Double Integral Calculator

Evaluate double integrals over rectangular, general, or polar regions with AI-powered step-by-step solutions

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double integral of x*y over [0,1]x[0,2]
double integral of x^2+y^2 over unit disk in polar
double integral of e^(x+y) over [0,1]x[0,1]
double integral of y dA over triangle with vertices (0,0),(1,0),(0,1)

What is a Double Integral?

A double integral computes the accumulation of a function f(x,y)f(x, y) over a two-dimensional region DD:

Df(x,y)dA\iint_D f(x,y)\,dA

where dAdA is the infinitesimal area element. In Cartesian coordinates dA=dxdydA = dx\,dy; in polar coordinates dA=rdrdθdA = r\,dr\,d\theta.

Common physical meanings:

  • f(x,y)=1f(x,y) = 1 gives the area of DD.
  • f(x,y)=h(x,y)f(x,y) = h(x,y) (height function) gives the volume under the surface z=h(x,y)z = h(x,y) above DD.
  • f=ρ(x,y)f = \rho(x,y) (surface density) gives the mass of a thin plate.

The key skills are: choose coordinates, set up bounds, and evaluate as iterated single integrals using Fubini's theorem.

How to Evaluate Double Integrals

Fubini's Theorem

For a continuous ff over a rectangle D=[a,b]×[c,d]D = [a, b] \times [c, d]:

DfdA=abcdf(x,y)dydx=cdabf(x,y)dxdy\iint_D f\,dA = \int_a^b \int_c^d f(x,y)\,dy\,dx = \int_c^d \int_a^b f(x,y)\,dx\,dy

Either order works, so pick the one that's easier to integrate.

Type I and Type II Regions

Type I (yy bounded by curves of xx):

D={(x,y):axb, g1(x)yg2(x)}D = \{(x,y) : a \leq x \leq b,\ g_1(x) \leq y \leq g_2(x)\}

DfdA=abg1(x)g2(x)f(x,y)dydx\iint_D f\,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx

Type II (xx bounded by curves of yy):

D={(x,y):cyd, h1(y)xh2(y)}D = \{(x,y) : c \leq y \leq d,\ h_1(y) \leq x \leq h_2(y)\}

DfdA=cdh1(y)h2(y)f(x,y)dxdy\iint_D f\,dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy

Polar Coordinates

For regions with circular symmetry, use x=rcosθx = r\cos\theta, y=rsinθy = r\sin\theta, dA=rdrdθdA = r\,dr\,d\theta:

Df(x,y)dA=Df(rcosθ,rsinθ)rdrdθ\iint_D f(x,y)\,dA = \iint_D f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta

The factor of rr from the Jacobian is essential — forgetting it is the most common error.

When to Switch Order of Integration

If an inner integral becomes intractable (e.g., ex2dx\int e^{x^2}\,dx has no elementary antiderivative), switching the order of integration often makes the problem solvable. Sketch the region first to find equivalent bounds in the other order.

Common Mistakes to Avoid

  • Wrong bounds order: Inner bounds may depend on outer variables, but outer bounds must be constants. Reversed = wrong answer.
  • Forgetting the polar Jacobian: dA=rdrdθdA = r\,dr\,d\theta, not drdθdr\,d\theta.
  • Not sketching the region: For non-rectangular DD, a sketch makes Type I vs Type II obvious.
  • Trying to integrate impossible inner functions: If you hit ex2dx\int e^{x^2}\,dx or similar non-elementary integrand, swap the order before giving up.
  • Sign errors with negative integrands: If ff changes sign over DD, the double integral can be zero — this is correct, not a mistake to 'fix'.

Examples

Step 1: Set up: 0101(x2+y2)dydx\int_0^1 \int_0^1 (x^2 + y^2)\, dy\, dx
Step 2: Integrate over yy: 01(x2+y2)dy=x21+13=x2+13\int_0^1 (x^2 + y^2)\,dy = x^2 \cdot 1 + \frac{1}{3} = x^2 + \frac{1}{3}
Step 3: Integrate over xx: 01(x2+13)dx=13+13=23\int_0^1 (x^2 + \frac{1}{3})\,dx = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}
Answer: 23\dfrac{2}{3}

Step 1: Switch to polar: x2+y2=r2x^2 + y^2 = r^2, dA=rdrdθdA = r\,dr\,d\theta
Step 2: Bounds: 0r10 \leq r \leq 1, 0θ2π0 \leq \theta \leq 2\pi
Step 3: Integral becomes: 02π01r2rdrdθ=02π01r3drdθ\int_0^{2\pi} \int_0^1 r^2 \cdot r\, dr\, d\theta = \int_0^{2\pi} \int_0^1 r^3\, dr\, d\theta
Step 4: Inner: 01r3dr=14\int_0^1 r^3\,dr = \frac{1}{4}
Step 5: Outer: 02π14dθ=π2\int_0^{2\pi} \frac{1}{4}\,d\theta = \frac{\pi}{2}
Answer: π2\dfrac{\pi}{2}

Step 1: Region: 0x10 \leq x \leq 1 and 0y1x0 \leq y \leq 1 - x (Type I)
Step 2: Set up: 0101xydydx\int_0^1 \int_0^{1-x} y\,dy\,dx
Step 3: Inner: 01xydy=(1x)22\int_0^{1-x} y\,dy = \frac{(1-x)^2}{2}
Step 4: Outer: 01(1x)22dx=12(1x)3301=16\int_0^1 \frac{(1-x)^2}{2}\,dx = \frac{1}{2} \cdot \frac{(1-x)^3}{-3}\Big|_0^1 = \frac{1}{6}
Answer: 16\dfrac{1}{6}

Frequently Asked Questions

Use polar when the region or integrand has circular symmetry — disks, annuli, sectors, or functions of x²+y². The Jacobian r often simplifies the integrand by canceling factors.

Fubini's theorem says that for a continuous function over a rectangle (or any region where the integral is absolutely convergent), the double integral equals an iterated integral, and the order of integration can be swapped without changing the result.

Sketch the region D. Find equivalent descriptions as Type I and Type II — that is, express the same region with x bounded by curves of y instead of y bounded by curves of x. Rewrite the integral with the new bounds.

The factor r comes from the Jacobian determinant of the transformation from (x,y) to (r,θ). Geometrically, a thin polar 'wedge' has area r·dr·dθ, not just dr·dθ.

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