The curl of the vector field is given by

\(\displaystyle \begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ y^3&y^2 &xyz \end{vmatrix}\)

Next, we take the cross product. One can do this by multiplying across from the top left to the lower right, and continuing downward, and then subtracting the terms multiplied from top right to the bottom left. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

\(\displaystyle \frac{\partial }{\partial y}xyz\hat{i}+\frac{\partial }{\partial x}y^2\hat{k}+\frac{\partial }{\partial z}y^3\hat{j}-(\frac{\partial }{\partial y}y^3\hat{k}+\frac{\partial }{\partial z}y^2\hat{i}+\frac{\partial }{\partial x}xyz\hat{j})= xz\hat{i}-3y^2\hat{k}-yz\hat{j}=\left \langle xz, -yz, -3y^2\right \rangle\)

A vector field is conservative if its curl produces the zero vector.

The curl of the vector field is given by

\(\displaystyle \begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ x^2&\ln(y^2) &z^3 \end{vmatrix}\)

Next, we take the cross product. One can do this by multiplying across from the top left to the lower right, and continuing downward, and then subtracting the terms multiplied from top right to the bottom left. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

\(\displaystyle \frac{\partial }{\partial y}z^3\hat{i}+\frac{\partial }{\partial x}\ln(y^2)\hat{k}+\frac{\partial }{\partial z}x^2\hat{j}-(\frac{\partial }{\partial y}x^2\hat{k}+\frac{\partial }{\partial z}\ln(y^2)\hat{i}+\frac{\partial }{\partial x}z^3\hat{j}) = 0\hat{i}+0\hat{j}+0\hat{k}=\vec{0}\)

The vector field is conservative.

\(\displaystyle f(x,y,z)=zy+x\)

Solution 1)

This probably was deceptively easy and could have been very quickly solved without doing any calculations. This problem involves first the basic definition of a conservative vector field, and a useful theorem on conservative vector fields. 

1) A vector field \(\displaystyle \vec{F}\) is conservative if there exists a scalar function \(\displaystyle f\) such that \(\displaystyle \vec{F}\) is its' gradient. 

2) If a vector field is conservative, its' curl must be zero. 

 In other words, the curl of the gradient is always zero for any scalar function. 

In this problem we were given a scalar function \(\displaystyle f(x,y,z)=zy+x\). If we now compute the gradient, we obtain a vector field we will call  \(\displaystyle \vec{F}=\vec{\triangledown} f\) (the gradient is our vector field). Automatically we know it fits the definition of a conservative vector field because we know there is a scalar function which has \(\displaystyle \vec{\triangledown}f\) as its' gradient. That function is \(\displaystyle f(x,y,z)=zy+x\).  

Now we know that since the gradient is a conservative vector field, and therefore the curl must be equal to zero. 

Solution 2) 

Just for fun, let's see if it works by doing the actual calculation. 

\(\displaystyle \vec{\triangledown}f=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}\)

\(\displaystyle \vec{\triangledown}f=\hat{i}+z\hat{j}+y\hat{k}\)

\(\displaystyle curl(\vec{\triangledown f}) = \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ \frac{\partial }{\partial x}&\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ 1\: \: &\:z &y \end{vmatrix}\)

\(\displaystyle =\begin{vmatrix} \frac{\partial}{\partial y}&\frac{\partial}{\partial z} \\ z&y \end{vmatrix}\hat{i} - \begin{vmatrix} \frac{\partial}{\partial x} &\frac{\partial}{\partial z} \\ y&1 \end{vmatrix}\hat{j}+\begin{vmatrix} \frac{\partial}{\partial x}&\frac{\partial}{\partial y} \\ 1&z \end{vmatrix}\hat{k}\)

\(\displaystyle =(1-1)\hat{i}-(0-0)\hat{j}+(0-0)\hat{k}\)

\(\displaystyle =0\)

Classify each vector field below as either conservative or non-conservative. 

