### The integrals of multivariable calculus

Multivariable calculus includes six different generalizations of the familiar one-variable integral of a scalar-valued function over an interval. One can integrate functions over one-dimensional curves, two dimensional planar regions and surfaces, as well as three-dimensional volumes. When integrating over curves and surfaces, one can integral vector fields, where the one integrates either the tangential (for curves) or the normal (for surfaces) component of the vector field.

In this page, we outline the various integrals, methods you can use to solve them, and their relationship to the fundamental theorems.

The integrals covered in this page are:

- the line integral of scalar-valued function,
- the line integral of vector fields,
- the surface integral of scalar-valued function,
- the surface integral of vector fields,
- double integrals, and
- triple integrals.

#### Line integral of scalar-valued function

The line integral of a scalar-valued function $\dlsi(\vc{x})$ over a curve $\dlc$ is written as \begin{align*} \dslint. \end{align*} One physical interpretation of this line integral is that it gives the mass of a wire from its density $\dlsi$.

The only way we've encountered to evaluate this integral is the direct method. We must parametrize $\dlc$ by some function $\dllp(t)$, for $a \le t \le b$. Then, we can calculate the line integral by turning it into a regular one-variable integral of the form \begin{align*} \dslint = \dpslint. \end{align*}

Note that the “length” $d\als$ became $\| \dllp'(t)\|dt$. This quantity $\| \dllp'(t)\|$ measures how $\dllp(t)$ stretches or shrinks the interval $[a,b]$ as it maps it onto $\dlc$.

#### Line integral of a vector field

The line integral of a vector field $\dlvf(\vc{x})$ over a curve $\dlc$ is written as \begin{align*} \dlint. \end{align*} One physical interpretation of this line integral is the calculation of work when a particle moves along a path. If $\dlvf$ were a force acting on a particle moving along $\dlc$, then the integral would be the total work performed by the force on the particle.

This integral is one of the most important of multivariable calculus. We have four alternatives to evaluate the integral, although most of the alternatives work only in special cases.

We can compute the integral directly. We parametrize $\dlc$ by some function $\dllp(t)$, for $a \le t \le b$. Then \begin{align*} \dlint = \dplint \end{align*}

This method always applies. Sometimes, though, the integral will be difficult or we won't even be able to evaluate it. Our lives can be made easier by using one of the fundamental theorems to convert the line integral into something else.

Since this integral is really a line integral of the scalar-valued function $\dlsi = \dlvf \cdot \vc{T}$ where $\vc{T}$ is the unit tangent vector \begin{align*} \vc{T} = \frac{ \dllp'(t)}{\| \dllp'(t)\|}, \end{align*} the formula for the direct method is the same as the formula for the scalar-valued path integral with $\dlsi = \dlvf \cdot \vc{T}$.

If the vector field $\dlvf$ happens to be conservative, then we could use the gradient theorem for line integrals. We reduce the problem from an integral over the curve $\dlc$ to something just depending on the “boundary” of $\dlc$, i.e., its endpoints.

We need to find a potential function $f$ so that $\nabla f = \dlvf$. Then, \begin{align*} \dlint = f(\vc{q}) - f(\vc{p}), \end{align*} where $\vc{p}$ and $\vc{q}$ are the endpoints of $\dlc$.

If $\dlc$ also happens to be a closed curve, then the integral of a conservative vector field $\dlvf$ will be zero. Also, that if you know $\dlvf$ is conservative, another thing you can do is just change the curve $\dlc$ to another curve that has the same endpoints as $\dlc$. In this case, the line integral of $\dlvf$ over $\dlc$ is the same as the line integral of $\dlvf$ over any other curve with the same endpoints.

If the vector field $\dlvf$ and the curve $\dlc$ happen to be in two dimensions and if $\dlc$ happens to be a simple closed curve, then we can use Green's theorem. Green's theorem converts the line integral over $\dlc$ to a double integral over the interior of $\dlc$, which we call $\dlr$, \begin{align*} \dlint = \iint_\dlr \left(\pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y} \right) \, dA. \end{align*} Note that $\dlvf$ must be continuously differentiable everywhere in $\dlr$ for this to work. Sometimes we write $\dlc= \partial \dlr$ to denote that $\dlc$ is the boundary of $\dlr$. $\dlc$ must be oriented in a counterclockwise fashion, otherwise, we'll be off by a minus sign.

If the vector field $\dlvf$ and the curve $\dlc$ happen to be in three dimensions and if $\dlc$ happens to be a simple closed curve, then we can use Stokes' theorem. Stokes' theorem converts the line integral over $\dlc$ to a surface integral over any surface $\dls$ for which $\dlc$ is a boundary, \begin{align*} \dlint = \sint{\dls}{\curl \dlvf}, \end{align*} and is valid for any surface over which $\dlvf$ is continuously differentiable. Sometimes we write $\dlc= \partial \dls$ to denote that $\dlc$ is the boundary of $\dls$. $\dlc$ must be a positively oriented boundary of $\dls$, otherwise, we'll be off by a minus sign.

