The Central Idea
Suppose a region is awkward in Cartesian coordinates \((x,y)\), but simple in another coordinate system \((u,v)\).
We introduce a transformation
\[ x=x(u,v), \qquad y=y(u,v). \]Then instead of integrating over a complicated \(xy\)-region, we integrate over a simpler \(uv\)-region.
The price of changing coordinates is that the area element must also change.
The Jacobian tells us exactly how.
Why Change Coordinates?
Coordinate changes are useful when they align the variables with the geometry or structure of a problem.
Common examples include:
- circles and disks;
- ellipses;
- rotated regions;
- wedges and sectors;
- cylinders and spheres;
- integrands depending on repeated combinations such as \(x+y\) and \(x-y\).
A Coordinate Transformation
A two-dimensional coordinate transformation has the form
\[ T(u,v) = \bigl(x(u,v),y(u,v)\bigr). \]It maps points in the \(uv\)-plane into points in the \(xy\)-plane.
Symbolically,
\[ \boxed{ (u,v)\longmapsto(x,y) }. \]Local Behavior of a Transformation
Near a point, a differentiable coordinate transformation behaves approximately like a linear transformation.
Small changes
\[ \Delta u, \qquad \Delta v \]produce approximate changes
\[ \Delta x \approx x_u\Delta u+x_v\Delta v \]and
\[ \Delta y \approx y_u\Delta u+y_v\Delta v. \]The Jacobian Matrix
The first derivatives of the transformation are organized into the matrix
\[ \boxed{ DT = \begin{bmatrix} x_u & x_v\\ y_u & y_v \end{bmatrix} }. \]This matrix is the derivative of the coordinate transformation.
The Jacobian Determinant
The determinant of the Jacobian matrix is
\[ \boxed{ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} x_u & x_v\\ y_u & y_v \end{vmatrix} }. \]In expanded form,
\[ \boxed{ J = x_u y_v-x_v y_u }. \]What the Jacobian Measures
The absolute value of the Jacobian determinant measures local area scaling.
A tiny rectangle of area
\[ du\,dv \]in the \(uv\)-plane becomes approximately a parallelogram of area
\[ \boxed{ |J|\,du\,dv }. \]Why the Absolute Value Appears
A determinant can be positive or negative.
The sign records orientation.
Area itself must be nonnegative, so the change-of-variables formula uses
\[ \boxed{|J|}. \]Orientation
If
\[ J>0, \]the transformation locally preserves orientation.
If
\[ J<0, \]it locally reverses orientation.
If
\[ J=0, \]the transformation locally collapses area in at least one direction.
The Two-Dimensional Change-of-Variables Formula
If a suitable one-to-one differentiable transformation maps a region \(S\) in the \(uv\)-plane onto a region \(R\) in the \(xy\)-plane, then
\[ \boxed{ \iint_R f(x,y)\,dA = \iint_S f(x(u,v),y(u,v)) \left| \frac{\partial(x,y)}{\partial(u,v)} \right| du\,dv }. \]Every Part Must Be Transformed
When changing variables, three things must change:
- the integrand;
- the region of integration;
- the area or volume element.
Forgetting any one of these pieces produces an incomplete transformation.
Example: A Simple Linear Transformation
Let
\[ x=2u, \qquad y=3v. \]Then
\[ x_u=2, \qquad x_v=0, \qquad y_u=0, \qquad y_v=3. \]Therefore,
\[ J = \begin{vmatrix} 2&0\\ 0&3 \end{vmatrix} = 6. \]Interpretation of the Linear Example
The transformation stretches lengths by a factor of \(2\) in one coordinate direction and by a factor of \(3\) in the other.
Areas are therefore multiplied by
\[ 2\cdot3=6. \]Thus,
\[ \boxed{ dA=6\,du\,dv }. \]Example: Shear Transformation
Consider
\[ x=u+v, \qquad y=v. \]Then
\[ J = \begin{vmatrix} 1&1\\ 0&1 \end{vmatrix} = 1. \]Therefore,
\[ \boxed{ dA=du\,dv }. \]The transformation changes shape but preserves area.
Example: Orientation Reversal
Let
\[ x=v, \qquad y=u. \]Then
\[ J = \begin{vmatrix} 0&1\\ 1&0 \end{vmatrix} = -1. \]The coordinates have been swapped, so orientation reverses.