 \(\displaystyle 1)\: \: \: \: \: \: \: \vec{v}=4xy\hat{i}+2x^2\hat{j}+\hat{k}\)

This is a conservative vector field. This can be easily determined by computing the curl:

\(\displaystyle curl(\vec{v}) = \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ \frac{\partial }{\partial x}&\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ 4xy\: \: &\: 2x^2 &1 \end{vmatrix}\)

\(\displaystyle =\begin{vmatrix} \frac{\partial}{\partial y}&\frac{\partial}{\partial z} \\ 2x^2&1 \end{vmatrix}\hat{i} - \begin{vmatrix} \frac{\partial}{\partial x} &\frac{\partial}{\partial z} \\ 4xy&1 \end{vmatrix}\hat{j}+\begin{vmatrix} \frac{\partial}{\partial x}&\frac{\partial}{\partial y} \\ 4xy&2x^2 \end{vmatrix}\hat{k}\) 

\(\displaystyle =(0-0)\hat{i}-(0-0)\hat{j}+(4x-4x)\hat{k}\)

\(\displaystyle =0\)

Because \(\displaystyle curl(\vec{v})=0\) the vector field \(\displaystyle \vec{v}\) is conservative. 

\(\displaystyle 2)\: \: \: \:\: \: \vec{F}=\frac{3}{r^2}\hat{r}\)

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Connection to Physics! 

This vector field can also be shown to be conservative. In fact, all inverse square vector fields are conservative.

Common examples are Newton's Universal Law of Gravitation: 

\(\displaystyle \vec{F}=\frac{GmM}{r^2 }\hat{r}\) and Coulombs Law in electrostatics which gives the force exerted by on point charge \(\displaystyle q_1\) onto another point charge \(\displaystyle q_2\): 

\(\displaystyle \vec{F_E}=\frac{kq_1q_2}{r^2}\hat{r}\)

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The easiest approach for a function defined in this way is to simply find a scalar function that has \(\displaystyle \vec{F}\) as its' gradient. 

\(\displaystyle \vec{\triangledown f}=\vec{F}\)

In general, the gradient of a vector field in terms of cylindrical coordinates \(\displaystyle (r, \phi, z)\) is written as: 

\(\displaystyle \vec{\triangledown}f=\frac{\partial f}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial f}{\partial \phi }\hat{\phi}+\frac{\partial f}{\partial z}\hat{z}\)

The vector field \(\displaystyle \vec{F}\) in this example does not have any \(\displaystyle \phi\) or \(\displaystyle z\) dependency, so the first therm is the only non-zero component. Therefore, we only need to integrate the partial derivative in the \(\displaystyle r\) component in order to find \(\displaystyle f.\)

\(\displaystyle f = \int \frac{df}{dr}dr=\int\frac{1}{r^2}dr=-\frac{1}{r}+C\)

Now check to see if it works: 

\(\displaystyle \vec{\triangledown}f = \frac{d}{dr}\left(-\frac{1}{r}+C\right)\hat{r}=\frac{1}{r^2}\hat{r}=\vec{F}\)

Therefore, we have shown that there exists some scalar function \(\displaystyle f\) such that the gradient of \(\displaystyle f\) gives the vector field \(\displaystyle \vec{F}\). Therefore, \(\displaystyle \vec{F}\) is a conservative vector field. 

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\(\displaystyle 3)\: \: \: \: \: \vec{u}=a\hat{i}+b\hat{j}+c\hat{k}\) 

In this case we must again conclude that vector field is conservative. All constant vector fields (vector fields for which every component is a constant) will always be conservative. To show this, start by integrating each component: 

 \(\displaystyle f=\int\frac{\partial f}{\partial x}dx=\int adx= ax+C_1\)

Similarly, we can integrate the \(\displaystyle \hat{j}\) and \(\displaystyle \hat{k}\) components to obtain: 

\(\displaystyle f =by+C_2\) 

\(\displaystyle f = cz+C_3\)

\(\displaystyle (ax+C_1)\hat{i}+(by+C_2)\hat{j}+(cz+C_3)\hat{k}\)

Note that each expression for \(\displaystyle f\) is distinct despite the fact they're supposed to be the same function. The reason being is that we have no reason to anticipate that each component of the gradient will have all the information about the original function.

