Double Integrals over General Regions

Integrate over a plane region by iterating two single integrals, with the inner limits describing the region. Choosing the order of integration is often the whole problem.

ages 19 to 2014 practice questionsvideo included

What a learner can do afterwards

Set up an iterated integral for a region bounded by two curves
Reverse the order of integration and redraw the region to find the new limits
Compute volume from a nonnegative height function and distinguish it from signed integrals of sign-changing functions

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For a region with one vertical span at each x, write lower y <= y <= upper y and put dy inside. The outer x-range is constant. Between y = x² and y = 3x on 0 <= x <= 3, the inner limits are x² and 3x. Regions with several spans may need separate integrals.

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Integrating 1 gives area. For -1 <= x <= 1 and 0 <= y <= 4, integrate 1 with respect to y to get 4. Then integrate 4 from x = -1 to 1, giving 4(1 - (-1)) = 8.

For a nonnegative height z = f(x, y), the double integral over the base gives volume above the xy-plane. A constant height multiplies the base area. Height 3 over a base of area 2 gives volume 6. A negative or sign-changing function gives a signed result instead.

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Reverse order by describing the same region with horizontal spans. The triangle 0 <= x <= 2, x <= y <= 2 becomes 0 <= y <= 2, 0 <= x <= y. Between y = x² and y = 3x, horizontal spans are y/3 <= x <= √y with 0 <= y <= 9.

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With functional inner limits, evaluate them before moving outward. For the integral of 1 with 0 <= x <= 3 and 0 <= y <= 2x, the inner integral gives 2x. The outer integral gives [x²] from 0 to 3 = 9.

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Hold x fixed while integrating with respect to y. On 0 <= x <= 2, 0 <= y <= 1, the inner integral of x + 2y is [xy + y²] from 0 to 1 = x + 1. Integrating x + 1 from 0 to 2 gives [x²/2 + x] = 4.

dy dx uses vertical slices with lower and upper y-boundaries. If those boundaries change formula, split the x-range. Alternatively, try dx dy, which uses horizontal slices with left and right x-boundaries. Rebuild the limits; changing order does not guarantee that one span will suffice.

Ordinary area uses lower-to-upper inner limits. Reversing those limits changes the sign of the integral. For a nonnegative integrand, a negative result is a reason to check whether a span was written backward.

A varying height must remain inside the integral. Over 0 <= x <= 2 and 0 <= y <= x, height z = x gives inner integral x × x = x². The outer integral is [x³/3] from 0 to 2 = 8/3.

Positive and negative heights can cancel in a signed double integral. To find geometric volume between a sign-changing surface and the xy-plane, split where its sign changes and integrate its magnitude instead.

To check whether a point belongs to a region, substitute its coordinates into every boundary inequality. It belongs only when all inequalities hold. For 0 <= y <= 3 and y <= x <= 5, check the y-range and then both x-boundaries.

Describe each span, match it to dy dx or dx dy, and integrate inside out. Split where boundaries change. A nonnegative height gives volume; a sign-changing height gives a signed result.

Lesson and assessment explanations written for Marble.

Additional questions and readings written for Marble. Videos credit Khan Academy.

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Where it sits

Learn first

Areas Between Two Curves

Optional connections

Riemann Sums and the Definite Integral as a Limit

This opens up

Product Measure and Fubini's Theorem Double Integrals in Polar Coordinates Green's Theorem in the Plane Joint Distributions, Covariance and Independence

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