For area,
\[ |J|=1. \]Polar Coordinates
Polar coordinates are adapted to circular geometry.
They are defined by
\[ \boxed{ x=r\cos\theta, \qquad y=r\sin\theta }. \]Here \(r\) measures distance from the origin and \(\theta\) measures angle from the positive \(x\)-axis.
Basic Polar Relationships
Because
\[ x=r\cos\theta \]and
\[ y=r\sin\theta, \]we have
\[ \boxed{ x^2+y^2=r^2 }. \]This is why expressions involving \(x^2+y^2\) often simplify dramatically in polar coordinates.
Polar Jacobian
Compute the partial derivatives:
\[ x_r=\cos\theta, \qquad x_\theta=-r\sin\theta, \] \[ y_r=\sin\theta, \qquad y_\theta=r\cos\theta. \]Therefore,
\[ J = \begin{vmatrix} \cos\theta&-r\sin\theta\\ \sin\theta&r\cos\theta \end{vmatrix}. \]Deriving the Polar Area Element
Expanding,
\[ J = r\cos^2\theta+r\sin^2\theta. \]Since
\[ \cos^2\theta+\sin^2\theta=1, \]we obtain
\[ \boxed{ J=r }. \]Thus,
\[ \boxed{ dA=r\,dr\,d\theta }. \]Why the Polar Factor \(r\) Makes Sense
A small polar rectangle has radial thickness
\[ dr \]and angular width
\[ d\theta. \]At radius \(r\), the angular side has approximate arc length
\[ r\,d\theta. \]Therefore the small area is approximately
\[ \boxed{ dr\cdot r\,d\theta = r\,dr\,d\theta }. \]Double Integrals in Polar Coordinates
The change-of-variables formula becomes
\[ \boxed{ \iint_R f(x,y)\,dA = \iint_S f(r\cos\theta,r\sin\theta) r\,dr\,d\theta }. \]Example: Area of a Disk
Consider the disk
\[ x^2+y^2\le a^2. \]In polar coordinates,
\[ 0\le r\le a, \qquad 0\le\theta\le2\pi. \]Therefore,
\[ \operatorname{Area} = \int_0^{2\pi}\int_0^a r\,dr\,d\theta. \]Complete the Disk Area
Integrate with respect to \(r\):
\[ \int_0^a r\,dr = \frac{a^2}{2}. \]Then
\[ \int_0^{2\pi}\frac{a^2}{2}\,d\theta = \pi a^2. \]Hence,
\[ \boxed{ \operatorname{Area}=\pi a^2 }. \]Example: Radially Symmetric Integrand
Consider
\[ \iint_R (x^2+y^2)\,dA \]over the disk
\[ x^2+y^2\le1. \]Since
\[ x^2+y^2=r^2, \]the integral becomes
\[ \int_0^{2\pi}\int_0^1 r^2\cdot r\,dr\,d\theta. \]Complete the Radial Example
Simplify:
\[ \int_0^{2\pi}\int_0^1 r^3\,dr\,d\theta. \]The radial integral gives
\[ \frac14. \]Therefore,
\[ \boxed{ \iint_R(x^2+y^2)\,dA = \frac{\pi}{2} }. \]Polar Description of an Annulus
The annulus
\[ a^2\le x^2+y^2\le b^2 \]becomes simply
\[ \boxed{ a\le r\le b }. \]If the full annulus is included,
\[ 0\le\theta\le2\pi. \]Polar Description of a Sector
A sector is naturally described by
\[ a\le r\le b \]and
\[ \alpha\le\theta\le\beta. \]This is one of the clearest examples of a region that is simpler in non-Cartesian coordinates.
General Linear Changes of Variables
Suppose
\[ x=au+bv, \qquad y=cu+dv. \]Then
\[ \boxed{ J = ad-bc }. \]The determinant of the transformation matrix is exactly the area-scaling factor.
Choosing Variables from Repeated Expressions
Sometimes the integrand or boundary equations repeatedly involve the same combinations.
