Chapter 13 Multiple Integrals

Author

Jiaye Xu

Published

February 14, 2026

13.1 Double Integrals over Rectangles

Double Integrals

Recall the Riemann integral introduced in single-variable calculus where we formed a partition \(P\) of the interval \([a, b]\) into subintervals of length \(\Delta x_k\), \(k=1,2,\ldots\), picked a sample point \(\bar x_k\) from the \(k\)th subinterval, and then wrote

In multivariable calculus, we proceed in a very similar fashion to define the integral for a function of two variables.

N.B. Rectangular grid partitioning the region \(R\) into small rectangles of area \(\Delta A_k = \Delta x_k \Delta y_k\). And the Riemann sum over the region \(R\) is \[\sum_{k=1}^nf(\bar x_k,\bar y_k)\Delta A_k\]

which corresponds to the sum of the volumes of boxes.

Comments:

  • For \(f(x)\geq 0\), \(\int_a^b f(x)dx\) represents the area of the region under the curve \(y=f(x)\) between \(a\) and \(b\).

  • For \(f(x,y)\geq 0\), \(\iint \limits_A f(x,y)dA\) represents the volume of the solid under the surface \(z=f(x,y)\) and above the rectangle \(R\).

The Existence Question

Out of the same reason as in the single-variable case, not every function of two variables is integrable on a given rectangle \(R\).

N.B. A function that is unbounded on \(R\) will always fail to be integrable.

As a consequence, most of the common functions (provided they are bounded) are integrable on every rectangle. For example,

is integrable on every rectangle.

However,

N.B. The staircase function is integrable on \(R\) because its discontinuities occur along two line segments.

Properties of the Double Integral

The double integral inherits most of the properties of the single integral, for example:

  1. linearity;

  2. additivity;

  3. comparison property.

Specifically,

Comment: All of these properties hold on more general sets than rectangles.

Evaluation of Double Integrals

First note that

13.2 Iterated Integrals

From double integrals to iterated integrals

Now we face the problem of evaluating \(\iint \limits_A f(x,y)dA\), where \(R\) is the rectangle

Previously, we learnt that we can interpret the double integral as the volume of the solid under the surface, that is

Another way to calculate the volume of this solid:

  • Slice the solid into thin slabs using the planes parallel to the \(xz\)-plane.

    N.B. The area of the face of this slab depends on how far it is from the \(xz\)-plane; that is, it depends on \(y\)

    • Denote the area of the face of this slab by \(A(y)\).

  • The volume \(\Delta V\) of the slab is given approximately by \[\Delta V \approx A(y)\Delta y\]

  • Recalling our old motto of three steps (slice, approximate, integrate), we write the integral \[V=\int_c^d A(y)dy\]

    where for fixed \(y\) we may calculate \(A(y)\) by means of an ordinary single integral, that is \[A(y)=\int_a^b f(x,y)dx\]

  • Now we obtain the conclusion of the expression

The last expression is called an iterated integral.

If we slice the solid with planes parallel to the \(yz\)-plane, we would have obtained another iterated integral, with the integrations occurring in the opposite order.

Comments:

While the two boxed results were derived under the assumption that \(f\) was nonnegative, they are valid in general. If \(f(x,y)\) is negative on part of \(R\), then

  1. \(\iint \limits_A f(x,y)dx\) gives the signed volume of the solid between the surface \(z=f(x,y)\) and the rectangle \(R\) of the \(xy\)-plane.

  2. The actual volume of this solid is \(\iint \limits_A |f(x,y)|dx\).

Evaluating Iterated Integrals

N.B. We shall usually omit the brackets in the iterated integral.

Calculating Volumes

13.3 Double Integrals over Nonrectangular Regions

Double Integral on a General Set

Now we consider an arbitrary closed bounded set \(S\) in the plane surround by a rectangle \(R\). We define \(f(x,y)\) on \(S\), and let \(f(x,y)=0\) outside of \(S\). We say that \(f\) is integrable on \(S\) if it is integrable on \(R\) and write \[\iint\limits_S f(x,y)dA=\iint\limits_R f(x,y)dA\]

In addition, we assert that the double integral on a general set \(S\) is

  1. linear,

  2. additive on sets that overlap only on smooth curves, and

  3. satisfies the comparison property.

