A PDE is, by definition, an equation built from partial derivatives. u_t, u_{xx}, and their relatives are the entire vocabulary this course writes in. Multiple integrals show up just as early. They appear the moment a flux-balance statement gets written down: a rate of change inside a region, balanced against a flux integrated over that region's boundary. The chain rule is what makes both of those survive a change of coordinates intact. And a change of coordinates only survives to begin with when its Jacobian is nonzero. The Jacobian is the last idea this section reviews.
A partial derivative asks one question: "how fast is this changing in one direction, holding everything else still?" Freeze every other variable. What's left is an ordinary single-variable derivative. The chain rule for a path adds up the change coming from each direction the path is moving in, all at once. Some of the total rate comes from x changing. Some comes from y changing. The two contributions add.
A single integral adds up a running total along a line. A double integral repeats that idea one notch up. It adds up a running total spread across a flat region, instead of along a line. Try adding up the value 1 everywhere across a rectangle two units by three units. That running total comes out to exactly six. That's the rectangle's plain area. A triple integral repeats the same idea once more. It spreads the running total through a solid, instead of a flat region.
Swapping (x,y) for a new pair of coordinates (\xi,\eta) (the Greek letters xi and eta) only makes sense if the swap can be undone. Every (\xi,\eta) pair has to trace back to exactly one nearby (x,y) point. Not several at once, and not none at all. The Jacobian is a single number that tests exactly this. It's built from the four partial derivatives of \xi and \eta, with respect to x and y. A nonzero value means the swap is safe to make, near that point. A zero value is a warning sign. It means the new coordinates have stopped being independent right there. That's like trying to pin down a location using two street names that happen to run parallel at that spot.
For u(x,y), the partial derivative u_x=\partial u/\partial x treats y as a fixed constant. It differentiates in x alone. Meanwhile u_y does the same thing, with the roles reversed. The multivariable chain rule composes these with a path. Say x=x(t) and y=y(t). Then \frac{d}{dt}u(x(t),y(t)) = u_x\,\frac{dx}{dt} + u_y\,\frac{dy}{dt}. A multiple integral totals a function across a region of any dimension. It's computed as one nested single-variable integral per dimension. A double integral \iint_D f(x,y)\,dA totals f over a planar region D. Taking f\equiv1 recovers D's plain area, the same way \int_a^b1\,dx recovers a length. For instance, take the rectangle D=[0,2]\times[0,3]. Then \iint_D1\,dA=\int_0^3\!\int_0^2 1\,dx\,dy =\int_0^3 2\,dy=6, exactly the rectangle's area. A triple integral \iiint_E f(x,y,z)\,dV repeats the same idea one dimension further, over a solid region E. Taking f\equiv1 recovers E's volume.
A change of coordinates (x,y)\mapsto(\xi,\eta) is only legitimate where it can actually be undone. Its Jacobian is the determinant J=\det\begin{pmatrix}\xi_x&\xi_y\\\eta_x&\eta_y\end{pmatrix} =\xi_x\eta_y-\xi_y\eta_x. Wherever J\neq0, the Inverse Function Theorem guarantees something useful: (\xi,\eta) can be solved back out for (x,y) near that point. So (\xi,\eta) is a genuine local coordinate system there. Wherever J=0, the two new coordinates fail to pin a point down uniquely. Then (\xi,\eta) collapses to fewer than two truly independent directions.
Two frozen directions, one surface
Toggle the trace planes -- the x = x0 and y = y0 slices are exactly what u_x and u_y measure: the slope of the surface along one frozen direction at a time.
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A coordinate grid, and where it folds
Drag the slider to slide m2 toward 2 -- watch the amber coordinate-line family rotate to meet the fixed indigo family, and the Jacobian readout above turn red the instant they stop crossing transversally.
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Worked example
Computing both partial derivatives
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1
For u(x,y)=x^2y+e^{xy}, find u_x and u_y.
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2
Freeze y and differentiate in x: u_x = 2xy + ye^{xy}. Freeze x and differentiate in y: u_y = x^2 + xe^{xy}.
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3\boxed{u_x = 2xy+ye^{xy},\qquad u_y = x^2+xe^{xy}}
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Worked example
A triple integral over a box
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1
Evaluate \displaystyle\iiint_E 2xy\,dV over the box 0\le x\le1, 0\le y\le2, 0\le z\le3.
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2
Nest the integrals: \int_0^3\!\int_0^2\!\int_0^1 2xy\,dx\,dy\,dz. The inner integral gives \int_0^1 2xy\,dx = y.
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3
\int_0^3\!\int_0^2 y\,dy\,dz = \int_0^3 2\,dz = 6. \boxed{6}
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Worked example
Checking a coordinate change for degeneracy
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1
For \xi=y-x and \eta=y+2x, find the Jacobian J=\xi_x\eta_y-\xi_y\eta_x and say whether (\xi,\eta) is a genuine coordinate system near the origin.
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2
\xi_x=-1, \xi_y=1, \eta_x=2, \eta_y=1, so J=(-1)(1)-(1)(2)=-1-2=-3.