For example, if the problem repeatedly contains
\[ x+y \]and
\[ x-y, \]a useful substitution may be
\[ \boxed{ u=x+y, \qquad v=x-y }. \]Solving for the Old Variables
From
\[ u=x+y, \qquad v=x-y, \]add the equations:
\[ u+v=2x. \]Therefore,
\[ x=\frac{u+v}{2}. \]Subtract:
\[ u-v=2y, \]so
\[ y=\frac{u-v}{2}. \]Jacobian for \(u=x+y,\ v=x-y\)
Using
\[ x=\frac{u+v}{2}, \qquad y=\frac{u-v}{2}, \]we have
\[ J = \begin{vmatrix} \frac12&\frac12\\ \frac12&-\frac12 \end{vmatrix}. \]Therefore,
\[ J=-\frac12. \]Hence,
\[ \boxed{ dA=\frac12\,du\,dv }. \]Inverse Jacobians
When both transformations are locally invertible,
\[ (u,v)\leftrightarrow(x,y), \]the Jacobians satisfy
\[ \boxed{ \frac{\partial(x,y)}{\partial(u,v)} = \frac{1} {\frac{\partial(u,v)}{\partial(x,y)}} } \]wherever the denominator is nonzero.
Why the Jacobian Must Be Nonzero
If
\[ J=0, \]the transformation locally collapses two-dimensional area into something lower-dimensional.
A local inverse generally cannot exist there.
Thus nonzero Jacobian is a central condition in local invertibility.
Jacobian and the Inverse Function Theorem
The inverse function theorem says, roughly, that if a differentiable transformation has a nonzero Jacobian determinant at a point, then under standard hypotheses it has a differentiable local inverse near that point.
The determinant therefore measures more than scaling: it also signals whether the coordinate map is locally usable as a genuine coordinate system.
Three-Dimensional Coordinate Changes
In three variables, suppose
\[ x=x(u,v,w), \qquad y=y(u,v,w), \qquad z=z(u,v,w). \]The Jacobian matrix is
\[ \begin{bmatrix} x_u&x_v&x_w\\ y_u&y_v&y_w\\ z_u&z_v&z_w \end{bmatrix}. \]Three-Dimensional Jacobian Determinant
The determinant is written
\[ \boxed{ \frac{\partial(x,y,z)} {\partial(u,v,w)} = \begin{vmatrix} x_u&x_v&x_w\\ y_u&y_v&y_w\\ z_u&z_v&z_w \end{vmatrix} }. \]Its absolute value gives the local volume-scaling factor.
Three-Dimensional Change-of-Variables Formula
For a suitable transformation,
\[ \boxed{ \iiint_E f(x,y,z)\,dV = \iiint_G f(x(u,v,w),y(u,v,w),z(u,v,w)) \left| \frac{\partial(x,y,z)} {\partial(u,v,w)} \right| du\,dv\,dw }. \]Cylindrical Coordinates
Cylindrical coordinates extend polar coordinates into three dimensions:
\[ \boxed{ x=r\cos\theta, \qquad y=r\sin\theta, \qquad z=z }. \]The volume element is
\[ \boxed{ dV=r\,dr\,d\theta\,dz } \]up to the chosen order of integration.
Why Cylindrical Coordinates Work Well
Cylindrical coordinates are useful when a solid has symmetry around the \(z\)-axis.
Equations such as
\[ x^2+y^2=a^2 \]become
\[ r=a. \]Cylinders and rotationally symmetric solids therefore become simpler.
Example: Volume of a Cylinder
Consider the solid
\[ x^2+y^2\le a^2, \qquad 0\le z\le h. \]In cylindrical coordinates,
\[ 0\le r\le a, \qquad 0\le\theta\le2\pi, \qquad 0\le z\le h. \]Thus,
\[ V = \int_0^h \int_0^{2\pi} \int_0^a r\,dr\,d\theta\,dz. \]Complete the Cylinder Volume
Evaluate:
\[ \int_0^a r\,dr = \frac{a^2}{2}. \]Then
\[ \int_0^{2\pi} \frac{a^2}{2}\,d\theta = \pi a^2. \]Finally,
\[ \int_0^h \pi a^2\,dz = \pi a^2h. \]Therefore,
\[ \boxed{ V=\pi a^2h }. \]Spherical Coordinates
One common spherical-coordinate convention is
\[ \boxed{ x=\rho\sin\phi\cos\theta, } \] \[ \boxed{ y=\rho\sin\phi\sin\theta, } \] \[ \boxed{ z=\rho\cos\phi. } \]Here \(\rho\) is radial distance, \(\theta\) is the azimuthal angle in the \(xy\)-plane, and \(\phi\) is measured downward from the positive \(z\)-axis.