Evaluation of Double Integrals on General Regions

  • A set (or region) \(S\) is \(\boldsymbol y\)-simple if there are functions \(\phi_1\) and \(\phi_2\) on \([a, b]\) such that \[S=\{(x,y): \phi_1(x)\leq y\leq \phi_2(x), a\leq x\leq b\}\]

  • A set \(S\) is \(\boldsymbol x\)-simple if there are functions \(\psi_1\) and \(\psi_2\) on \([c, d]\) such that \[S=\{(x,y): \psi_1(y)\leq x\leq \psi_2(y), c\leq y\leq d\}\]

Comments:

  1. A set (or region) \(S\) is \(y\)-simple (or\(x\)-simple) if it is simple in the \(y\)-direction (or \(x\)-direction), meaning that a line in this direction intersects \(S\) in a single interval (or point or not at all).

  2. A set can be neither \(x\)-simple nor \(y\)-simple.

To evaluate the double integral of a function over a \(y\)-simple set \(S\). For example, in the following figure

where we enclose \(S\) in a rectangle \(R\) and let \(f(x,y)=0\) outside \(S\), then,

N.B. In the inner integration, \(x\) is held fixed; thus, this integration is along the heavy vertical line of the figure above. This integration yields the area \(A(x)\) of the cross section shown in Figure 7.

For a \(x\)-simple set \(S\), we can evaluate the double integral with the similar reasoning. Therefore, we summarize the formulae for both cases,

  • \(y\)-simple set

  • \(x\)-simple set

  • neither \(x\)-simple nor \(y\)-simple

    If it is the union of the two \(x\)-simple (or \(y\)-simple) sets, calculate each piece and add them together to obtain the integral. For example,

    this region is the union of two \(y\)-simple sets.

Basic Examples

The iterated integrals where the limits of the inner integral are variables.

N.B. For iterated integrals, the outer integral cannot have limits that depend on either variable of integration.

Further Examples

Calculate volumes by means of iterated integrals by the following steps:

  1. Visualize the solid and specify the region \(S\).

  2. Figure our our goal volume in terms of a double integral.

  3. Write the double integral as an iterated integral. (You may have to specify a \(x\)-simple or \(y\)-simple set. )

  4. Evaluate the iterated integral. (Techniques used in the basic examples)

Our goal is to seek the volume of the solid under the surface \(z=\frac{3}{4}(4-x-2y)\) and above the region \(S\).

Then, write the double integral as an iterated integral,

  • Step 1.

  • Step 2. Our goal is to find

  • Step 3. Since region is

  • Step 4.

N.B. The technique of changing the order of integration. We use it when the inner integral does not have an antiderivative in terms of elementary functions. For example,

13.4 Double Integrals in Polar Coordinates

Certain curves in the plane, such as circles, cardioids, and roses, are easier to describe in terms of polar coordinates than in Cartesian coordinates.

Polar Coordinates

In section 13.2 we learnt that if we have a surface determined by \(z=f(x,y)\) over a region \(R\), assuming that \(f(x,y)\) is continuous and nonnegative, then, the volume \(V\) of the solid under this surface and above \(R\) is given by \[V=\iint\limits_R f(x,y)dA\]

Now let us take a close look at the shape of this region \(R\).

Suppose we have a polar rectangle \(R\) in Figure 1.

In polar coordinates,

  • a polar rectangle \(R\) has the form \[R=\{(r,\theta): a\leq r \leq b, \alpha\leq\theta\leq \beta )\}\]

    where \(a\geq 0\) and \(\beta-\alpha\leq 2\pi\).

  • The equation of the surface can be written as \[z=f(x,y)=f(r\cos \theta,r\sin\theta)=F(r,\theta)\]

Now we are prepared to calculate the volume \(V\) using polar coordinates.

Integrals in Polar Coordinates

  • Partition \(R\) into smaller polar rectangles \(R_1, R_2, \ldots, R_k\) by means of a polar grid and let \(\Delta r_k\) and \(\Delta \theta_k\) denote the dimensions of the typical piece \(R_k\).

    • Take a closer look at this typical piece \(R_k\),

      and the area of this piece \(A(R_k)\) (denoted \(\Delta A_k\) in the zoom-in figure above) can be obtained by \[\text{area of large sector}-\text{area of small sector}\]

      leading to the formula \[A(R_k)=\Delta A_k=r_k\Delta r\Delta \theta\]

      where \(r_k\) is the average radius of \(A_k\) in the zoom-in figure.

