Computing Curl in Cartesian Coordinates

Introduction

A vector field in three-dimensional space assigns a vector to each point. Its curl is another vector field, computed from how the original field’s components vary across space. In Cartesian coordinates, this calculation reduces to six partial derivatives arranged in three signed pairs.

The main challenge is not usually differentiating the component functions. It is matching each derivative to the correct component and keeping the signs in the right order. A reliable formula and a consistent calculation process make both tasks manageable.

Definition and Cartesian Formula

Let \(\mathbf{v}\) be a vector field on a region in \(\mathbb{R}^3\), written in Cartesian coordinates as \(\mathbf{v}(x,y,z)=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle\). Assume the component functions have continuous first partial derivatives on the region. This regularity condition is sufficient for the curl formula below to be defined throughout the region.

Definition: The curl of \(\mathbf{v}\) in Cartesian coordinates is \(\nabla\times\mathbf{v}\), with components \[ \nabla\times\mathbf{v} = \left\langle \frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right\rangle. \]

The partial derivatives are taken with respect to the indicated variable, holding the other two variables fixed. The first component uses derivatives of \(R\) and \(Q\); the second uses \(P\) and \(R\); the third uses \(Q\) and \(P\). Each component is a difference, and reversing either pair changes the sign.

A common mnemonic for the same formula is

\nabla\times\mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ P & Q & R \end{vmatrix}.

Here \(\mathbf{i},\mathbf{j},\mathbf{k}\) are the standard unit vectors. This determinant is a mnemonic for expanding the cross product of the differential operator \(\nabla\) with the field, not an ordinary determinant of numbers.

Why the Component Formula Has These Signs

In Cartesian coordinates, the operator \(\nabla\) is \(\mathbf{i}\frac{\partial}{\partial x}+\mathbf{j}\frac{\partial}{\partial y}+\mathbf{k}\frac{\partial}{\partial z}\). To form \(\nabla\times\mathbf{v}\), each derivative acts on the component functions, and the unit vectors combine by the usual cross-product rules. For example, \(\mathbf{i}\times\mathbf{j}=\mathbf{k}\), while \(\mathbf{j}\times\mathbf{i}=-\mathbf{k}\).

Expanding by these rules gives

\begin{aligned} \nabla\times\mathbf{v} ={}& \mathbf{i}\times \left( \frac{\partial P}{\partial x}\mathbf{i} +\frac{\partial Q}{\partial x}\mathbf{j} +\frac{\partial R}{\partial x}\mathbf{k} \right)\\ &+\mathbf{j}\times \left( \frac{\partial P}{\partial y}\mathbf{i} +\frac{\partial Q}{\partial y}\mathbf{j} +\frac{\partial R}{\partial y}\mathbf{k} \right)\\ &+\mathbf{k}\times \left( \frac{\partial P}{\partial z}\mathbf{i} +\frac{\partial Q}{\partial z}\mathbf{j} +\frac{\partial R}{\partial z}\mathbf{k} \right). \end{aligned}

Terms involving the cross product of a unit vector with itself vanish. The remaining terms are

\begin{aligned} \nabla\times\mathbf{v} ={}& \left(\frac{\partial Q}{\partial x}\mathbf{k} -\frac{\partial R}{\partial x}\mathbf{j}\right) +\left(-\frac{\partial P}{\partial y}\mathbf{k} +\frac{\partial R}{\partial y}\mathbf{i}\right)\\ &+\left(\frac{\partial P}{\partial z}\mathbf{j} -\frac{\partial Q}{\partial z}\mathbf{i}\right). \end{aligned}

Collect the coefficients of \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\), respectively. They are \(R_y-Q_z\), \(P_z-R_x\), and \(Q_x-P_y\), which gives the stated formula. Thus the signs come from the orientation of the Cartesian unit vectors in the cross product.

A Reliable Calculation Process

For a field \(\mathbf{v}=\langle P,Q,R\rangle\), it is useful to keep the component order visible until the final line. First identify \(P,Q,R\); then calculate the six partial derivatives that appear in the formula; finally form the three differences in order. If the question asks for the curl at a particular point, evaluate the resulting vector field there only after simplifying the derivatives.

Label the components.
Write the field as \(\langle P,Q,R\rangle\), in \(x,y,z\) component order.

Differentiate the required pairs.
Compute \(R_y,Q_z,P_z,R_x,Q_x,P_y\), taking each partial derivative with the other coordinates fixed.

Subtract in the prescribed order.
Form \(\langle R_y-Q_z,\ P_z-R_x,\ Q_x-P_y\rangle\).

Evaluate if needed.
Substitute a specified point into the simplified curl vector field, not into the original field in place of taking derivatives.

Worked Examples

Worked Example: A Polynomial Vector Field

Compute the curl of \(\mathbf{v}(x,y,z)=\langle xy+z^2,\ x^2z+3y,\ yz+x\rangle\). Then evaluate it at \((1,2,-1)\).

The components are \(P=xy+z^2\), \(Q=x^2z+3y\), and \(R=yz+x\). The needed partial derivatives are

\begin{aligned} R_y&=z, & Q_z&=x^2,\\ P_z&=2z, & R_x&=1,\\ Q_x&=2xz, & P_y&=x. \end{aligned}

Substituting them into the three component differences gives

\nabla\times\mathbf{v} = \langle z-x^2,\ 2z-1,\ 2xz-x\rangle.