Spherical Radius
Spherical coordinates satisfy
\[ \boxed{ x^2+y^2+z^2=\rho^2 }. \]This makes them especially effective for spheres and radially symmetric three-dimensional integrands.
Spherical Volume Element
The Jacobian determinant leads to the volume element
\[ \boxed{ dV = \rho^2\sin\phi\, d\rho\,d\phi\,d\theta } \]up to the chosen order of integration.
Standard Spherical Ranges
For a full sphere,
\[ \rho\ge0, \] \[ 0\le\theta\le2\pi, \]and
\[ 0\le\phi\le\pi. \]Individual problems may restrict these ranges.
Example: Volume of a Sphere
For the sphere
\[ x^2+y^2+z^2\le a^2, \]spherical coordinates give
\[ 0\le\rho\le a, \qquad 0\le\phi\le\pi, \qquad 0\le\theta\le2\pi. \]Therefore,
\[ V = \int_0^{2\pi} \int_0^\pi \int_0^a \rho^2\sin\phi \,d\rho\,d\phi\,d\theta. \]Complete the Sphere Volume
The three factors separate:
\[ \int_0^a\rho^2\,d\rho = \frac{a^3}{3}, \] \[ \int_0^\pi\sin\phi\,d\phi = 2, \] \[ \int_0^{2\pi}d\theta = 2\pi. \]Therefore,
\[ \boxed{ V=\frac43\pi a^3 }. \]Coordinate Conventions Matter
Different textbooks may use different symbols for spherical angles.
Always verify:
- which angle is measured in the \(xy\)-plane;
- which angle is measured from the \(z\)-axis;
- the corresponding ranges;
- the correct Jacobian factor.
Jacobian as Local Geometry
The Jacobian is not merely a correction factor inserted by rule.
It describes how the coordinate grid itself deforms.
Straight coordinate rectangles in the new variables may become curved quadrilateral regions in the original variables, but at sufficiently small scales they are approximated by parallelograms.
Jacobian and Basis Vectors
In two variables, the vectors
\[ \frac{\partial T}{\partial u} = \langle x_u,y_u\rangle \]and
\[ \frac{\partial T}{\partial v} = \langle x_v,y_v\rangle \]describe how the coordinate curves move.
The magnitude of their two-dimensional determinant gives the area of the small parallelogram they span.
Jacobian and Scaling
If
\[ |J|>1, \]the transformation locally expands area.
If
\[ 0<|J|<1, \]it locally contracts area.
If
\[ |J|=1, \]it locally preserves area.
Transforming Boundaries
A useful coordinate transformation often turns complicated boundaries into coordinate lines.
For example, if
\[ u=x+y \]and the original region is bounded by
\[ x+y=1 \]and
\[ x+y=3, \]then those boundaries become simply
\[ u=1 \qquad\text{and}\qquad u=3. \]Transforming the Integrand
Suppose
\[ f(x,y)=e^{x+y}. \]If
\[ u=x+y, \]then the integrand becomes
\[ \boxed{ e^u }. \]A good change of variables often simplifies both the region and the integrand at the same time.
Recognizing a Useful Transformation
Look for:
- repeated linear combinations;
- quadratic forms such as \(x^2+y^2\);
- families of parallel boundary lines;
- elliptic or circular symmetry;
- products or ratios that repeatedly occur;
- coordinate patterns already suggested by the geometry.
Coordinate Changes and Ellipses
Consider the ellipse
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} \le1. \]Set
\[ x=au, \qquad y=bv. \]Then the region becomes
\[ u^2+v^2\le1. \]The Jacobian is
\[ \boxed{ ab }. \]Area of an Ellipse by Coordinate Change
Using the transformation
\[ x=au, \qquad y=bv, \]the ellipse maps to the unit disk.
Therefore,
\[ \operatorname{Area} = \iint_{u^2+v^2\le1} ab\,du\,dv. \]Since the unit disk has area \(\pi\),
\[ \boxed{ \operatorname{Area}=\pi ab }. \]Jacobian and Probability Density
If random variables are transformed, probability density must adjust for the local stretching of coordinates.
The same Jacobian factor that preserves area in integration also preserves total probability under a change of variables.
This is a major application in probability and statistics.