    • Then, using the notation in Figure 3, we obtain an equivalent equation \[A(R_k)=\bar r_k\Delta r_k\Delta \theta_k \]

      where \(\bar r_k\) is the average radius of \(R_k\) in Figure 3.

  • Therefore, the volume have the following approximation equation

  • Taking the limit as the norm of the partition approaches zero, we obtain the formula for the volume in the form of a double integral, \[V=\iint\limits_R F(r,\theta)rdrd\theta=\iint\limits_R f(r\cos \theta,r\sin\theta)rdrd\theta\]

In summary, we have two equivalent expressions for \(V\),

Comment:

The boxed result was derived under the assumption that \(f\) was nonnegative, but it is valid for general functions, in particular for continuous functions of arbitrary sign.

Iterated Integrals

N.B. One technique for evaluating the integrals for the volume is writing the polar double integral as an iterated integral.

Comment:

Without the help of polar coordinates, we could not have done this problem. Because the extra factor of \(r\) was just what we needed in order to antidifferentiate \(e^{r^2}\).

General Regions

Recall how we extended the double integral over an rectangular region \(R\) to the integral over a general set \(S\).

We can do the same thing for double integrals in polar coordinates by introducing the idea of \(r\)-simple and \(\theta\)-simple set.

  • A set \(S\) is an \(\boldsymbol r\)-simple set if it has the form

  • A set \(S\) is an \(\boldsymbol \theta\)-simple set if it has the form

Note that \(S\) is an \(r\)-simple set.

Comments: Polar Equations for Cardioids:

We consider equations of the form

with \(a\) and \(b\) positive. Their graphs are called limaçons, with the special cases in which \(a=b\) referred to as cardioids.

N.B. Polar Equations for general Circles with radius \(a\):

  • If a circle of radius \(a\) is centered at the origin (or pole), its polar equation is simply \(r=a\).

  • If a circle of radius \(a\) is centered at \((r_0, \theta_0)\) its equation is quite complicated unless we choose \(r_0=a\). By the Law of Cosines, we obtain the polar equation for general case, which is

    • when \(\theta_0=0\), \(r=2a\cos\theta\)

    • when \(\theta_0=\pi/2\), \(r=2a\sin\theta\)

A Probability Integral

  • The standard normal probability density function is

  • Another result related to this density function is \[\int_{-\infty}^\infty f(x)dx=1\]

    In the next two examples, we will prove this result.

13.6 Surface Area

In this section, we develop a formula for the area of a surface defined by \(z=f(x,y)\) over a specified region – a surface \(G\) over the closed and bounded region \(S\) in the \(xy\)-plane.

  • Partition \(P\) of the region \(S\) with lines parallel to the \(x\)- and \(y\)-axes.

    • Let \(R_m\), \(m=1,2,\ldots,n\), denote the resulting rectangles that lie completely within \(S\),

    • Let \(G_m\) be that part of the surface that projects onto \(R_m\), and let \(P_m\) be the point of \(G_m\) that projects onto the corner of \(R_m\) with the smallest \(x\)- and \(y\)-coordinates.

    • Let \(T_m\) denote the parallelogram from the tangent plane at \(P_m\) that projects onto \(R_m\).

  • Find the area of \(T_m\).

    • Let \(\mathbf u_m\) and \(\mathbf v_m\) denote the vectors that form the sides of \(T_m\).

    • The area of \(T_m\) is \(\|\mathbf u_m\times \mathbf v_m\|\), where

    • Therefore, the area of \(T_m\) is

      where we assume that \(f\) has continuous first partial derivatives \(f_x\) and \(f_y\).

  • Take the limit to obtain the surface area of \(G\).

    In summary,

Comment:

We make the derivation as if the region \(S\) in the \(xy\)-plane is a rectangle; this need not be the case. The formula is applicable for a general region \(S\). Figure 3 shows what happens when \(S\) is not a rectangle.

Some Examples

Properties of the right circular cylinder

A right circular cylinder (with height equal to diameter) and an inscribed sphere have the remarkable property that

the surfaces between two parallel planes (perpendicular to the axis of the cylinder) have equal area.

Comment:

Our next example demonstrates this property for a hemisphere and these steps easily extend to show that the property holds for a sphere.

