Double and Triple Integrals: Integrating Over Regions in Space
Published:
By: Rajib Belbase
For students in Calculus III. This post covers the essential ideas behind double and triple integrals — what they mean geometrically, how to set them up, and how to compute them. All material follows Chapter 15 of Calculus, 9th edition, by James Stewart.
1. From Single to Double Integrals
You already know how to integrate a function of one variable over an interval \([a, b]\):
\[\int_a^b f(x)\, dx\]Geometrically, this computes the signed area under the curve \(y = f(x)\) above the \(x\)-axis.
Now suppose \(f(x, y)\) is a function of two variables defined over a region \(D\) in the \(xy\)-plane. The double integral of \(f\) over \(D\) is:
\[\iint_D f(x, y)\, dA\]Geometrically, when \(f(x, y) \geq 0\), this gives the volume of the solid that lies above \(D\) and below the surface \(z = f(x, y)\).
1.1 The Big Idea: Slicing and Summing
The double integral is built the same way as the single integral — by dividing, approximating, and taking a limit.
Divide the region \(D\) into small rectangles of area \(\Delta A = \Delta x \cdot \Delta y\). Over each small rectangle, approximate the solid by a thin rectangular box of height \(f(x_{ij}^*, y_{ij}^*)\). The volume of that box is \(f(x_{ij}^*, y_{ij}^*)\,\Delta A\). Summing over all rectangles and taking the limit as they shrink:
\[\iint_D f(x, y)\, dA = \lim_{m,n \to \infty} \sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*)\,\Delta A\]This is the Riemann sum definition of the double integral [Stewart §15.1].
2. Iterated Integrals and Fubini’s Theorem
In practice, we never compute the limit above directly. Instead, we use Fubini’s Theorem, which converts a double integral into two successive single integrals — called an iterated integral.
2.1 Fubini’s Theorem
Theorem (Fubini [Stewart §15.2]). If \(f\) is continuous on the rectangle \(R = [a,b] \times [c,d]\), then:
\[\iint_R f(x,y)\, dA = \int_a^b \int_c^d f(x,y)\, dy\, dx = \int_c^d \int_a^b f(x,y)\, dx\, dy\]Both orders of integration give the same answer. The key idea: integrate with respect to one variable while treating the other as a constant, exactly like partial differentiation in reverse.
2.2 Example: Double Integral Over a Rectangle
Compute \(\displaystyle\iint_R (x + 2y)\, dA\) where \(R = [0,3] \times [1,2]\).
\[\int_0^3 \int_1^2 (x + 2y)\, dy\, dx\]Inner integral (fix \(x\), integrate over \(y\)):
\[\int_1^2 (x + 2y)\, dy = \left[xy + y^2\right]_1^2 = (2x + 4) - (x + 1) = x + 3\]Outer integral:
\[\int_0^3 (x + 3)\, dx = \left[\frac{x^2}{2} + 3x\right]_0^3 = \frac{9}{2} + 9 = \frac{27}{2}\]3. Double Integrals Over General Regions
Rectangular regions are easy, but most interesting problems involve non-rectangular domains. Stewart §15.3 classifies these into two types.
3.1 Type I Regions
A region \(D\) is Type I if it lies between two functions of \(x\):
\[D = \{(x, y) \mid a \leq x \leq b,\; g_1(x) \leq y \leq g_2(x)\}\]The double integral becomes:
\[\iint_D f(x,y)\, dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\, dx\]The limits on the inner integral depend on \(x\) — this is what makes it different from the rectangular case.
3.2 Type II Regions
A region \(D\) is Type II if it lies between two functions of \(y\):
\[D = \{(x, y) \mid c \leq y \leq d,\; h_1(y) \leq x \leq h_2(y)\}\] \[\iint_D f(x,y)\, dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x,y)\, dx\, dy\]3.3 Example: Integral Over a Triangular Region
Compute \(\displaystyle\iint_D xy\, dA\) where \(D\) is the triangle with vertices \((0,0)\), \((1,0)\), \((1,1)\).
The region is bounded by \(x = 1\) on the right, \(y = 0\) below, and \(y = x\) above — a Type I region with \(0 \leq x \leq 1\) and \(0 \leq y \leq x\).
\[\int_0^1 \int_0^x xy\, dy\, dx\]Inner integral:
\[\int_0^x xy\, dy = x \cdot \left[\frac{y^2}{2}\right]_0^x = \frac{x^3}{2}\]Outer integral:
\[\int_0^1 \frac{x^3}{2}\, dx = \frac{1}{2} \cdot \frac{x^4}{4}\Bigg|_0^1 = \frac{1}{8}\]3.4 Choosing the Right Order
Sometimes one order of integration leads to an integral that is impossible (or very hard) to evaluate in closed form, while the other is straightforward. A classic example is:
\[\int_0^1 \int_x^1 \sin(y^2)\, dy\, dx\]Integrating \(\sin(y^2)\) with respect to \(y\) first has no elementary antiderivative. But switching to Type II (integrating \(x\) first, with \(0 \leq x \leq y\)):
\[\int_0^1 \int_0^y \sin(y^2)\, dx\, dy = \int_0^1 y\sin(y^2)\, dy = \left[-\frac{\cos(y^2)}{2}\right]_0^1 = \frac{1 - \cos 1}{2}\]Moral: always sketch the region and consider both orders before committing.
