial}{\partial x} \left( 2x\right)+ \dfrac{\partial }{\partial y} \left( 3y\right) +\dfrac{\partial }{\partial z} \left( 4z\right)=2+3+4=9\)

(c) \(\text{curl}( \text{curl}(\vecs{A}))=\text{curl}(\vecs{0})=\vecs{0}\)

(d) Since \(\text{div}(\vecs{A})\) is not a vector, we cannot take its curl. This does not exist.

Exercise \(\PageIndex{2b}\)

Find the expression denoted in terms of divergence and curl (if possible) for \(\vecs{B}= \left\langle e^{yz+x},e^{xz+y},e^{xy+z} \right\rangle \)

(a) \(\text{curl}(\vecs{B})\)

(b) \(\text{div}(\vecs{B})\)

(c) \(\text{div}(\text{curl}(\vecs{B}))\)

(d) \(\nabla \text{div}(\vecs{B})\)

Answer

(a) \(\text{curl}(\vecs{B}) = \left| \begin{array} {ccc} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ e^{yz+x} & e^{xz+y}& e^{xy+z} \end{array} \right| = \left\langle xe^{xy+z}-xe^{xz+y}, ye^{yz+x}-ye^{xy+z}, ze^{xz+y}-ze^{yz+x}\right\rangle \)

(b) \(\text{div}(\vecs{B})=\dfrac{\partial}{\partial x} \left( e^{yz+x}\right)+ \dfrac{\partial }{\partial y} \left(e^{xz+y}\right) +\dfrac{\partial }{\partial z} \left( e^{xy+z}\right)= e^{yz+x}+e^{xz+y}+e^{xy+z}\)

(c) \(\text{div}( \text{curl}(\vecs{B}))=0\) (since the divergence of a curl is always zero)

(d) To find its gradient, we find each partial derivative of the divergence of \(\vecs{B}\):

\( \dfrac{\partial}{\partial x} \left( e^{yz+x}+e^{xz+y}+ e^{xy+z}\right) =e^{yz+x}+ze^{xz+y}+ye^{xy+z}\)

\(\dfrac{\partial}{\partial y} \left( e^{yz+x}+e^{xz+y}+e^{xy+z}\right) =ze^{yz+x}+e^{xz+y}+xe^{xy+z}\)

\( \dfrac{\partial}{\partial z} \left( e^{yz+x}+e^{xz+y}+ e^{xy+z}\right) =ye^{yz+x}+xe^{xz+y}+e^{xy+z}\)

Therefore, \(\nabla \text{div}(\vecs{B})=\left\langle e^{yz+x}+ze^{xz+y}+ye^{xy+z}, ze^{yz+x}+e^{xz+y}+xe^{xy+z},ye^{yz+x}+xe^{xz+y}+e^{xy+z} \right\rangle\)

Exercise \(\PageIndex{2c}\)

Find the expression denoted in terms of divergence and curl (if possible) for \(\vecs{C}=\left\langle \dfrac{y}{x}, \dfrac{z}{y},\dfrac{x}{z}\right\rangle \)

(a) \(\text{curl}(\vecs{C})\)

(b) \(\text{div}(\vecs{C})\)

(c) \(\text{div}(\text{div}(\vecs{C}))\)

(d) \(\text{curl}\left( \left\langle \dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}\right\rangle+ \vecs{C}\right)\)

Answer

(a) \(\text{curl}(\vecs{C}) = \left| \begin{array} {ccc} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ \dfrac{y}{x} & \dfrac{z}{y}& \dfrac{x}{z} \end{array} \right| = \left\langle 0-\dfrac{1}{y}, 0-\dfrac{1}{z}, 0-\dfrac{1}{x} \right\rangle =\langle -y^{-1},-z^{-1},-x^{-1}\rangle \)

(b) \(\text{div}(\vecs{C})=\dfrac{\partial}{\partial x} \left( \dfrac{y}{x}\right)+ \dfrac{\partial }{\partial y} \left( \dfrac{z}{y}\right) +\dfrac{\partial }{\partial z} \left( \dfrac{x}{z}\right)= -\dfrac{y}{x^2}-\dfrac{z}{y^2}-\dfrac{x}{z^2}\)

(c) Since \(\text{div}(\vecs{C})\) is not a vector, we cannot take its divergence. This does not exist.

