Double Integral Calculator

About the Double Integral Calculator

A single integral adds up values along an interval; a double integral adds up values over a two-dimensional region. It gives the volume under a surface, the area of a region, the mass of a thin plate with variable density, or the average value of a function of two variables.

In practice a double integral is computed as two ordinary integrals, one inside the other. This calculator performs the inner integral symbolically while treating the outer variable as a constant, evaluates it at the inner limits, and then integrates the result. Each answer is compared with an independent two-dimensional quadrature.

What is a double integral?

Split a region R into many tiny rectangles of area ΔA, multiply the value of f at a point in each rectangle by its area, and add everything up. As the rectangles shrink, these sums approach a limit, written $\iint_R f(x,y)\,dA$ . When f is positive, each term is the volume of a thin column, so the integral is the volume of the solid under the surface z = f(x, y).

Fubini’s theorem says that for continuous functions this limit equals an iterated integral: integrate first in one variable with the other held fixed, then integrate the result in the remaining variable. On a rectangle either order gives the same number.

How to use the Double Integral Calculator

Type the integrand f(x, y), for example x*y or x^2 + y^2.
Choose dy dx (y inside) or dx dy (x inside).
Enter the inner limits; they may depend on the outer variable.
Enter the outer limits as numbers, pi or e.
Press Evaluate integral and follow the inner and outer steps.

Formulas

Rectangle

\iint_R f\,dA = \int_a^b\int_c^d f(x,y)\,dy\,dx

Type I region

\int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx

Type II region

\int_c^d\int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy

Area of a region

A = \iint_R 1\,dA

Average value

f_{\text{avg}} = \frac{1}{A}\iint_R f\,dA

Worked example: ∬ xy over [0,1]×[0,2]

Take $\int_0^1\int_0^2 xy\,dy\,dx$ . Holding x fixed, the inner integral is $\int_0^2 xy\,dy = x\left[\tfrac{y^2}{2}\right]_0^2 = 2x$ .

The outer integral is then $\int_0^1 2x\,dx = 1$ . Integrating in the other order, $\int_0^2\int_0^1 xy\,dx\,dy = \int_0^2 \tfrac{y}{2}\,dy = 1$ as well, as Fubini’s theorem promises on a rectangle.

The solution below is generated by the same engine that powers the calculator above.

Final answer

\int_{0}^{1}\int_{0}^{2} x y\,dy\,dx = 1

How we read your problem

Evaluate the double integral of xy over the region, integrating y first.

\int_{0}^{1}\int_{0}^{2} x y\,dy\,dx

Step-by-step solution

Set up the iterated integral
The region is a rectangle, so the limits are all constants. Integrate with respect to y first, holding x constant, then with respect to x.
$\int_{0}^{1}\int_{0}^{2} x y\,dy\,dx$
Power rule: ∫ xⁿ dx = xⁿ⁺¹/(n+1)
$\int y\,dy = \frac{{y}^{2}}{2}$
Constant multiple rule: ∫ k·f dx = k∫ f dx
$\int x y\,dy = \frac{x {y}^{2}}{2}$
Antiderivative with respect to y (x held constant)
$F = \frac{x {y}^{2}}{2}$
Evaluate the inner integral at its limits
$\int_{0}^{2} x y\,dy = F\big|_{y=0}^{y=2} = 2x$
Integrate the result with respect to x
$\int_{0}^{1} 2x\,dx$
Power rule: ∫ xⁿ dx = xⁿ⁺¹/(n+1)
$\int x\,dx = \frac{{x}^{2}}{2}$
Constant multiple rule: ∫ k·f dx = k∫ f dx
$\int 2x\,dx = {x}^{2}$
Antiderivative
$F(x) = {x}^{2}$
Evaluate F at the limits
$F(1) = 1,\quad F(0) = 0$
Subtract: F(upper) − F(lower)
$1 - \left(0\right) = 1$

Formulas used

Iterated integral (Fubini)
$\iint_R f\,dA = \int_a^b\left(\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\right)dx$
Area of a region
$A = \iint_R 1\,dA$

A double integral adds up f(x, y)·dA over the region. When f is positive it is the volume under the surface z = f(x, y) and above the region.

Verification

Two-dimensional Gauss–Legendre quadrature gives 1
Inner antiderivative verified by differentiation ✓
Exact value 1 agrees with the numerical value ✓
Fubini's theorem: on a rectangle the order dx dy or dy dx gives the same value.

Detailed explanation

Setting up the limits

Sketch the region first. For dy dx order, draw a vertical line through the region at a typical x: it enters at the lower curve $y = g_1(x)$ and leaves at the upper curve $y = g_2(x)$ . Those curves are the inner limits, and the outer limits are the smallest and largest x in the region.

For the region between $y = x^2$ and $y = x$ , the curves meet at x = 0 and x = 1, and on that interval the line y = x lies above the parabola. The area is $\int_0^1\int_{x^2}^{x} 1\,dy\,dx = \int_0^1 (x - x^2)\,dx = \tfrac16$ .

Switching the order

Sometimes one order leads to an integral with no elementary antiderivative while the other is easy. To reverse the order, describe the same region with horizontal lines instead: for each y, find where the line enters and leaves the region, and use the extreme y values as the new outer limits.

The same region between the parabola and the line can be written as $0 \le y \le 1$ and $y \le x \le \sqrt{y}$ , so $\int_0^1\int_{y}^{\sqrt y} 1\,dx\,dy$ also equals 1/6. Comparing the two set-ups is a good way to check limits.

Curved boundaries and polar regions

Over the quarter of the unit disc in the first quadrant, a vertical line at x runs from y = 0 to $y = \sqrt{1 - x^2}$ . With f = xy the inner integral is $\tfrac{x(1 - x^2)}{2}$ , and the outer integral gives exactly 1/8.

Discs, rings and sectors are often easier in polar coordinates, where $dA = r\,dr\,d\theta$ . You can still enter them here in Cartesian form, as above, or substitute x = r cos θ and y = r sin θ yourself and enter the polar integrand with x playing the role of r and y the role of θ.

Common mistakes

Putting a variable in the outer limits; they must be constants.
Using the outer variable’s name in the inner integrand differential.
Listing the lower and upper curves in the wrong order.
Forgetting to find where the boundary curves intersect.
Reversing the order of integration without redrawing the region.

Tips

Sketch the region and shade it before writing any limits.
If the inner integral looks impossible, try the other order.
Check that f = 1 gives the area you expect.
Use symmetry: an odd function over a symmetric region integrates to 0.
Use the definite integral calculator to check the outer step on its own.

Frequently asked questions

How do I evaluate a double integral?

Integrate the inner integral with the outer variable held constant, substitute the inner limits, then integrate the result between the outer limits.

Does the order of integration matter?

Not for the value when f is continuous, but the limits must describe the same region, and one order may be much easier to compute.

Can the inner limits be functions?

Yes. For dy dx the inner limits may contain x, such as y from x² to x; the outer limits must be numbers.

How do I find the area of a region with a double integral?

Integrate the constant function 1 over the region; the result is its area.

What does a double integral represent geometrically?

For positive f, the volume of the solid under the surface z = f(x, y) and above the region in the xy-plane.

What if there is no exact answer?

The calculator falls back to two-dimensional Gauss–Legendre quadrature and reports a numerical value, as with e^(x + y) on some regions.

What is Fubini’s theorem?

It states that for a continuous function on a rectangle the double integral equals either iterated integral, so the order can be swapped.

Evaluate a double integral

Enter your integrand and limits above to see the inner and outer integrals worked out.

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