Multiple integrals
The idea
A double integral accumulates a quantity over a two-dimensional region: volume under a surface, total mass from an area density, average temperature over a plate. The Riemann logic is unchanged from one variable — tile the region with small rectangles of area dA, multiply each by a function value, sum, refine. Fubini's theorem makes it computable: integrate one variable at a time, as an iterated integral, holding the outer variable fixed during the inner pass.
All the craft lives in setting limits. For a rectangle the bounds are constants, but for a general region the inner limits are functions of the outer variable: you sweep a line across the region and record where it enters and exits. Always draw the region first; the sketch dictates the limits, not the other way around.
The classic misconception is that swapping the order of integration just swaps the limit labels. Inner limits must depend only on the outer variable, and reversing the order over a non-rectangular region means re-describing the region from scratch — sometimes that re-description is the only way to make an integral doable.
Worked example
Evaluate the double integral of f(x, y) = x + y over the triangular region bounded by y = 0, y = x, and x = 2.
- Sketch the region: a triangle with vertices (0, 0), (2, 0), and (2, 2). For each fixed x between 0 and 2, y runs from the floor y = 0 up to the slanted edge y = x.
- Set up the iterated integral with those limits: the outer integral takes x from 0 to 2, the inner takes y from 0 to x — inner limits depending on x, exactly as a non-rectangular region demands.
- Do the inner pass with x frozen: the integral of (x + y) dy from 0 to x is xy + y²/2 evaluated at y = x, giving x² + x²/2 = 3x²/2.
- Do the outer pass: the integral of 3x²/2 from 0 to 2 is x³/2 evaluated from 0 to 2, which is 8/2 = 4.
- Sanity-check the size: the triangle has area 2, and on it x + y ranges from 0 to 4 with the larger values near (2, 2), so an average height of 2 — total 4 — sits right in the believable band.
Answer. The double integral of x + y over the triangle equals 4.
Check your understanding
- Why must the inner limits of an iterated integral be free of the inner variable, and what would it even mean if they were not?
- How would you re-describe this triangle to integrate in the order dx dy instead, and what limits result?
- What does the double integral compute when f(x, y) = 1, and why is that a useful special case to remember?
- When would switching the order of integration turn an impossible inner antiderivative into an easy one?
Build the foundations first
Multiple integrals builds on these concepts. If any feel shaky, start there.