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Double Integral Calculator

Evaluate ∬ f(x, y) dA as an iterated integral, inner layer first, with each step shown. Bounds can be curves, and the region is sketched so you can check it.

Reads as xy2x y^{2}
yy fromto
xx fromto

Inner bounds may use the outer variables, e.g. y from 0 to x.

Examples:

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How an iterated integral works

Hold xx fixed and integrate across the region in the yy direction: that gives the area of one slice of the solid under z=f(x,y)z = f(x,y). Then integrate those slice areas along xx. That is exactly the order the steps follow: the inner integral’s result is a function of xx only, and the outer integral turns it into a number.

∬Rf(x,y) dA=∫ab(∫g1(x)g2(x)f(x,y) dy)dx\iint_R f(x,y)\,dA = \int_a^b \left( \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy \right) dx

Setting up the bounds

  1. Sketch the region. Decide which variable runs between two curves (inner) and which runs between two numbers (outer).
  2. For “dy dx”: yy goes from the bottom curve to the top curve, and xx from the leftmost to the rightmost point.
  3. Check the sketch above matches what you drew. If the inner bounds swap places partway along, split the region into two integrals.

One-variable version: the integral calculator. Going up a dimension: the triple integral calculator. The inner steps use the same rules as partial derivatives in reverse: the other variable acts as a constant.

Questions students ask

How do you evaluate a double integral?

As an iterated integral: integrate the inner variable first while treating the outer one as a constant, then integrate the result over the outer range. For ∫02∫01xy2 dy dx\int_0^2\int_0^1 xy^2\,dy\,dx: the inner integral gives x3\frac{x}{3}, and ∫02x3dx=23\int_0^2 \frac{x}{3}dx = \frac23.

What does a double integral represent?

For f≥0f \ge 0 it is the volume under the surface z=f(x,y)z = f(x,y) above the region RR. With f=1f = 1 it is simply the area of RR. It also gives mass from a density, averages and probabilities.

Does the order of integration matter?

Not for the value (Fubini’s theorem, for continuous ff), but it changes the bounds and can make one order much easier. On non-rectangular regions you must rewrite the bounds when you switch. For example, 0≤y≤x, 0≤x≤10 \le y \le x,\ 0 \le x \le 1 becomes y≤x≤1, 0≤y≤1y \le x \le 1,\ 0 \le y \le 1.

How do I enter variable bounds?

Inner bounds may contain the outer variable: with order dy dx, y can run from x2x^2 to x\sqrt{x}. The region sketch updates so you can check it is the region you meant.

Can it do double integrals in polar coordinates?

Enter rr and θ\theta as xx and yy and remember the extra factor rr: ∬f dA=∫∫f(rcos⁡θ,rsin⁡θ) r dr dθ\iint f\,dA = \int\int f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta. For example, the area of a unit disc is ∫02π∫01r dr dθ=π\int_0^{2\pi}\int_0^1 r\,dr\,d\theta = \pi (enter x·1 with x from 0 to 1 and y from 0 to 2π).

What if there is no antiderivative?

If a layer has no elementary antiderivative (like ex2e^{x^2}), the calculator switches to nested Gauss–Legendre quadrature and gives a numerical value to about 9 significant digits.