N.B. For most surface area problems it is easy to set up the double integral. However, it is often difficult or impossible to evaluate these integrals because of the difficulty of finding antiderivatives.

Let’s see the following example.

13.7 Triple Integrals in Cartesian Coordinates

Triple Integrals – Rectangular Box

Assumptions:

  • a function \(f\) of three variables defined over a box-shaped region \(B\);

  • a partition \(P\) of \(B\) (with planes parallel to the coordinate planes), thus obtaining sub-boxes \(B_1, B_2, \dots, B_n\);

  • a sample point \((\bar x_k,\bar y_k, \bar z_k)\) on \(B_k\).

The triple integral of \(f\) over \(B\) is defined by

where \(\Delta V_k =\Delta x_k\Delta y_k \Delta z_k\) is the volume of \(B_k\), provided that this limit exists.

Comments:

  • The norm of partition \(\lVert P\rVert\) is the length of the longest diagonal of all the subboxes.

  • When a single limiting value is attained, no matter how the partitions and sample points are chosen, we say that \(f\) is integrable over \(B\).

  • When \(f\) is continuous on \(B\), \(f\) is integrable. In fact, discontinuities on a finite number of smooth surfaces can be allowed.

  • Standard properties of triple integrals:

    • linearity;

    • additivity;

    • comparison property;

    • triple integrals can be written as triple iterated integrals.

Comment: For the triple iterated integrals, there are six possible orders of integration, and all of them will lead to the same answer.

General Solid Regions

Now we consider a closed bounded set \(S\) enclosed in a box \(B\) in three-space and a function \(f\) defined on \(S\), assuming the value of \(f\) is zero outside \(S\), then by definition

We can evaluate the triple integral defined above, if \(S\) is not complicated and has good properties:

  • \(S\) be a \(\boldsymbol z\)-simple set, that is, vertical lines intersect \(S\) in a single line segment;

  • \(S_{xy}\) be the projection of \(S\) in \(xy\)-plane, and \(S_{xy}\) is a \(\boldsymbol y\)-simple set.

we can write

and then

Comments:

Other orders of integration may be possible, but in each case we should expect that

  1. the limits on the inner integral are functions of two variables, e.g., \(\psi_1(x,y), \psi_2(x,y)\);

  2. the limits on the middle integral to be functions of one variable, e.g., \(\phi_1(x),\phi_2(x)\);

  3. the limits on the outer integral to be constants, e.g., \(a_1,a_2\).

Evaluation of Triple Integrals

Alternative order of integration to evaluate the same integral, in Example 4.

Application: Mass and Center of Mass

Mass

If the symbol \(\delta (x,y,z)\) is the density (mass per unit volume) of an object occupying a solid region \(S\), the integral of \(\delta\) over \(S\) gives the mass of the object.

By the method of slicing, approximating, integrating, the mass \(m\) of the solid \(S\) is the limit \[m=\lim_{n\to \infty}\sum_{k=1}^n \Delta m_{k}=\lim_{n\to \infty}\sum_{k=1}^n \delta (\bar x_k,\bar y_k,\bar z_k)\Delta V_k=\underset{S}\iiint\delta(x,y,z)dV\]

Moment

The first moment of a solid region \(S\) about a coordinate plane is defined as the triple integral over \(S\) of the distance from a point \((x, y, z)\) in \(S\) to the plane multiplied by the density of the solid at that point.

For instance, the (first) moment about the \(yz\)-plane is the integral \[M_{yz}=\underset{S}\iiint x\delta(x,y,z)dV\]

Center of Mass

The center of mass is found from the first moments. For instance, the \(x\)-coordinate of the center of mass is \(\bar x=M_{yz}/m\).

Similar formulas are given in the textbook.

Application: Probability

In Section 6.7, we learned that probabilities for random variables can be computed as areas under the probability density function and expectations can be computed like moments. Now we generalize these concepts to the case of multiple random variables.

Joint Probability Density

A function \(f(x,y,z)\) is a joint probability density function (PDF) for the random variables \((X,Y,Z)\) if \(f(x,y,z) \geq 0\) for all \((x,y,z)\) in \(S\) and \[\underset{S}\iiint f(x,y,z) dz dy dx=1\] where \(S\) is the region of all possible values for \((X,Y,Z)\).