#### Surface integral of a scalar-valued function

The surface integral of a scalar-valued function $\dlsi(\vc{x})$ over a surface $\dls$ is written as \begin{align*} \dssint \end{align*} One physical interpretation of the integral is the mass of a sheet with density given by $\dlsi$.

The only way we've encountered to evalute this integral is the direct method. We must parametrize $\dls$ by some function $\dlsp(\spfv,\spsv)$, for $(\spfv,\spsv) \in \dlr$. Then, \begin{align*} \dssint = \dpssint \end{align*}

Note that the “area” $d\sas$ became $\left\| \pdiff{\dlsp}{\spfv} \times \pdiff{\dlsp}{\spsv}\right\| \,d\spfv\,d\spsv$. The quantity $d\sas$ became $\left\| \pdiff{\dlsp}{\spfv} \times \pdiff{\dlsp}{\spsv}\right\|$ measures how $\dlsp(\spfv,\spsv)$ stretches or shrinks the region $\dlr$ as it maps it onto $\dls$.

#### Surface integral of a vector field

The surface integral over surface $\dls$ of a vector field $\dlvf(\vc{x})$ is written as \begin{align*} \dsint. \end{align*} A physical interpretation is the flux of a fluid through $\dls$ whose velocity is given by $\dlvf$. For this reason, we sometimes refer to the integral as a “flux integral.”

Like the line integral of a vector field, this integral plays a big role in this course. We have three alternatives to evaluate the integral, although most of the alternatives work only in special cases.

We can compute the integral directly. We parametrize $\dls$ by some function $\dlsp(\spfv,\spsv)$, for $(\spfv,\spsv) \in \dlr$. Then, \begin{align*} \dsint = \dpsint. \end{align*}

This method always applies. Sometimes, though, the integral will be difficult or we won't even be able to evaluate it. Our lives can be made easier by using one of the fundamental theorems to convert the surface integral into something else.

Since this integral is really a surface integral of the scalar-valued function $\dlsi = \dlvf \cdot \vc{n}$ where $\vc{n}$ is the unit normal vector \begin{align*} \vc{n} = \frac{\displaystyle\pdiff{\dlsp}{\spfv} \times \pdiff{\dlsp}{\spsv}} {\displaystyle\left\|\pdiff{\dlsp}{\spfv} \times \pdiff{\dlsp}{\spsv}\right\|}, \end{align*} the formula for the direct method is the same as the formula for the scalar-valued surface integral with $\dlsi = \dlvf \cdot \vc{n}$.

If the vector field $\dlvf$ happens to be the curl of another vector field $\vc{G}$, i.e., $\dlvf = \curl \vc{G}$, then we can apply Stokes' theorem to convert the surface integral of $\curl \vc{G}$ into the line integral of $\vc{G}$ around the positively oriented boundary of $\dls$, which we denote $\partial \dls$, \begin{align*} \dsint = \sint{\dls}{\curl \vc{G}} =\lint{\dlc}{\vc{G}} \end{align*}

Multivariable calculus student's don't typically learn methods to find $\vc{G}$ from $\dlvf$. We can use Stokes' theorem to convert a surface integral into a line integral only if we are told outright that $\dlvf = \curl \vc{G}$ and are given what $\vc{G}$ is. But, if given the surface integral that looks like $\sint{\dls}{\curl \vc{G}}$, we can immediately recognize that Stokes' theorem is an option.

Stokes' theorem allows us to do one more thing to the integral $\sint{\dls}{\curl \vc{G}}$. We can switch the surface $\dls$ to any other surface $\dls'$ as long as the boundaries of $\dls$ and $\dls'$ are the same, i.e., $\partial \dls = \partial \dls'$ (assuming both boundaries are positively oriented). If $\dls$ is a complicated surface, we could feasibly save ourselves some work by integrating over another surface $\dls'$ if that surface is simpler than $\dls$.

If the surface $\dls$ happens to be a closed surface so that it is the boundary of some solid $\dlv$, i.e., $\dls = \partial \dlv$, then we can use the divergence theorem to convert the surface integral into the triple integral of $\div \dlvf$ over $\dlv$, \begin{align*} \dsint = \iiint_\dlv \div \dlvf \, dV, \end{align*} where we orient $\dls$ so that it has an outward pointing normal vector. This works only if $\dlvf$ is continuously differentiable everywhere in the solid $\dlv$.

#### Double integrals

The double integral of a (scalar-valued) function $\dlsi(\vc{x})$ over a two-dimensional region $\dlr$ is written as \begin{align*} \iint_\dlr \dlsi \, dA. \end{align*} One physical interpretation is the mass of a region with density \dlsi$.

We have encountered three alternatives to evaluate the integral.

We can compute the integral directly in terms of the original variables $x$ and $y$. In this case, $dA = dx\,dy$.

