33
6
sq units
6
sq units
5.5
18
6
-9
18
33
6
sq units
6
sq units
5.5
18
6
-9
18
The Double Integral Calculator computes the exact double integral of a bilinear function $$f(x,y) = axy + bx + cy + d$$ over a rectangular region $$[x_1, x_2] \times [y_1, y_2]$$. Double integrals extend the concept of single-variable integration to two dimensions, computing the volume between a surface and the xy-plane over a specified region.
Double integration is a cornerstone of multivariable calculus, introduced formally in the 19th century through the work of mathematicians such as Augustin-Louis Cauchy and Bernhard Riemann. The double integral $$\iint_R f(x,y)\,dA$$ generalizes the single integral by summing infinitesimal volumes $$f(x,y)\,dx\,dy$$ over a two-dimensional region $$R$$.
For rectangular regions and polynomial integrands, Fubini's Theorem allows us to compute double integrals as iterated single integrals. We first integrate with respect to one variable (treating the other as a constant), then integrate the result with respect to the second variable. For the bilinear function $$f(x,y) = axy + bx + cy + d$$, every step uses the power rule, and the final answer is a closed-form expression in terms of the coefficients and bounds.
This calculator decomposes the double integral into four separate contributions — from the $$xy$$ term, the $$x$$ term, the $$y$$ term, and the constant — allowing you to see how each part of the integrand contributes to the total volume. The closed-form formulas are:
$$\iint_R axy\,dA = a \cdot \frac{x_2^2 - x_1^2}{2} \cdot \frac{y_2^2 - y_1^2}{2}$$
$$\iint_R bx\,dA = b \cdot \frac{x_2^2 - x_1^2}{2} \cdot (y_2 - y_1)$$
Applications of double integrals are extensive. In physics, they compute mass, center of mass, and moments of inertia for planar regions with variable density. In engineering, double integrals determine forces on surfaces under pressure, heat flow through plates, and electric charge distributions. In probability, the joint distribution of two random variables involves double integration. Enter your function coefficients and rectangular bounds to compute the exact volume.
Using Fubini's Theorem, the double integral over a rectangle is computed as iterated integrals:
$$\iint_R f(x,y)\,dA = \int_{x_1}^{x_2} \int_{y_1}^{y_2} (axy + bx + cy + d)\,dy\,dx$$
Inner integral (integrate with respect to y):
$$\int_{y_1}^{y_2} (axy + bx + cy + d)\,dy = ax\frac{y_2^2 - y_1^2}{2} + bx(y_2 - y_1) + c\frac{y_2^2 - y_1^2}{2} + d(y_2 - y_1)$$
Outer integral (integrate with respect to x):
$$\int_{x_1}^{x_2} \left[ax\frac{y_2^2 - y_1^2}{2} + bx(y_2 - y_1) + c\frac{y_2^2 - y_1^2}{2} + d(y_2 - y_1)\right]dx$$
Each term integrates independently using the power rule, yielding the four contributions shown in the output. The total volume is their sum.
The Double Integral (Volume) is the net signed volume between the surface $$z = f(x,y)$$ and the xy-plane over the rectangular region. Positive values indicate the surface is predominantly above the xy-plane; negative values indicate it is predominantly below.
The four contribution terms show how each component of the integrand affects the total. This decomposition is useful for understanding which terms dominate and for verifying the computation.
The Region Area is simply $$(x_2 - x_1)(y_2 - y_1)$$, the area of the rectangular domain of integration. Dividing the volume by the region area gives the average value of $$f(x,y)$$ over the rectangle.
Inputs
Results
xy term: 2·(4/2)·(9/2) = 2·2·4.5 = 18. x term: 1·(4/2)·3 = 6. y term: (−1)·2·(9/2) = −9. Correction: y term = −1·2·4.5 = −9. Wait: c_coeff·dx·ySqDiff = −1·2·4.5 = −9. Let me recheck. Actually −1 × 2 × 4.5 = −9. Hmm, the closed form gives −4.5? No: ySqDiff = (9−0)/2 = 4.5, dx = 2. So yContrib = −1 × 2 × 4.5 = −9. Const = 3 × 2 × 3 = 18. Total = 18 + 6 − 9 + 18 = 33.
Inputs
Results
xy term: 1·((9−1)/2)·((4−1)/2) = 4·1.5 = 6. Constant term: 1·2·1 = 2. Total = 6 + 0 + 0 + 2 = 8. Average value = 8/2 = 4.
A double integral $$\iint_R f(x,y)\,dA$$ sums the values of $$f(x,y)$$ over a two-dimensional region $$R$$. Geometrically, when $$f(x,y) \geq 0$$, it gives the volume under the surface $$z = f(x,y)$$ above the region $$R$$ in the xy-plane.
Fubini's Theorem states that if $$f(x,y)$$ is continuous on a rectangular region $$[a,b] \times [c,d]$$, the double integral can be computed as an iterated integral: $$\iint_R f\,dA = \int_a^b \int_c^d f(x,y)\,dy\,dx = \int_c^d \int_a^b f(x,y)\,dx\,dy$$. The order of integration can be reversed.
This calculator is designed for rectangular regions $$[x_1, x_2] \times [y_1, y_2]$$. For non-rectangular regions (bounded by curves), the integration limits become functions of the other variable, requiring different computational approaches.
Just as a single integral gives signed area (negative where $$f < 0$$), a double integral gives signed volume. Where $$f(x,y) > 0$$, the contribution is positive (volume above the xy-plane). Where $$f(x,y) < 0$$, the contribution is negative (volume below the xy-plane).
The bilinear function $$axy + bx + cy + d$$ is a sum of four terms, and integration is linear, so the double integral splits into four independent integrals. This decomposition aids understanding and makes verification straightforward.
The average value is the double integral divided by the area of the region: $$\bar{f} = \frac{1}{\text{Area}(R)} \iint_R f(x,y)\,dA$$. For a rectangle, $$\text{Area}(R) = (x_2 - x_1)(y_2 - y_1)$$.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
How helpful was this calculator?
Be the first to rate!
