A double integral adds up values of a function over a two-dimensional region. For a function z = f(x,y), it can represent the volume between the surface and the xy-plane over a chosen area. This idea matters because many physical quantities depend on two variables, such as height over land, temperature on a plate, or density across a thin sheet.
Double integrals turn a surface or field into one accumulated number.
Key Facts
- For a rectangular region R = [a,b] x [c,d], the double integral is ∫∫_R f(x,y) dA = ∫_a^b ∫_c^d f(x,y) dy dx.
- If f(x,y) ≥ 0, then ∫∫_R f(x,y) dA gives the volume under z = f(x,y) above region R.
- The small area element on a rectangle is dA = dx dy or dA = dy dx.
- Fubini's Theorem says ∫_a^b ∫_c^d f(x,y) dy dx = ∫_c^d ∫_a^b f(x,y) dx dy when f is continuous on the rectangle.
- A Riemann sum for a double integral is ∫∫_R f(x,y) dA ≈ ΣΣ f(x_i,y_j) ΔA.
- For constant height f(x,y) = k over a region of area A, ∫∫_R k dA = kA.
Vocabulary
- Double integral
- A double integral is an integral that accumulates a function over a two-dimensional region.
- Region of integration
- The region of integration is the set of points in the xy-plane where the function is being added.
- Iterated integral
- An iterated integral is a double integral evaluated as two single-variable integrals in sequence.
- Area element
- The area element dA represents a tiny piece of area in the plane, often written as dx dy or dy dx for rectangles.
- Fubini's Theorem
- Fubini's Theorem states that a continuous function over a rectangular region can be integrated in either order.
Common Mistakes to Avoid
- Forgetting the region limits, which is wrong because a double integral has no numerical meaning until the domain of x and y is specified.
- Treating the inner variable as constant, which is wrong because the inner integral must be evaluated with respect to its own differential while the other variable is held constant.
- Switching the order of integration without changing the limits, which is wrong for non-rectangular regions and can describe a different area.
- Assuming every double integral is a volume, which is wrong because it represents signed accumulation and gives physical volume only when f(x,y) is nonnegative over the region.
Practice Questions
- 1 Evaluate ∫_0^2 ∫_0^3 (x + y) dy dx.
- 2 Find the volume under z = 4 - x over the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 5.
- 3 Explain why ∫_0^1 ∫_0^2 f(x,y) dy dx and ∫_0^2 ∫_0^1 f(x,y) dx dy represent the same accumulation when f is continuous on the rectangle.