A probability involving \((X,Y,Z)\) can then be computed as the triple integral over the appropriate region. That is, \[\underset{R}\iiint f(x,y,z) dz dy dx\]

Expected Value

The expected value of some function \(g(X,Y,Z)\) is defined to be \[E(g(X,Y,Z))=\underset{S}\iiint g(x,y,z)f(x,y,z) dz dy dx\]

13.8 Triple Integrals in Cylindrical and Spherical Coordinates

When the evaluation of triple integrals over a solid region \(S\) in three-space involving a cylinder, cone, or sphere, we can often simplify our work by using cylindrical or spherical coordinates:

  • When \(S\) has an axis of symmetry, cylindrical coordinates may be helpful,

  • when \(S\) is symmetric with respect to a point, spherical coordinates may be helpful.

Triple Integrals in Cylindrical Coordinates

In Section 11.9, we introduced the equations relating the Cartesian and Cylindrical coordinates

as cylindrical coordinates illustrated in Figure 1.

Written in cylindrical coordinates, the function \(f(x,y,z)\) can be transformed to \[f(x,y,z)=f(r\cos\theta,r\sin\theta,z)=F(r,\theta,z)\]

Consider a partition \(P\) of solid region \(S\) by means of a cylindrical grid, which is analogous to using regular grid in Section 13.7. The typical volume element is a cylindrical wedge, rather than a rectangular box, has volume \(\Delta V_k=\bar r_k \Delta r_k \Delta \theta_k \Delta z_k\).

Therefore, a Riemann sum for \(f\) over \(S\) has the form \[R_n=\sum_{k=1}^n f(\bar r_k\cos\bar\theta_k,\bar r_k\sin\bar\theta_k,\bar z_k)\bar r_k \Delta z_k\Delta r_k \Delta \theta_k \]

The triple integral of a function \(f\) over \(S\) is obtained by taking a limit of such Riemann sums with partitions whose norms approach zero, that is, \[\lim_{n\to \infty}R_n=\underset{S}\iiint f(x,y,z)dV=\underset{S}\iiint f(r\cos\theta,r\sin\theta,z)rdzdrd\theta\] If \(S\) is a \(z\)-simple solid, its projection in the \(xy\)-plane is \(r\)-simple, as shown in Figure 3, and \(f\) is continuous on \(S\),

then

Since the density function has the form \(\delta(x,y,z)=kz\),

Triple Integrals in Spherical Coordinates

Recall the equations relating the Cartesian and Spherical coordinates introduced in Section 11.9,

as spherical coordinates illustrated in Figure 7.

When evaluating triple integrals over a solid region \(S\) in spherical coordinates, we partition the region by means of an appropriate spherical grid into \(n\) spherical wedges. The volume of a typical spherical wedge has the form \(\Delta V_k=\bar \rho^2_k \sin\bar\phi_k\Delta \rho_k \Delta \theta_k \Delta \phi_k\).

The corresponding Riemann sum for \(f\) over \(S\) has the form \[R_n=\sum_{k=1}^n f(\bar\rho_k\sin\bar\phi_k\cos\bar\theta_k,\bar\rho_k\sin\bar\phi_k\sin\bar\theta_k,\bar\rho_k\cos\bar\phi_k)\bar \rho^2_k \sin\bar\phi_k\Delta \rho_k \Delta \theta_k \Delta \phi_k\] As the norm of a partition approaches zero, the appropriate Riemann sum have a limit when \(f\) is continuous

  • Find the volume:

  • Find the center of mass:

13.9 Change of Variables in Multiple Integrals

This section is an extension of method of substitution for single integrals to multiple integrals. For a single variable one-to-one function, \(g\), who has an inverse \(g^{-1}\), we know from the method of substitution in Chapter 5 that

Interchanging the role of \(x\) and \(u\) allows us to write out the result formula of making the substitution \(x=g(u)\)

as illustrated in Figure 1.

Transformation from \(\boldsymbol{uv}\)-Plane to the \(\boldsymbol{xy}\)-Plane

If we let \(x=x(u,v)\), \(y=y(u,v)\), and \(G(u,v)=(x(u,v),y(u,v))\), where the transformation \(G\) is a vector-valued function with a vector input, pair \((x,y)\) is called the image of \((u,v)\), and \((u,v)\) is called the preimage of \((x,y)\).