The differentiation with respect to \(\displaystyle x\) for instance to obtain the \(\displaystyle \hat{i}\) component will "delete," all constant terms, and all terms consisting of variables being held constant. The trick is to include all unique terms from each of the three calculations and to simply add them and combine their constants of integration into one constant.  \(\displaystyle C = C_1 +C_2 +C_3\)

\(\displaystyle f(x,y,z)=ax+by+cz+C\)

As an exercise, you can test this to compute the gradient.  

You can also prove the vector field is conservative by showing that the curl is zero. This is obvious since each derivative in the curl will be equal to zero because all terms are constant. 

\(\displaystyle 4)\: \: \: \: \: \vec{q} = \cos(x)\hat{i}+e^x\hat{j}-\sin(x)\hat{k}\)

 \(\displaystyle curl(\vec{q}) = \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ \frac{\partial }{\partial x}&\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ \cos(x)\: \: &\: e^x &-sin(x) \end{vmatrix}\)

 \(\displaystyle =\begin{vmatrix} \frac{\partial}{\partial y}&\frac{\partial}{\partial z} \\ e^x &-\sin(x) \end{vmatrix}\hat{i} - \begin{vmatrix} \frac{\partial}{\partial x} &\frac{\partial}{\partial z} \\cos(x)&-\sin(x) \end{vmatrix}\hat{j}+\begin{vmatrix} \frac{\partial}{\partial x}&\frac{\partial}{\partial y} \\ \cos(x)&e^x \end{vmatrix}\hat{k}\)

 The derivatives in the \(\displaystyle \vec{i}\) component all vanish do due to the fact that neither partial derivative operators in the first row differentiate with respect to \(\displaystyle x\).

 \(\displaystyle =(0-0)\hat{i}-\left[-\cos(x)-0\right]\hat{j}+(e^x-0)\hat{k}\)

\(\displaystyle =\cos(x)\hat{j}+e^x\hat{k}\)

This shows that the curl is non-zero and therefore the vector field \(\displaystyle \vec{q}\) is not conservative. 

 \(\displaystyle curl(\vec{q})\neq 0\)

First we need to evaluate the vector field evaluated along the curve. 

\(\displaystyle \vec{F}(\vec{r}(t))=10t^2(t^2)^2(t^3)^3\vec{i}+5t^3\vec{j}+2t^2t^3\vec{k}\)

\(\displaystyle \vec{F}(\vec{r}(t))=10t^{15}\vec{i}+5t^3\vec{j}+2t^5\vec{k}\)

Now we need to find the derivative of \(\displaystyle \vec{r}(t)\)

\(\displaystyle \vec{r}\ '(t)=\vec{i}+2t\vec{j}+2t^2\vec{k}\)

Now we can do the product of \(\displaystyle \vec{F}(\vec{r}(t))\) and \(\displaystyle \vec{r}(t)\).

\(\displaystyle \vec{F}(\vec{r}(t))\cdot \vec{r}(t)=(10t^{15}, 5t^3, 2t^5)\cdot(1, 2t, 2t^2)\)

\(\displaystyle =10t^{15}+10t^5+4t^7\)

Now we can put this into the integral and evaluate it.

\(\displaystyle \int_{0}^{1} 10t^{15}+10t^5+4t^7 \ dt\)

\(\displaystyle =\Big(\frac{5}{8}t^{16}+\frac{5}{3}t^6+\frac{1}{2}t^8\Big)\Big|_{0}^{1}\)

\(\displaystyle =\Big(\frac{5}{8}(1)^{16}+\frac{5}{3}(1)^6+\frac{1}{2}(1)^8\Big)\)

\(\displaystyle =\frac{5}{8}+\frac{5}{3}+\frac{1}{2}=\frac{13}{6}+\frac{5}{8}=\frac{30}{48}+\frac{104}{48}=\frac{134}{48}=\frac{67}{24}\)

The formula for work is given by

\(\displaystyle W = \int_C F(\mathbf{r}(t)) \cdot \mathbf{r}'(t) dt\).