(d) We see that \( \text{curl}\left( \left\langle \dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}\right\rangle\right)=\left| \begin{array} {ccc} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ \dfrac{1}{x} & \dfrac{1}{y}& \dfrac{1}{z} \end{array} \right| =\langle 0-0,0-0,0-0\rangle =\vecs{0}\)

Therefore \(\text{curl}\left( \left\langle \dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}\right\rangle+ \vecs{C}\right)=\vecs{0}+\text{curl}(\vecs{C})=\langle -y^{-1},-z^{-1},-x^{-1}\rangle \)

Exercise \(\PageIndex{2d}\)

Find the expression denoted in terms of divergence and curl (if possible) for \(f(x,y,z)=x^2+2y^2-z^3\)

(a) \(\text{curl}( \nabla f )\)

(b) \(\text{div}( \nabla f )\)

(c) \(\text{curl}(\langle x,y,z\rangle\cdot \nabla f)\)

(d) \(\nabla \text{div}(\nabla f)\)

Answer

Each of these involve the gradient of \(f: \quad \nabla f=\langle 2x,4y,-3z^2\rangle\).

(a) \(\text{curl}(\nabla f) = \left| \begin{array} {ccc} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ 2x & 4y & -3z^2 \end{array} \right| = \left\langle 0-0, 0-0, 0-0\right\rangle =\vecs{0} \)

(b) \(\text{div}(\nabla f)=\dfrac{\partial}{\partial x} \left( 2x\right)+ \dfrac{\partial }{\partial y} \left( 4y\right) +\dfrac{\partial }{\partial z} \left( -3z^2\right)= 2+4-6z=6-6z\)

(c) \(\langle x,y,z\rangle\cdot \nabla f = 2x^2+4y^2-3z^3\) is not a vector, so we cannot take its curl. This does not exist.

(d) \(\nabla \text{div}(\nabla f)=\left\langle \dfrac{\partial}{\partial x}\left(6-6z\right), \dfrac{\partial}{\partial y}\left(6-6z\right), \dfrac{\partial}{\partial z}\left(6-6z\right) \right\rangle = \langle 0,0,-6\rangle \)

Exercise \(\PageIndex{3}\)

If \(\vecs{F}(x,y,z)=\langle |x|,|y|,|z|\rangle\) and \(\vecs{G}(x,y,z)=\langle |y|,|z|,|x|\rangle\), at a point \((x,y,z)\) where \(x,y,z \ne 0\), find:

(a) \(\text{curl}(\vecs{F})\)

(b) \(\text{div}(\vecs{F})\)

(c) \(\text{curl}(\vecs{G})\)

(d) \(\text{div}(\vecs{G})\)

Answer

(a) \(\text{curl}(\vecs{F}) = \left| \begin{array} {ccc} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ |x| & |y| & |z| \end{array} \right| = \left\langle 0-0, 0-0, 0-0 \right\rangle =\vecs{0} \)

(b) \(\text{div}(\vecs{F})=\dfrac{\partial}{\partial x} \left( |x|\right)+ \dfrac{\partial }{\partial y} \left( |y|\right) +\dfrac{\partial }{\partial z} \left( |z|\right)= \left\{ \begin{array} {rl} 1+1+1 & x>0,\, y>0, \, z>0 \\
-1+1+1 & x<0,\, y>0, \, z>0 \\
1-1+1 & x>0,\, y<0, \, z>0 \\
-1-1+1 & x<0,\, y<0, \, z>0 \\
1+1-1 & x>0,\, y>0, \, z<0 \\
-1+1-1 & x<0,\, y>0, \, z<0 \\
1-1-1 & x>0,\, y<0, \, z<0 \\
-1-1-1 & x<0,\, y<0, \, z<0 \end{array} \right.= \left\{ \begin{array} {rl} 3 & x>0,\, y>0, \, z>0 \\
1 & x<0,\, y>0, \, z>0 \\
1 & x>0,\, y<0, \, z>0 \\
-1 & x<0,\, y<0, \, z>0 \\
1 & x>0,\, y>0, \, z<0 \\
-1 & x<0,\, y>0, \, z<0 \\
-1 & x>0,\, y<0, \, z<0 \\
-3 & x<0,\, y<0, \, z<0 \end{array} \right.\)