Accordingly, for a grid in the \(uv\)-plane with lines parallel to the \(u\)-axes and \(v\)-axes, the images of vertical lines in the \(uv\)-plane are called \(\boldsymbol u\)-curves of \(G\), and the images of horizontal lines are called \(v\)-curves of \(G\).

  • Step 1: Solve for \(u,v\) in terms of \(x,y\).

  • Step 2: Substitute given values for \(u\) and \(v\), respectively.

  • Step 1: Solve for \(u,v\) in terms of \(x,y\).

  • Step 2: Substitute given values for \(u\) and \(v\), respectively.

The Change of Variable Formula for Double Integrals

  • Step 1: Specify the region in \(uv\)-plane.

  • Step 2: Specify the Jacobian for the transformation.

  • Step 3: Evaluation the integrals after transformation.

Comments: Ideally, the region of integration are expected to be a regular grid, therefore, some “irregular” region often suggest a transformation.

  • Step 1: Specify the region in \(uv\)-plane.

Substitute \(u=x^2+v^2\) and \(v=y^2-x^2\), we transform the region \(R\) to region \(S\), which is just a simple rectangle,

Solving for \(x\) and \(y\) in terms of \(u\) and \(v\),

  • Step 2: Specify the Jacobian for the transformation.

  • Step 3: Find the mass, the moments, and therefore, the center of mass.

The Change of Variable Formula for Triple Integrals

Transformation from \(\mathbb{R}^3\) to \(\mathbb{R}^3\) mapping region \(S\) in \(uvw\)-space onto \(R\) in the \(xyz\)-space:

If \(G(u,v,w)=(x(u,v,w),y(u,v,w),z(u,v,w))\), then

14.1 Vector Fields

Suppose a region in space (or in the plane) is occupied by a moving fluid, such as air or water flowing in a curved pipe in Figure 2. The fluid is made up of a large number of particles, and at any instant of time, a particle has a velocity \(v\). At different points of the region at a given (same) time, these velocities can vary.

We can think of a velocity vector being attached to each point of the fluid representing the velocity of a particle at that point. Such a fluid flow is an example of a vector field.

Vector Field

A vector field is a function \(\mathbf F\) that assigns a vector \(\mathbf{F(p)}\) to each point \(\mathbf p\) in its domain. A vector field in three-space has the form \[\mathbf{F(p)}=\mathbf{F}(x,y,z)=M(x,y,z)\mathbf i+N(x,y,z)\mathbf j+P(x,y,z)\mathbf k\]

An example of a vector field in two-space is the velocity filed of a wheel spinning at a constant rate of \(\frac{1}{2}\) radian per time unit, having the formula \[\mathbf{F(p)}=\mathbf{F}(x,y)=M(x,y)\mathbf i+N(x,y)\mathbf j=-\frac{1}{2}y\mathbf i+\frac{1}{2}x\mathbf j\]

Other examples of vector fields in science include electric fields, magnetic fields, force fields, and gravitational fields.

Inverse-Square Law of Gravitational Attraction (Newton’s Law of Gravitation)

According to Isaac Newton, the magnitude of the force of attraction (in newtons, N) between objects of mass \(M\) and \(m\) in kilograms (kg), respectively, is given by \[\lVert \mathbf F \rVert=G\frac{Mm}{d^2}\] where \(d\) is the distance between the centers of mass of the bodies in meters (m) and \(G\) is the gravitational constant, \(6.674\times 10^{-11}Nm^2/kg^2\).

Since the vectors represent forces, we will call such a field a \(force field\).

The Gradient Field

Scalar Field

In contrast to a vector field, a function \(f\) that attaches a number to each point in space is called a scalar field.

Gradient Field (Gradient of a Scalar Field)

Let \(f(x,y,z)\) determine a scalar field and suppose that \(f\) is differentiable. Then the gradient of \(f\), denoted by \(\nabla f\), is the vector field given by \[\mathbf{F}(x,y,z)=\nabla f(x,y,z)=\frac{\partial f}{\partial x}\mathbf i + \frac{\partial f}{\partial y}\mathbf j+ \frac{\partial f}{\partial z}\mathbf k\] A vector field \(\mathbf F\) that is the gradient of a scalar field \(f\) is called a conservative vector field, and \(f\) is its potentialfunction.

Divergence and Curl

Relating divergence and curl to the gradient operator \(\nabla\), recall that \(\nabla\) is the operator