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3\boxed{J=-3\neq0}
Nonzero, so (\xi,\eta) is a valid local coordinate system everywhere, not only near the origin. Had \eta instead been chosen as y-x (the same combination as \xi), J would come out 0. That's exactly the degenerate case the widget above lets you dial in to.
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Where this goes next
Conservation-law derivations
Every conservation-law derivation ahead in this course starts the same way. That's true for heat, mass, or momentum. It starts from a triple integral of a density over an arbitrary region. It also uses a surface integral of the flux crossing that region's boundary. Turning that statement into a single pointwise PDE takes the divergence theorem, reviewed next. But the raw material on both sides of the equation is exactly the multiple integrals and partial derivatives reviewed here. Ch.01, "Where PDEs come from," is where that argument runs in full, for the heat and wave equations. A genuine double integral shows up by ch.05, "Separation of variables and the heat equation." There, it matches two-dimensional data against a double Fourier series. The Jacobian reviewed above resurfaces by name too. It appears in ch.02's reduction of a second-order PDE to canonical form. There, a proposed coordinate change is only accepted once its Jacobian is checked.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Ordinary Differential Equations
For g(x,y)=x^3-4xy^2, find g_y.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Treating x as a constant, \dfrac{\partial}{\partial y}[x^3]=0 and \dfrac{\partial}{\partial y}[-4xy^2]=-8xy.
Correct answer
Also accepted:
Why
From Calculus III — Multivariable Calculus
For f(x,y,z) = x^2yz^3, find f_z(x,y,z).
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Freeze x and y; the power rule on z^3 gives 3z^2.
Correct answer
Also accepted:
Why
This works the same way with three variables -- freeze all but the one you're differentiating with respect to.
From Calculus III — Multivariable Calculus
For f(x,y) = \dfrac{x}{x+y}, find f_x(1,1).
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Why
Quotient rule with y fixed: f_x = \dfrac{(x+y)(1) - x(1)}{(x+y)^2} = \dfrac{y}{(x+y)^2}, so f_x(1,1) = \dfrac{1}{4} = 0.25.
Correct answer
Also accepted:
Why
Use the quotient rule treating y as a constant.
f_x = y / (x+y)^2.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
For z=x^2y^3, x=2s, y=s, find dz/ds at s=1 (using the single-variable-path chain rule with x, y both functions of s).
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Why
z_x=2xy^3, z_y=3x^2y^2, dx/ds=2, dy/ds=1. At s=1: x=2, y=1. dz/ds=2(2)(1)(2)+3(4)(1)(1)=8+12=20.
Correct answer
Also accepted:
Why
dz/ds = z_x*(dx/ds) + z_y*(dy/ds), evaluated at x=2, y=1.
From Calculus III — Multivariable Calculus
For z=xy, x=\cos t, y=\sin t, find dz/dt at t=0.
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Why
z_x=y, z_y=x, dx/dt=-sin t, dy/dt=cos t. dz/dt = y(-sin t) + x(cos t). At t=0: x=1,y=0, so dz/dt = 0(0)+1(1)=1.
Correct answer
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Why
At t=0, x=cos(0)=1 and y=sin(0)=0.
From Calculus III — Multivariable Calculus
For z=x^2+y^2, x=t, y=t^2, find dz/dt at t=1.
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Why
z_x=2x, z_y=2y, dx/dt=1, dy/dt=2t. dz/dt=2x(1)+2y(2t). At t=1: x=1,y=1, so dz/dt=2(1)+2(1)(2)=2+4=6.
Correct answer
Also accepted:
Why
dz/dt = z_x*(dx/dt) + z_y*(dy/dt), evaluated at t=1 (so x=1, y=1).
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
If f(x,y,z) represents mass density and E is a solid object, \iiint_E f\,dV computes:
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Volume alone ignores density entirely -- that's just \iiint_E 1\,dV.
The centroid requires dividing weighted moment integrals by mass, not this single integral alone.
Why
Density times tiny volume, summed over the solid, is exactly total mass -- the three-dimensional analogue of a lamina's mass integral.
Recall how a lamina's mass integral combines density and area in two variables.
Density times a small volume element, summed over the whole solid, gives a total amount of what quantity?
From Calculus III — Multivariable Calculus
For a box E=[a,b]\times[c,d]\times[p,q], Fubini's Theorem for triple integrals guarantees:
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Why
Just as in two variables, continuity of f on the box guarantees every one of the six possible iteration orders produces the identical value.
Think about what Fubini's Theorem guarantees for double integrals over a rectangle, then extend that to three variables.
The guarantee doesn't depend on which order you pick, as long as f is continuous on the box.
From Calculus III — Multivariable Calculus
Evaluate \int_0^1\int_0^1\int_0^1 xyz\,dz\,dy\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Since xyz factors, the triple integral is the product of three identical integrals \int_0^1 t\,dt=\tfrac12, giving (\tfrac12)^3=\tfrac18.
Correct answer
Also accepted:
Why
The integrand xyz separates into a product x times y times z.
Each single integral \int_0^1 t,dt = 1/2, so the answer is (1/2)^3.