Writing our path in parametric equation form, we have

\(\displaystyle x=t, y=t^2, 2< t< 5\).

Hence

\(\displaystyle \mathbf{r}(t) = < t,t^2>, F(\mathbf{r}(t))=< t^2,t^4>, \mathbf{r}'(t)= < 1,2t>\)

Plugging this into our work equation, we get

\(\displaystyle W = \int_C F(\mathbf{r}(t)) \cdot \mathbf{r}'(t) dt\)

\(\displaystyle = \int_{2}^{5}< t^2,t^4> \cdot < 1,2t> dt\)

\(\displaystyle = \int_2^5 t^2+2t^5 dt\).

\(\displaystyle = 5226\)

The line integral of a vector field is given by

\(\displaystyle \int \vec{F}\cdot d\vec{r} = \int [\vec{F}(\vec{r}(t))\cdot \vec{r'}(t)]dt\)

So, we must evaluate the vector field on the curve:

\(\displaystyle \vec{F}(\vec{r}(t))=\left \langle 2(2t)(t^3), 9t^2, t^6\right \rangle\)

Then, we take the derivative of the curve with respect to t:

\(\displaystyle \vec{r'}(t)=\left \langle 2, 3, 3t^2\right \rangle\)

Taking the dot product of these two vectors, we get

\(\displaystyle 8t^4+27t^2+3t^8\)

This is the integrand of our integral. Integrating, we get

\(\displaystyle \int_{0}^{1}(8t^4+27t^2+3t^8 )dt = \frac{8t^5}{5}+\frac{27t^3}{3}+\frac{3t^9}{9}\Big|^{1}_{0}= \frac{164}{15}\)

The line integral of the vector field is equal to

\(\displaystyle \int_{0}^{1}[\vec{F}(\vec{r}(t))\cdot \vec{r'}(t)]dt\)

The parameterization (using the corresponding elements of the curve) of the vector field is

\(\displaystyle \vec{F}(\vec{r}(t))= \left \langle t^4+2t^3, t^3+2t^2, t^5\right \rangle\)

The derivative of the parametric curve is

\(\displaystyle \vec{r'}(t)=\left \langle 3t^2, 2t, 1 \right \rangle\)

Taking the dot product of the two vectors, we get

\(\displaystyle \vec{F}(\vec{r}(t))\cdot \vec{r'(t)}= 3t^6+6t^5+2t^4+4t^3+t^5= 3t^6+7t^5+2t^4+4t^3\)

Integrating this with respect to t on the given interval, we get

\(\displaystyle \int_{0}^{1}(3t^6+7t^5+2t^4+4t^3)dt=\frac{3t^3}{7}+\frac{7t^6}{6}+\frac{2t^5}{5}+t^4\left.\begin{matrix} 1\\0 \end{matrix}\right|= \frac{629}{210}\)

To calculate the line integral of the vector field, we must evaluate the vector field on the curve, take the derivative of the curve, and integrate the dot product on the given interval.

The vector field evaluated on the given curve is

\(\displaystyle \vec{F}(\vec{r}(t))=\left \langle 2t^2, 2t^3+4t^2, t^2\right \rangle\)

The derivative of the curve is given by

\(\displaystyle \vec{r'}(t)=\left \langle 4t, 1, 1\right \rangle\)

The dot product of these is

\(\displaystyle \vec{F}(\vec{r(t)})\cdot \vec{r'}(t)=11t^3+5t^2\)

Integrating this over our given t interval, we get

\(\displaystyle \int_{0}^{1}(10t^3+5t^2)dt=\frac{5t^4}{2}+\frac{5t^3}{3}\Big|^{1}_{0}= \frac{25}{6}\)