(c) \(\text{curl}(\vecs{G}) = \left| \begin{array} {ccc} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ |y| & |z| & |x| \end{array} \right| = \left\{ \begin{array} {cl} \langle 0-1,0-1,0-1\rangle & x>0,\, y>0, \, z>0 \\
\langle 0-1,0-(-1),0-1\rangle & x<0,\, y>0, \, z>0 \\
\langle 0-1,0-1,0-(-1)\rangle & x>0,\, y<0, \, z>0 \\
\langle 0-1,0-(-1),0-(-1)\rangle & x<0,\, y<0, \, z>0 \\
\langle 0-(-1),0-1,0-1\rangle & x>0,\, y>0, \, z<0 \\
\langle 0-(-1),0-(-1),0-1\rangle & x<0,\, y>0, \, z<0 \\
\langle 0-(-1),0-1,0-(-1)\rangle & x>0,\, y<0, \, z<0 \\
\langle 0-(-1),0-(-1),0-(-1)\rangle & x<0,\, y<0, \, z<0 \end{array} \right.= \left\{ \begin{array} {cl} \langle -1,-1,-1\rangle & x>0,\, y>0, \, z>0 \\
\langle -1,1,-1\rangle & x<0,\, y>0, \, z>0 \\
\langle -1,-1,1\rangle & x>0,\, y<0, \, z>0 \\
\langle -1,1,1\rangle & x<0,\, y<0, \, z>0 \\
\langle 1,-1,-1\rangle & x>0,\, y>0, \, z<0 \\
\langle 1,1,-1\rangle & x<0,\, y>0, \, z<0 \\
\langle 1,-1,1\rangle & x>0,\, y<0, \, z<0 \\
\langle 1,1,1\rangle & x<0,\, y<0, \, z<0 \end{array} \right.\)

(d) \(\text{div}(\vecs{G})=\dfrac{\partial}{\partial x} \left( |y|\right)+ \dfrac{\partial }{\partial y} \left( |z|\right) +\dfrac{\partial }{\partial z} \left( |x|\right)=0+0+0=0\)

Exercise \(\PageIndex{4}\)

Find some vector field \(\vecs{F}\) with \(\text{curl}(\vecs{F})=\langle 1,2,3\rangle \)

Answer

We need \(\vecs{F}=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle\) to satisfy: \(\left\{ \begin{array} {l} R_y-Q_z=1 \\ P_z-R_x=2 \\ Q_x-P_y=3 \end{array}\right.\)

Pick \(R_y=1\) and \(Q_z=0\) so that \(\displaystyle R(x,y,z)=\int R_y\,dy = \int 1\,dy =y+r(x,z)\) and \(Q(x,y,z)=q(x,y)\)

Then our system of equations above becomes: \(\left\{ \begin{array} {l} P_z-r_x(x,z)=2 \\ q_x(x,y)-P_y=3 \end{array}\right.\)

Letting \(-r_x(x,z)=0\) we can see that \(r(x,z)=\hat{r}(z)\), and lock in \(R(x,y,z)=y+\hat{r}(z)\)

This forces \(P_z=2\) so we have \(\displaystyle P(x,y,z)=\int P_z\,dz = \int 2\,dz =2z+p(x,y)\)

This turns our final equation \(q_x(x,y)-P_y=3\) into \(q_x(x,y)-p_y(x,y)=3\).

