Double and Triple Integrals Explained

A single integral adds up values along a line. Double and triple integrals do the same thing over a region in a plane or a solid in space. They let you find volumes, masses, centres of mass and probabilities, and they are a core part of university multivariable calculus. The good news: you evaluate them using the single-variable skills you already have.

What Is a Double Integral?

For a function \( f(x, y) \) over a region \( R \) in the \( xy \)-plane, the double integral is written:

$$ \iint_R f(x, y)\,dA $$

If \( f(x, y) \ge 0 \), it gives the volume under the surface \( z = f(x, y) \) and above \( R \). If \( f(x, y) = 1 \), it simply gives the area of \( R \).

Iterated Integrals: How to Evaluate Them

We evaluate a double integral as two single integrals, one inside the other:

$$ \int_a^b \int_c^d f(x, y)\,dy\,dx $$

1. Inner integral first: integrate with respect to \( y \), treating \( x \) as a constant.
2. Outer integral next: integrate the result with respect to \( x \).

This is just like partial differentiation in reverse: hold one variable fixed while you work on the other.

Fubini's Theorem

For a rectangular region and a continuous function, the order does not matter:

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

Worked Example 1: A Rectangular Region

Evaluate \( \displaystyle\int_0^2 \int_0^3 (x + 2y)\,dy\,dx \).

Step 1: Inner integral (with respect to \( y \), \( x \) constant):

$$ \int_0^3 (x + 2y)\,dy = \Big[xy + y^2\Big]_0^3 = 3x + 9 $$

Step 2: Outer integral (with respect to \( x \)):

$$ \int_0^2 (3x + 9)\,dx = \left[\frac{3x^2}{2} + 9x\right]_0^2 = 6 + 18 = 24 $$

Check with the other order:

$$ \int_0^3 \int_0^2 (x + 2y)\,dx\,dy = \int_0^3 \left[\frac{x^2}{2} + 2xy\right]_0^2 dy = \int_0^3 (2 + 4y)\,dy = \Big[2y + 2y^2\Big]_0^3 = 6 + 18 = 24 $$

Both orders give 24, as Fubini's theorem guarantees.

Non-Rectangular Regions

When the region is bounded by curves, the inner limits become functions of the outer variable.

Type I region: \( a \le x \le b \) and \( g_1(x) \le y \le g_2(x) \):

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

The outer limits must always be constants.

Worked Example 2: Region Between Two Curves

Evaluate \( \displaystyle\iint_R xy\,dA \), where \( R \) is the region between \( y = x^2 \) and \( y = x \).

Step 1: Find the intersections. \( x^2 = x \Rightarrow x = 0 \) or \( x = 1 \).

Step 2: Set up the limits. For \( 0 \le x \le 1 \), \( x \ge x^2 \), so \( y \) runs from \( x^2 \) (bottom) to \( x \) (top):

$$ \int_0^1 \int_{x^2}^{x} xy\,dy\,dx $$

Step 3: Inner integral.

$$ \int_{x^2}^{x} xy\,dy = x\left[\frac{y^2}{2}\right]_{x^2}^{x} = \frac{x}{2}\left(x^2 - x^4\right) = \frac{x^3 - x^5}{2} $$

Step 4: Outer integral.

$$ \int_0^1 \frac{x^3 - x^5}{2}\,dx = \frac{1}{2}\left[\frac{x^4}{4} - \frac{x^6}{6}\right]_0^1 = \frac{1}{2}\left(\frac{1}{4} - \frac{1}{6}\right) = \frac{1}{2}\cdot\frac{1}{12} = \frac{1}{24} $$

Double Integrals in Polar Coordinates

For circles and sectors, switch to polar coordinates with \( x = r\cos\theta \), \( y = r\sin\theta \) and:

$$ dA = r\,dr\,d\theta $$

Do not forget the extra \( r \). For example, the area of a circle of radius \( a \) is:

$$ \int_0^{2\pi} \int_0^a r\,dr\,d\theta = \int_0^{2\pi} \frac{a^2}{2}\,d\theta = 2\pi \cdot \frac{a^2}{2} = \pi a^2 $$

Triple Integrals

A triple integral extends the idea to a solid region \( E \) in 3D:

$$ \iiint_E f(x, y, z)\,dV $$

With \( f = 1 \), it gives the volume of \( E \). If \( f \) is a density, it gives the mass. You evaluate it as three nested single integrals, working from the inside out.

Worked Example 3: A Rectangular Box

Evaluate \( \displaystyle\int_0^1 \int_0^2 \int_0^3 xyz\,dz\,dy\,dx \).

Step 1: Innermost (with respect to \( z \)):

$$ \int_0^3 xyz\,dz = xy\left[\frac{z^2}{2}\right]_0^3 = \frac{9}{2}xy $$

Step 2: Middle (with respect to \( y \)):

$$ \int_0^2 \frac{9}{2}xy\,dy = \frac{9}{2}x\left[\frac{y^2}{2}\right]_0^2 = \frac{9}{2}x \cdot 2 = 9x $$

Step 3: Outer (with respect to \( x \)):

$$ \int_0^1 9x\,dx = \left[\frac{9x^2}{2}\right]_0^1 = \frac{9}{2} $$

Worked Example 4: Volume of a Tetrahedron

Find the volume of the solid in the first octant under the plane \( x + y + z = 1 \).

Step 1: Set up the limits. \( z \) runs from 0 to \( 1 - x - y \). The shadow on the \( xy \)-plane is the triangle with \( 0 \le y \le 1 - x \) and \( 0 \le x \le 1 \):

$$ V = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} 1\,dz\,dy\,dx $$

Step 2: Innermost. \( \int_0^{1-x-y} dz = 1 - x - y \).

Step 3: Middle.

$$ \int_0^{1-x} (1 - x - y)\,dy = \left[(1 - x)y - \frac{y^2}{2}\right]_0^{1-x} = (1 - x)^2 - \frac{(1 - x)^2}{2} = \frac{(1 - x)^2}{2} $$

Step 4: Outer.

$$ V = \int_0^1 \frac{(1 - x)^2}{2}\,dx = \frac{1}{2}\left[-\frac{(1 - x)^3}{3}\right]_0^1 = \frac{1}{2}\cdot\frac{1}{3} = \frac{1}{6} $$

This matches the formula \( \frac{1}{3} \times \text{base area} \times \text{height} = \frac{1}{3} \times \frac{1}{2} \times 1 = \frac{1}{6} \).

Tips for Setting Up Limits

• Sketch the region before writing any limits.
• Outer limits are always constants; inner limits may depend on outer variables.
• Draw a test line through the region in the direction of the inner variable to see where it enters and leaves.
• Switch the order if the inner integral looks impossible; a different order can make it easy.
• Use polar, cylindrical or spherical coordinates for circular or spherical shapes, remembering the extra factors \( r \) or \( \rho^2\sin\phi \).

Key Takeaways

• Double and triple integrals are evaluated as nested single integrals, from the inside out.
• With integrand 1, a double integral gives area and a triple integral gives volume.
• For non-rectangular regions, inner limits are functions of the outer variables, and outer limits are constants.
• In polar coordinates, always use \( dA = r\,dr\,d\theta \).

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