We can compute the integral by changing to the variables $\cvarfv$ and $\cvarsv$ by finding a function $(x,y) = \vc{T}(\cvarfv,\cvarsv)$. Then the integral is \begin{align*} \iint_\dlr \dlsi\, dA = \left.\iint_{\dlr^{\textstyle *}}\right. \dlsi(\vc{T}(\cvarfv,\cvarsv)) \left| \det \jacm{\vc{T}}(\cvarfv,\cvarsv) \right| d\cvarfv\,d\cvarsv, \end{align*} where $\dlr$ is parametrized by $(x,y)=\vc{T}(\cvarfv,\cvarsv)$ for $(\cvarfv,\cvarsv)$ in $\dlr^{\textstyle *}$. We sometimes write the determinant of the matrix of partial derivatives of $\vc{T}(\cvarfv,\cvarsv)$ as $\displaystyle \det \jacm{\vc{T}}(\cvarfv,\cvarsv) = \pdiff{(x,y)}{(\cvarfv,\cvarsv)}$.

If $f$ happens to be equal to $\pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y}$ for some vector field $\dlvf$, then we could use Green's theorem to convert the double integral into the integral of $\dlvf$ around the boundary of $\dlr$, which we denote $\partial \dlr$, \begin{align*} \iint_\dlr f \, dA = \iint_\dlr \left(\pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y}\right) dA = \lint{\partial \dlr}{\dlvf}. \end{align*} To orient the boundary properly, outside boundaries must be counterclockwise and inner boundaries must be clockwise.

We usually think of Green's theorem going the other way, i.e., converting a line integral into a double integral. One reason for this is that multivariable calculus students don't usually learn methods to find $\dlvf$ from $f$. We can use Green's theorem to convert a double integral into a line integral only if we are told outright that $f =\pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y}$ and are given what $\dlvf$ is. But, if given the double integral that looks like $\iint_\dlr \left(\pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y}\right) dA$, we can immediately recognize that Green's theorem is an option.

As a special case, if we are given an integral $\iint_\dlr dA$ (i.e., finding the area), we can let $\dlvf(x,y) = (-y,x)/2$ so that $\pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y} =1$ and $\iint_\dlr dA = \lint{\partial \dlr}{\dlvf}$.

#### Triple integrals

The triple integral of a (scalar-valued) function $\dlsi(\vc{x})$ over a three-dimensional solid $\dlv$ is written as \begin{align*} \iiint_\dlv \dlsi \, dV. \end{align*} One physical interpretation is the mass of a solid with density given by $\dlsi$.

We have encountered three alternatives to evaluate the integral.

We can compute the integral directly in terms of the original variables $x$, $y$, and $z$. In this case, $dV = dx\,dy\,dz$.

We can compute the integral by changing to the variables $\cvarfv$, $\cvarsv$, and $\cvartv$ by finding a function $(x,y,z) = \vc{T}(\cvarfv,\cvarsv,\cvartv)$. Then the integral is \begin{align*} \iiint_\dlv \dlsi\, dV = \left.\iiint_{\dlv^{\textstyle *}}\right. \dlsi(\vc{T}(\cvarfv,\cvarsv,\cvartv)) \left| \det \jacm{\vc{T}}(\cvarfv,\cvarsv,\cvartv) \right| d\cvarfv\,d\cvarsv\,d\cvartv, \end{align*} where $\dlv$ is parametrized by $(x,y,z)=\vc{T}(\cvarfv,\cvarsv,\cvartv)$ for $(\cvarfv,\cvarsv,\cvartv)$ in $\dlv^{\textstyle *}$. We sometimes write the determinant of the matrix of partial derivatives of $\vc{T}(\cvarfv,\cvarsv,\cvartv)$ as $\displaystyle \det \jacm{\vc{T}}(\cvarfv,\cvarsv,\cvartv) = \pdiff{(x,y,z)}{(\cvarfv,\cvarsv,\cvartv)}$.

If $\dlsi$ happens to be equal to $\div \dlvf$ for some vector field $\dlvf$, then we could use the divergence theorem to convert the triple integral into the surface integral of $\dlvf$ around the boundary of $\dlv$, which we denote $\partial \dlv$, \begin{align*} \iiint_\dlv \dlsi\, dV = \iiint_\dlv \div \dlvf\, dV = \sint{\partial \dlv}{\dlvf}. \end{align*}

We usually think of the divergence theorem going the other way, i.e., converting a surface integral into a triple integral. One reason for this is that multivariable calculus student don't typically learn methods to find $\dlvf$ from $f$. We can use the divergence theorem to convert a triple integral into a surface integral only if we are told outright that $\dlsi = \div \dlvf$ and are given what $\dlvf$ is. But, if given a triple integral that looks like $\iiint_\dlv \div\dlvf \, dV$, we can immediately recognize that the divergence theorem is an option.

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