### Length, area, and volume factors

Many of the integrals of multivariable calculus are integrals over mappings from a domain onto another set (the image). In line integrals over parametrized curves, for example, one integrates over a curve that is the image of an interval; a vector-valued function maps the interval onto the image curve. The functions that perform the mappings stretch or shrink parts of the domain as they transform the domain onto the image. Hence, one part of calculating these integrals is accounting for the expansions or contraction of these maps.

In what follows, we outline the different types of “expansion factors” that appears in the integrals. There are three types of expansion factors, depending on the dimension of the underlying integral.

The types of expansions factors are:

- length expansion factors (for one-dimensional integrals)
- area expansion factors (for two-dimensional integrals), and
- volume expansion factors (for three-dimensional integrals).

#### Length expansion factors

One dimensional integrals are based on the familiar one-variable integral of a function $\dlsi(t)$ from $t=a$ to $t=b$, \begin{align*} \int_a^b \dlsi(t) dt. \end{align*} In the one-variable integral, we can think of $dt$ as being a length measurement. Other one-dimensional integrals will include an “length expansion factor” that multiplies $dt$ in order to accurately account for length.

The one-dimensional integral of multivariable calculus is the line integral. However, before dicussing line integrals, we examine changing variables in one-dimensional integrals to show that the “$u$-substitution” of one-variable calculus contains another example of a length expansion factor. Both one-variable integrals under a change of variables and line integrals are based on length-changing maps that transform an interval to another interval or a curve.

##### Length when change variables in one-variable integrals

The following is an attempt to tie one-variable change of variables to multivariable change of variables. If it is too confusing, just skip it and move on.

When you perform a “$u$-substitution” in one-variable calculus, you are changing variables. To help you link one-variable $u$-substitution to multivariable change of variables, we can write a $u$-substitution in the same language as multivariable calculus.

Let's say you are given some integral $\int_a^b \dlsi(x) dx$ which you want to compute with a “$u$-substitution.” In a $u$-substitution, you write $u$ as a function of $x$, but to make the notation match that of a multivariable change of variables, let's do the reverse and write $x = T(u)$, where $T$ is an invertible “change of variables” function. Then the $u$-substitution is $u = T^{-1}(x)$, where $T^{-1}(x)$ is the inverse of $T(u)$.

To perform the $u$-substitution, you replace $x$ with $T(u)$, integrate from $T^{-1}(a)$ to $T^{-1}(b)$, and replace $dx$ with $T'(u)du$. In this notation, a $u$-substitution looks like: \begin{align*} \int_a^b \dlsi(x) dx = \int_{T^{-1}(a)}^{T^{-1}(b)} \dlsi(T(u)) T'(u) du. \end{align*}

We could go a little further and make this formula even closer to what we write in multivariable calculus. We could write the interval $[a,b]$ as $I$. The original integral $\int_a^b f(x)dx$ is over the interval $I=[a,b]$, so we could write the integral as \begin{align*} \int_I \dlsi(x) dx. \end{align*}

If $x=T(u)$ is our change of variables, then $T$ maps an interval $I^{*}$ in “$u$-space” to the interval $I$ in “$x$-space.” If $T^{-1}(b)$ is greater than $T^{-1}(a)$, then $I^*$ is the interval $[T^{-1}(a), T^{-1}(b)]$. Otherwise, $I^*$ is the interval $[T^{-1}(b), T^{-1}(a)]$. Our change of variables formula is then \begin{align*} \int_I \dlsi(x) dx = \int_{I^{ *}} \dlsi(T(u))|T'(u)|du. \end{align*}

Note that in this case, the change of variables “length expansion factor” is $| T'(u)|$. We need the absolute value because of how we defined $I^*$. The case when $T'(u) <0$ corresponds to $T^{-1}(b)< T^{-1}(a)$, i.e., the case where we flipped the order of the endpoints in our definition of $I^*= [T^{-1}(b), T^{-1}(a)]$. This flipping changes the sign of the integral; adding the absolute value $|T'(u)|$ changes the sign back to the correct sign.

The length expansion factor $|T'(u)|$ indicates how much $T$ expands or contracts $I^*$ when it maps $I^*$ onto $I$.

##### Length in line integrals

In line integrals, a curve $\dlc$ is parametrized by a function $\dllp(t)$, which maps on interval $t \in [a,b]$ onto the curve. In this case, the length measure on the curve is $d\als = \|\dllp\,'(t)\|dt$. The length expansion factor $\|\dllp\,'(t)\|$ accounts for expansion or contraction by $\dllp$ when it maps the interval $I=[a,b]$ onto $\dlc$. Hence, the integral of a scalar-valued function $\dlsi(\vc{x})$ is \begin{align*} \dslint = \dpslint. \end{align*}

For line integrals of vector fields, we integrate $\dlsi = \dlvf \cdot \vc{T}$, where $\vc{T}$ is the unit tangent vector of the curve: \begin{align*} \vc{T} = \frac{\dllp'(t)}{\|\dllp'(t)\|}. \end{align*} In this case, the denominator cancels the $\|\dllp\,'(t)\|$ factor, \begin{align*} \dlint = \slint{\dlc}{\dlvf \cdot \vc{T}} = \dplint, \end{align*} but the expansion or contraction of $\dllp(t)$ is still included in the $\dllp'(t)$ factor.