This is simplest when one of them is zero, so \( p_y(x,y)=0\longrightarrow p(x,y)=\hat{p}(x)\)

This allows us to lock in \(P(x,y,z)=2z+\hat{p}(x)\)

Which forces \(q_x(x,y)=3\) and gives \(\displaystyle q(x,y)=\int q_x\,dx = \int 3\,dx =3x+\hat{q}(y)\)

This allows us to lock in \(Q(x,y,z)=3x+\hat{q}(y)\), so

Therefore one family of solutions is \(\vecs{F}=\left\langle 2z+\hat{p}(x), 3x+\hat{q}(y), y+\hat{r}(z)\right\rangle \)

This gives infinitely many (but not all such) vector fields \(\vecs{F}\), but for one example use \(\hat{p}(x)=0,\hat{q}(y)=0,\hat{r}(z)=0\):

\(\vecs{F}_1=\left\langle 2z,3x, y\right\rangle \longrightarrow \text{curl}(\vecs{F})= \left| \begin{array} {ccc} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ 2z & 3x & y\end{array} \right|=\langle 1-0,2-0, 3-0\rangle = \langle 1,2,3\rangle \) as desired.

Exercise \(\PageIndex{5}\)

Find two different vector fields \(\vecs{F}\) with \(\text{curl}(\vecs{F})=\langle x,y,-2z\rangle \)

Answer

We need \(\vecs{F}=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle\) to satisfy: \(\left\{ \begin{array} {l} R_y-Q_z=x \\ P_z-R_x=y \\ Q_x-P_y=-2z \end{array}\right.\)

Pick \(R_y=x\) and \(Q_z=0\) so that \(\displaystyle R(x,y,z)=\int R_y\,dy = \int x\,dy =xy+r(x,z)\) and \(Q(x,y,z)=q(x,y)\)

Then our system of equations above becomes: \(\left\{ \begin{array} {l} P_z-(y+r_x(x,z))=y \\ q_x(x,y)-P_y=-2z \end{array}\right. \longrightarrow \left\{ \begin{array} {l}P_z-r_x(x,z)=2y \\ q_x(x,y)-P_y=-2z \end{array}\right. \)

Letting \(-r_x(x,z)=0\) we can see that \(r(x,z)=\hat{r}(z)\), and lock in \(R(x,y,z)=xy+\hat{r}(z)\)

This forces \(P_z=2y\) so we have \(\displaystyle P(x,y,z)=\int P_z\,dz = \int 2y\,dz =2yz+p(x,y)\)

This turns our final equation \(q_x(x,y)-P_y=-2z\) into \(q_x(x,y)=p_y(x,y)\).

This is simplest when both are zero, so \( p(x,y)=\hat{p}(x)\) and \(q(x,y)=\hat{q}(y)\)

This allows us to lock in \(P(x,y,z)=2yz+\hat{p}(x)\) and \(Q(x,y,z)=\hat{q}(y)\), so

Therefore one family of solutions is \(\vecs{F}=\left\langle 2yz+\hat{p}(x), \hat{q}(y), xy+\hat{r}(z)\right\rangle \)

This gives infinitely many vector fields \(\vecs{F}\), but for one example use \(\hat{p}(x)=0,\hat{q}(y)=0,\hat{r}(z)=0\):

\(\vecs{F}_1=\left\langle 2yz,0, xy\right\rangle \longrightarrow \text{curl}(\vecs{F})= \left| \begin{array} {ccc} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ 2yz & 0 & xy\end{array} \right|=\langle x-0,2y-y, 0-2z\rangle = \langle x,y,-2z\rangle \) as desired.