Jacobian and Conservation
Coordinate changes should not alter the actual accumulated quantity.
The Jacobian compensates for changes in coordinate cell size so that
\[ \boxed{ \text{physical or geometric total} } \]remains invariant under the choice of coordinates.
Common Error: Forgetting the Jacobian
In polar coordinates,
\[ dA \neq dr\,d\theta. \]The correct area element is
\[ \boxed{ dA=r\,dr\,d\theta }. \]Omitting \(r\) changes the value of the integral.
Common Error: Using the Jacobian without Absolute Value
For ordinary area and volume integrals under a change of variables, use
\[ \boxed{|J|}. \]A negative determinant signals orientation reversal, not negative physical area.
Common Error: Transforming Only the Integrand
A complete coordinate change transforms:
\[ f(x,y), \]the region \(R\), and
\[ dA. \]Changing only the formula but leaving the old bounds is generally invalid.
Common Error: Keeping Old Variables in the New Integral
Once an integral has been rewritten in \(u\) and \(v\), the final integrand and bounds should be expressed consistently in those variables.
Mixed-variable expressions often indicate that the substitution is not yet complete.
Common Error: Using the Wrong Jacobian Direction
If the new integral is written in \(du\,dv\), the needed factor is
\[ \left| \frac{\partial(x,y)}{\partial(u,v)} \right|. \]If instead you compute
\[ \frac{\partial(u,v)}{\partial(x,y)}, \]you must invert it when appropriate.
Common Error: Ignoring Singular Points
Some coordinate systems have locations where the Jacobian vanishes.
In polar coordinates,
\[ J=r, \]which vanishes at
\[ r=0. \]The origin is a coordinate singularity because the angle is not uniquely determined there.
Common Error: Assuming a Coordinate Map Is One-to-One Everywhere
Polar coordinates satisfy
\[ (r,\theta) \]and
\[ (r,\theta+2\pi) \]for the same physical point.
Practical bounds are chosen to avoid unnecessary multiple coverage of the region.
A Reliable Change-of-Variables Procedure
- Study the geometry of the region and the algebra of the integrand.
- Choose new variables that simplify both when possible.
- Solve for the old variables in terms of the new variables if needed.
- Transform the boundaries and determine the new region.
- Compute the appropriate Jacobian determinant.
- Take its absolute value for area or volume scaling.
- Rewrite the integrand entirely in the new variables.
- Replace the differential element.
- Evaluate the transformed integral.
- Check geometry, dimensions, and reasonableness.
A Reliable Polar-Coordinate Procedure
- Replace \(x\) by \(r\cos\theta\).
- Replace \(y\) by \(r\sin\theta\).
- Replace \(x^2+y^2\) by \(r^2\) whenever useful.
- Describe the region with radial and angular bounds.
- Replace \(dA\) by \(r\,dr\,d\theta\).
- Evaluate the resulting iterated integral.
Example: Full Polar Workflow
Evaluate
\[ \iint_R (x^2+y^2)\,dA \]over the disk
\[ x^2+y^2\le4. \]In polar coordinates,
\[ x^2+y^2=r^2, \] \[ 0\le r\le2, \qquad 0\le\theta\le2\pi, \]and
\[ dA=r\,dr\,d\theta. \]Complete the Full Polar Workflow
Therefore,
\[ \iint_R (x^2+y^2)\,dA = \int_0^{2\pi} \int_0^2 r^3\,dr\,d\theta. \]The radial integral is
\[ \left[\frac{r^4}{4}\right]_0^2 = 4. \]Thus,
\[ \boxed{ \iint_R (x^2+y^2)\,dA = 8\pi }. \]Practice 1: Simple Jacobian
Find
\[ \frac{\partial(x,y)}{\partial(u,v)} \]for
\[ x=3u, \qquad y=2v. \]Practice 2: Shear Jacobian
Find the Jacobian determinant for
\[ x=u+2v, \qquad y=v. \]Practice 3: Polar Jacobian
Verify that
\[ x=r\cos\theta, \qquad y=r\sin\theta \]has Jacobian
\[ r. \]Practice 4: Disk Area
Use polar coordinates to evaluate
\[ \iint_R1\,dA \]over
\[ x^2+y^2\le9. \]Practice 5: Radial Integrand
Use polar coordinates to evaluate
\[ \iint_R (x^2+y^2)\,dA \]over the unit disk.
