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.
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.
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.
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.
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.
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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