If we instead started off by picking \(R_y=0\) and \(Q_z=-x\), then \(\displaystyle Q(x,y,z)=\int Q_z\,dz = \int -x\,dz =-xz+q(x,y)\) and \(R(x,y,z)=r(x,z)\)

Then our system of equations above becomes: \(\left\{ \begin{array} {l} P_z-r_x(x,z)=y \\ (q_x(x,y)-z)-P_y=-2z \end{array}\right. \longrightarrow \left\{ \begin{array} {l}P_z-r_x(x,z)=y \\ q_x(x,y)-P_y=-z \end{array}\right. \)

Letting \(-r_x(x,z)=0\) we can see that \(r(x,z)=\hat{r}(z)\), and lock in \(R(x,y,z)=\hat{r}(z)\)

This forces \(P_z=y\) so we have \(\displaystyle P(x,y,z)=\int P_z\,dz = \int y\,dz =yz+p(x,y)\)

This turns our final equation \(q_x(x,y)-P_y=-z\) into \(q_x(x,y)=p_y(x,y)\).

This is simplest when both are zero, so \( p(x,y)=\hat{p}(x)\) and \(q(x,y)=\hat{q}(y)\)

This allows us to lock in \(P(x,y,z)=yz+\hat{p}(x)\) and \(Q(x,y,z)=-xz+\hat{q}(y)\), so

Therefore one family of solutions is \(\vecs{F}=\left\langle yz+\hat{p}(x), -xz+\hat{q}(y), \hat{r}(z)\right\rangle \)

This gives infinitely many more vector fields \(\vecs{F}\), but another example use \(\hat{p}(x)=0,\hat{q}(y)=0,\hat{r}(z)=0\):

\(\vecs{F}_2=\left\langle yz,-xz, 0\right\rangle \longrightarrow \text{curl}(\vecs{F})= \left| \begin{array} {ccc} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ yz & -xz & 0\end{array} \right|=\langle 0-(-x),y-0, (-z)- z\rangle = \langle x,y,-2z\rangle \) as desired.

Notice that there are infinitely many other such solutions, but they all differ by a vector field with curl zero:

\(\text{curl}\left(\vecs{F}_1-\vecs{F}_2\right)=\text{curl}\left(\langle yz,xz, xy\rangle\right) =\left| \begin{array} {ccc} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ yz & xz & xy\end{array} \right|=\langle x-x,y-(-y),z-z \rangle=\vecs{0} \)

Exercise \(\PageIndex{6}\)

Find three different vector fields \(\vecs{H}\) with \(\text{div}(\vecs{H})=xyz\)

Answer

There are again infinitely many ways to find such an \(\vecs{H}=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\). Here are three simple ones:

Setting \(P_x=xyz\) gives \(\displaystyle P(x,y,z)= \int xyz\, dx=\dfrac{1}{2}x^2yz+p(y,z)\)

Letting \(Q_y=0\) and \(R_z=0\) yield \(Q(x,y,z)=q(x,z)\) and \(R(x,y,z)=r(x,y)\)

Thus \(\vecs{H}_1=\left\langle \dfrac{1}{2}x^2yz+p(y,z),q(x,z),r(x,y)\right\rangle\) is one such answer.

Setting \(Q_y=xyz\) gives \(\displaystyle Q(x,y,z)= \int xyz\, dy=\dfrac{1}{2}xy^2z+q(x,z)\)

Letting \(P_x=0\) and \(R_z=0\) yield \(P(x,y,z)=p(y,z)\) and \(R(x,y,z)=r(x,y)\)

Thus \(\vecs{H}_2=\left\langle p(y,z),\dfrac{1}{2}xy^2z+q(x,z),r(x,y)\right\rangle\) is another such answer.

Setting \(R_z=xyz\) gives \(\displaystyle R(x,y,z)= \int xyz\, dz=\dfrac{1}{2}xyz^2+r(x,y)\)

Letting \(P_x=0\) and \(Q_y=0\) yield \(P(x,y,z)=p(y,z)\) and \(Q(x,y,z)=q(x,z)\)

Thus \(\vecs{H}_3=\left\langle p(y,z),q(x,z),\dfrac{1}{2}xyz^2+r(x,y)\right\rangle\) is a third such answer.

This is not an exhaustive list!