At \((1,2,-1)\), the first component is \(-1-1=-2\), the second is \(2(-1)-1=-3\), and the third is \(2(1)(-1)-1=-3\). Therefore,

(\nabla\times\mathbf{v})(1,2,-1)=\langle -2,-3,-3\rangle.

As a check, the three derivative pairs at that point are \(R_y=-1\) and \(Q_z=1\), \(P_z=-2\) and \(R_x=1\), and \(Q_x=-2\) and \(P_y=1\). Their prescribed differences are \(-2,-3,-3\), matching the evaluated curl.

Worked Example: A Trigonometric Vector Field

Find the curl of \(\mathbf{v}(x,y,z)=\langle\sin y,\cos z,\sin x\rangle\).

Here \(P=\sin y\), \(Q=\cos z\), and \(R=\sin x\). Differentiating with respect to the indicated variables gives

\begin{aligned} R_y&=0, & Q_z&=-\sin z,\\ P_z&=0, & R_x&=\cos x,\\ Q_x&=0, & P_y&=\cos y. \end{aligned}

The first curl component is \(0-(-\sin z)=\sin z\). The second is \(0-\cos x=-\cos x\). The third is \(0-\cos y=-\cos y\). Hence

\nabla\times\mathbf{v} = \langle\sin z,-\cos x,-\cos y\rangle.

For instance, at \((0,\pi/3,\pi/2)\), the components are \(\sin(\pi/2)=1\), \(-\cos(0)=-1\), and \(-\cos(\pi/3)=-1/2\). Thus the curl there is \(\langle 1,-1,-1/2\rangle\). Directly substituting this point into the three derivative differences gives the same values.

Worked Example: A Curl with All Three Components Nonzero

Compute the curl of \(\mathbf{v}(x,y,z)=\langle x^2y,\ y^2z,\ z^2x\rangle\).

The components are \(P=x^2y\), \(Q=y^2z\), and \(R=z^2x\). The required derivatives are

\begin{aligned} R_y&=0, & Q_z&=y^2,\\ P_z&=0, & R_x&=z^2,\\ Q_x&=0, & P_y&=x^2. \end{aligned}

Therefore, the curl is

\nabla\times\mathbf{v} = \langle -y^2,-z^2,-x^2\rangle.

Checking each component against its derivative pair confirms the signs: \(0-y^2=-y^2\), \(0-z^2=-z^2\), and \(0-x^2=-x^2\). At \((1,2,3)\), the curl is \(\langle-4,-9,-1\rangle\), since \(y^2=4\), \(z^2=9\), and \(x^2=1\).

Reading the Result as a Vector

The curl is not a single number: it is a vector-valued function. If \(\mathbf{v}\) varies over a region, its curl assigns a vector to each point in that region. For example, the result \(\langle z-x^2,2z-1,2xz-x\rangle\) in the first example gives the curl at any point \((x,y,z)\); the specific vector \(\langle-2,-3,-3\rangle\) is its value at one chosen point.

The three entries of the answer are its \(x\)-, \(y\)-, and \(z\)-components, in that order. A zero component means that the curl vector has no component along that coordinate axis at the point. If all three entries are zero, the curl vector at that point is the zero vector. These are statements about the computed vector; they do not replace the component calculation.

Common Sign Error

A frequent mistake is to reverse the order of the subtractions. For example, take \(\mathbf{v}=\langle 0,0,xy\rangle\). Then \(P=0\), \(Q=0\), and \(R=xy\). The correct first component is \(R_y-Q_z=x-0=x\); the second is \(P_z-R_x=0-y=-y\); and the third is \(Q_x-P_y=0-0=0\). Thus

\nabla\times\mathbf{v}=\langle x,-y,0\rangle.

A wrong calculation that reverses the differences gives \(\langle -x,y,0\rangle\), the negative of the correct answer. At \((2,3,0)\), this mistake would produce \(\langle-2,3,0\rangle\) instead of \(\langle2,-3,0\rangle\). To correct it, return to the ordered formula and check each component separately: \(R_y-Q_z\), then \(P_z-R_x\), then \(Q_x-P_y\). Do not infer a sign from memory without checking which derivative comes first.

Summary

For \(\mathbf{v}=\langle P,Q,R\rangle\), Cartesian curl is \(\langle R_y-Q_z,\ P_z-R_x,\ Q_x-P_y\rangle\). Each entry uses two partial derivatives, and the order of subtraction matters. The result is another vector field; evaluating it at a point gives the curl vector at that point. Keeping the components labelled, calculating the required derivatives, and applying the differences in order provides a consistent way to avoid sign errors.

Check Your Understanding

Use the Cartesian component formula to answer each question. Keep the derivative pairs and their order visible in your working.

For \(\mathbf{v}=\langle yz,xz,xy\rangle\), write the three partial-derivative pairs that determine the curl.
Find the curl of \(\mathbf{v}=\langle x^2,y^2,z^2\rangle\).
For \(\mathbf{v}=\langle\cos y,\sin z,\cos x\rangle\), compute the curl as a vector field.
If the curl is \(\langle R_y-Q_z,P_z-R_x,Q_x-P_y\rangle\), which two partial derivatives determine its second component?
Explain why reversing every subtraction in the curl formula changes the result to its negative.