#### Area expansion factors

Two-dimensional integrals are based on the double integral of a function $\dlsi(x,y)$ over a planar region $\dlr$, \begin{align*} \iint_\dlr \dlsi\, dA = \iint_\dlr \dlsi(x,y) \, dx\,dy. \end{align*} In the double integral, we can think of $dA=dx\,dy$ as being an area measurement. Other two-dimensional integrals will include an “area expansion factor” that multiplies $dx\,dy$ in order to accurately account for area.

Both double integrals under a change of variables and surface integrals involve area-changing maps of a planar region onto another planar region or surface. To correctly compute these integrals, one must account for how the maps change area.

##### Area when changing variables in double integrals

To change variables in a double integral, we find a function $(x,y) = \cvarf(\cvarfv,\cvarsv)$ that maps some new region $\dlr^*$ in $(\cvarfv,\cvarsv)$-space to the original region $\dlr$ in $(x,y)$-space. We then need a factor that accounts for the expansion or contraction of $\cvarf$ as it maps $\dlr^*$ onto $\dlr$. The area expansion factor is the absolute value of the determinant of the matrix of partial derivatives of $\cvarf$: \begin{align*} \left| \det \jacm{\cvarf}(\cvarfv,\cvarsv) \right|. \end{align*} We sometimes write this is \begin{align*} \left| \det \jacm{\cvarf}(\cvarfv,\cvarsv) \right| = \left| \pdiff{(x,y)}{(\cvarfv,\cvarsv)} \right|. \end{align*}

With the proper area expansion factor, the formula for changing variables in double integrals is \begin{align*} \iint_\dlr \dlsi\, dA = \iint_{\dlr^{\textstyle *}} \dlsi(\cvarf(\cvarfv,\cvarsv)) \left| \det \jacm{\cvarf}(\cvarfv,\cvarsv) \right| d\cvarfv\,d\cvarsv. \end{align*}

##### Area in surface integrals

In surface integrals, a surface $\dls$ is parametrized by a function $\dlsp(\spfv,\spsv)$ for $(\spfv,\spsv) \in \dlr$, where $\dlr$ is a planar region. In this case, the area measure on the surface is \begin{align*} d\sas = \left\| \pdiff{\dlsp}{\spfv} \times \pdiff{\dlsp}{\spsv}\right\| \,d\spfv\,d\spsv. \end{align*} The area expansion factor $\left\| \pdiff{\dlsp}{\spfv} \times \pdiff{\dlsp}{\spsv}\right\|$ accounts for expansion or contraction by $\dlsp$ when it maps $\dlr$ onto $\dls$. Hence, the integral of a scalar-valued function $\dlsi(\vc{x})$ is \begin{align*} \dssint = \dpssint. \end{align*}

For surface integrals of vector fields, we integrate $\dlsi = \dlvf \cdot \vc{n}$, where $\vc{n}$ is the unit normal vector of the surface: \begin{align*} \vc{n} = \frac{\pdiff{\dlsp}{\spfv} \times \pdiff{\dlsp}{\spsv}} {\left\|\pdiff{\dlsp}{\spfv} \times \pdiff{\dlsp}{\spsv}\right\|}. \end{align*} In this case, the denominator cancels the $\left\|\pdiff{\dlsp}{\spfv} \times \pdiff{\dlsp}{\spsv}\right\|$ factor, so the integral becomes \begin{align*} \dsint = \dpsint. \end{align*} The expansion or contraction of $\dlsp(\spfv,\spsv)$ is still included in the $\left(\pdiff{\dlsp}{\spfv} \times \pdiff{\dlsp}{\spsv}\right)$ factor.

#### Volume expansion factors

Three-dimensional integrals are based on the triple integral of a function $\dlsi(x,y,z)$ over a three-dimensional region $\dlv$, \begin{align*} \iiint_\dlv \dlsi\, dV = \iiint_\dlv \dlsi(x,y,z) \, dx\,dy\,dz. \end{align*} In the triple integral, we can think of $dV=dx\,dy\,dz$ as being a volume measurement.

Triple integrals under a change of variables involve volume-changing maps of one three-dimensional region onto another. To correctly compute triple integrals, one must account for how the maps change volume. The integrals will include an “volume expansion factor” that multiplies $dx\,dy\,dz$ in order to accurately account for volume.

##### Volume when change variables in triple integrals

To change variables in a triple integral, we find a function $(x,y,z) = \cvarf(\cvarfv,\cvarsv,\cvartv)$ that maps some new solid $\dlv^*$ in $(\cvarfv,\cvarsv,\cvartv)$-space to the original solid $\dlv$ in $(x,y,z)$-space. We then need a factor that accounts for the expansion or contraction of $\cvarf$ as it maps $\dlv^*$ onto $\dlv$. The volumne expansion factor is the absolute value of the determinant of the matrix of partial derivatives of $\cvarf$: \begin{align*} \left| \det \jacm{\cvarf}(\cvarfv,\cvarsv,\cvartv) \right|. \end{align*} We sometimes write this is \begin{align*} \left| \det \jacm{\cvarf}(\cvarfv,\cvarsv,\cvartv) \right| = \left| \pdiff{(x,y,z)}{(\cvarfv,\cvarsv,\cvartv)} \right|. \end{align*}

With the proper volume expansion factor, the formula for changing variables in triple integrals is \begin{align*} \iiint_\dlv \dlsi\, dV = \iiint_{\dlv^{\textstyle *}} \dlsi(\cvarf(\cvarfv,\cvarsv,\cvartv)) \left| \det \jacm{\cvarf}(\cvarfv,\cvarsv,\cvartv) \right| d\cvarfv\,d\cvarsv\,d\cvartv. \end{align*}

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