Practice 6: Linear Combination Variables
Let
\[ u=x+y, \qquad v=x-y. \]Express \(x\) and \(y\) in terms of \(u\) and \(v\).
Practice 7: Jacobian of the Linear Combination Change
For
\[ x=\frac{u+v}{2}, \qquad y=\frac{u-v}{2}, \]find
\[ \left| \frac{\partial(x,y)}{\partial(u,v)} \right|. \]Practice 8: Cylindrical Coordinates
Write the volume integral for the cylinder
\[ x^2+y^2\le4, \qquad 0\le z\le5 \]in cylindrical coordinates.
Practice 9: Spherical Coordinates
Write the volume integral for the ball
\[ x^2+y^2+z^2\le4 \]in spherical coordinates.
Practice 10: Identify the Error
A student changes a disk integral to polar coordinates and writes
\[ \int_0^{2\pi}\int_0^a f(r\cos\theta,r\sin\theta) \,dr\,d\theta. \]Explain what is missing.
Practice Answers
Practice 1
\[ J = \begin{vmatrix} 3&0\\ 0&2 \end{vmatrix} = \boxed{6}. \]Practice 2
\[ J = \begin{vmatrix} 1&2\\ 0&1 \end{vmatrix} = \boxed{1}. \]Practice 3
We have
\[ x_r=\cos\theta, \qquad x_\theta=-r\sin\theta, \] \[ y_r=\sin\theta, \qquad y_\theta=r\cos\theta. \]Therefore,
\[ J = r\cos^2\theta+r\sin^2\theta = \boxed{r}. \]Practice 4
The disk has
\[ 0\le r\le3, \qquad 0\le\theta\le2\pi. \]Therefore,
\[ \operatorname{Area} = \int_0^{2\pi}\int_0^3 r\,dr\,d\theta = \boxed{9\pi}. \]Practice 5
\[ \int_0^{2\pi}\int_0^1 r^3\,dr\,d\theta = \boxed{\frac{\pi}{2}}. \]Practice 6
Add and subtract the equations:
\[ \boxed{ x=\frac{u+v}{2}, \qquad y=\frac{u-v}{2} }. \]Practice 7
\[ J = \begin{vmatrix} \frac12&\frac12\\ \frac12&-\frac12 \end{vmatrix} = -\frac12. \]Therefore,
\[ \boxed{ |J|=\frac12 }. \]Practice 8
The cylindrical bounds are
\[ 0\le r\le2, \qquad 0\le\theta\le2\pi, \qquad 0\le z\le5. \]Thus,
\[ \boxed{ \int_0^5 \int_0^{2\pi} \int_0^2 r\,dr\,d\theta\,dz }. \]Practice 9
The spherical bounds are
\[ 0\le\rho\le2, \qquad 0\le\phi\le\pi, \qquad 0\le\theta\le2\pi. \]Therefore,
\[ \boxed{ \int_0^{2\pi} \int_0^\pi \int_0^2 \rho^2\sin\phi \,d\rho\,d\phi\,d\theta }. \]Practice 10
The polar area element is missing its Jacobian factor.
The correct integral is
\[ \boxed{ \int_0^{2\pi}\int_0^a f(r\cos\theta,r\sin\theta) r\,dr\,d\theta }. \]What to Remember
Coordinate changes reorganize multivariable problems so that geometry and algebra align more naturally.
- A transformation replaces old coordinates with new ones.
- The Jacobian matrix contains the first derivatives of the transformation.
- The Jacobian determinant measures local area or volume scaling.
- Use the absolute value of the Jacobian for ordinary area and volume integration.
- Polar coordinates use \(dA=r\,dr\,d\theta\).
- Cylindrical coordinates use the same radial factor \(r\).
- Spherical coordinates use \(dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\).
- A successful substitution transforms the integrand, the region, and the differential element consistently.
The core structural idea is
\[ \boxed{ \text{new coordinates} + \text{Jacobian scaling} = \text{same accumulated quantity} }. \]Where We Go Next
Coordinate changes help us integrate and describe multivariable geometry. Another major multivariable task is optimization when the inputs are not free to vary independently.
The next lesson, Constrained Extrema, develops methods for finding maxima and minima subject to equations or other restrictions, including the geometric logic behind Lagrange multipliers.