Multiple integrals extend an ordinary integral to more than one variable at once. The core idea does not change: add up many small pieces. What changes is the shape being added over — a flat region for a double integral, or a solid for a triple integral, instead of a single line. Sometimes it helps to change coordinates before adding up those pieces. In the plane, a common choice is polar coordinates. In space, common choices are cylindrical and spherical coordinates, depending on the shape of the region. Making that switch needs a correction factor, so that area or volume is never over-counted or under-counted. That correction factor is the Jacobian. Chapter 04 reopens exactly this idea and asks where the correction factor actually comes from.
A multiple integral uses the same idea as an ordinary integral: add up infinitely many tiny pieces. The only change is that it runs over an area or a volume, not a length. Picture a rubber grid stamped with unit squares, then stretched over a curved region. Near any one point, the stretch looks almost flat and even. Some squares get bigger. Some get skewed into slanted shapes. The Jacobian is the local "how much bigger" factor for whichever tiny square sits at that point. It is exactly the fix this running total needs, to stay honest after the stretch.
A double integral \iint_R f\,dA sums f over every tiny piece of area dA in the region R. It is computed as two nested single-variable integrals. A triple integral \iiint_E f\,dV does the same over a solid E, with volume piece dV, nested three deep. The Jacobian is not a special multivariable-calculus invention. It answers the same "how does area or volume scale" question that the 1D substitution rule dx = \frac{dx}{du}\,du already asked. Here, though, the answer is a determinant instead of a single derivative. That's because more than one coordinate direction can stretch and shear at once, and one derivative can't capture that alone. In polar coordinates, the Jacobian works out to the single factor r, so dA=r\,dr\,d\theta. In cylindrical and spherical coordinates, the same determinant produces the extra r or \rho^2\sin\phi factor that every table of multiple integrals already lists.
Formula
The Jacobian
also known as: the local area-scaling factor of a change of variables
Recalled from Calculus III — Multivariable Calculus · The Jacobian — first taught there.
For a transformation x=x(u,v),\ y=y(u,v), the Jacobian is the determinant of the matrix of first partial derivatives. \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[4pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix} = \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u} Its absolute value \left|\dfrac{\partial(x,y)}{\partial(u,v)}\right| is the factor by which the transformation stretches or shrinks area near (u,v). A small rectangle of area du\,dv in the uv-plane maps to a region of area approximately \left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|du\,dv in the xy-plane. The same construction extends to three variables as a 3\times3 determinant of first partials, \dfrac{\partial(x,y,z)}{\partial(u,v,w)}.
Near (u_0,v_0), the transformation's best linear approximation sends the edge vector (du,0) to \left(\frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}\right)du and the edge vector (0,dv) to \left(\frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}\right)dv — the two columns of the Jacobian matrix, scaled. The image of the small du\times dv rectangle is (to first order) the parallelogram spanned by those two vectors, and the area of a parallelogram spanned by (a,b) and (c,d) is |ad-bc|, exactly the absolute value of that 2\times2 determinant.
A radial field: magnitude grows with distance from the origin, just like the Jacobian factor r
Compare the arrow lengths near the origin to the arrow lengths further out -- both grow linearly with distance, the same r dependence the polar Jacobian carries.
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Worked example
Converting a double integral to polar
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1
Rewrite \displaystyle\iint_R (x^2+y^2)\,dA over the disk x^2+y^2\le 4 in polar coordinates.
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2
x^2+y^2 = r^2, and the Jacobian for polar coordinates contributes an extra factor of r: dA = r\,dr\,d\theta.
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3
The disk becomes 0\le r\le 2, 0\le\theta\le 2\pi.
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4
\boxed{\displaystyle\int_0^{2\pi}\!\!\int_0^2 r^2\cdot r\,dr\,d\theta = \int_0^{2\pi}\!\!\int_0^2 r^3\,dr\,d\theta}
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Where this goes next
Chapter 4 of this course
Every coordinate system this course adds — general curvilinear coordinates in chapter 04 — carries its own Jacobian, computed the same determinant way. The scale factors h_u, h_v, h_w chapter 04 introduces are literally the Jacobian's diagonal entries for an orthogonal system.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
For a box E=[a,b]\times[c,d]\times[p,q], Fubini's Theorem for triple integrals guarantees:
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
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.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
The Jacobian is the multivariable generalization of which single-variable concept?
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
Just as single-variable substitution trades dx for g'(u) du, the Jacobian is the exact factor a coordinate change trades dA or dV for.
Recall what single-variable u-substitution replaces dx with.
Both formulas trade one differential for another, multiplied by a derivative-based factor.
From Calculus III — Multivariable Calculus
For u=x-y,\ v=x+y (solved as x=(u+v)/2, y=(v-u)/2), find the Jacobian \left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|.
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
\dfrac{\partial x}{\partial u}=\tfrac12,\ \dfrac{\partial x}{\partial v}=\tfrac12,\ \dfrac{\partial y}{\partial u}=-\tfrac12,\ \dfrac{\partial y}{\partial v}=\tfrac12, giving determinant \tfrac14+\tfrac14=\tfrac12.
Correct answer
Also accepted:
Why
Compute the four partials of x and y with respect to u and v, each equal to plus or minus 1/2.
The determinant is (1/2)(1/2) - (1/2)(-1/2).