4. Double Integrals in Polar Coordinates
When the region \(D\) is a disk, ring, or sector, polar coordinates \(x = r\cos\theta\), \(y = r\sin\theta\) often simplify the computation dramatically [Stewart §15.4].
The key substitution for the area element is:
\[dA = r\, dr\, d\theta\]The extra factor of \(r\) — the Jacobian — is essential and must not be omitted.
\[\iint_D f(x,y)\, dA = \int_\alpha^\beta \int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta, r\sin\theta)\cdot r\, dr\, d\theta\]4.1 Example: Area of a Disk
Compute \(\displaystyle\iint_D 1\, dA\) where \(D\) is the disk \(x^2 + y^2 \leq a^2\).
In polar: \(0 \leq r \leq a\), \(0 \leq \theta \leq 2\pi\).
\[\int_0^{2\pi}\int_0^a r\, dr\, d\theta = \int_0^{2\pi} \frac{a^2}{2}\, d\theta = \pi a^2\]Reassuringly, the double integral recovers the familiar formula for the area of a circle.
4.2 Example: Gaussian Integral
The famous integral \(\int_{-\infty}^\infty e^{-x^2}\, dx = \sqrt{\pi}\) is proved using polar coordinates. Let \(I = \int_{-\infty}^\infty e^{-x^2}\, dx\). Then:
\[I^2 = \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)}\, dx\, dy = \int_0^{2\pi}\int_0^\infty e^{-r^2} r\, dr\, d\theta = 2\pi \cdot \frac{1}{2} = \pi\]So \(I = \sqrt{\pi}\) — a result that single-variable calculus alone cannot prove.
5. Applications of Double Integrals
Double integrals compute more than volumes [Stewart §15.5].
| Quantity | Formula |
|---|---|
| Area of \(D\) | \(A(D) = \iint_D 1\, dA\) |
| Volume below \(z = f(x,y)\) | \(V = \iint_D f(x,y)\, dA\) |
| Mass of a lamina with density \(\rho(x,y)\) | \(m = \iint_D \rho(x,y)\, dA\) |
| Center of mass | \(\bar{x} = \frac{1}{m}\iint_D x\,\rho\, dA,\quad \bar{y} = \frac{1}{m}\iint_D y\,\rho\, dA\) |
| Moment of inertia about \(x\)-axis | \(I_x = \iint_D y^2\,\rho(x,y)\, dA\) |
| Average value of \(f\) over \(D\) | \(f_{\text{avg}} = \dfrac{1}{A(D)}\iint_D f(x,y)\, dA\) |
6. Triple Integrals
The step from double to triple integrals is conceptually identical — we integrate over a 3D solid region \(E\) instead of a 2D region.
\[\iiint_E f(x, y, z)\, dV\]When \(f(x,y,z) = 1\), the triple integral gives the volume of \(E\). More generally it computes mass, charge, or any quantity distributed throughout a solid [Stewart §15.7].
6.1 Fubini’s Theorem for Triple Integrals
If \(f\) is continuous on the box \(B = [a,b]\times[c,d]\times[r,s]\):
\[\iiint_B f(x,y,z)\, dV = \int_r^s\int_c^d\int_a^b f(x,y,z)\, dx\, dy\, dz\]There are \(3! = 6\) possible orders of integration, and any valid order gives the same result.
6.2 General Solid Regions
For a Type 1 solid (the most common setup), \(E\) lies between two surfaces above a 2D region \(D\):
\[E = \{(x,y,z) \mid (x,y) \in D,\; u_1(x,y) \leq z \leq u_2(x,y)\}\] \[\iiint_E f\, dV = \iint_D \left[\int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z)\, dz\right] dA\]The strategy: integrate in \(z\) first (between the two bounding surfaces), then handle the resulting double integral over \(D\) by Type I or Type II methods.
6.3 Example: Volume of a Tetrahedron
Find the volume of the tetrahedron bounded by the planes \(x = 0\), \(y = 0\), \(z = 0\), and \(x + y + z = 1\).
The solid: \(0 \leq x \leq 1\), \(0 \leq y \leq 1-x\), \(0 \leq z \leq 1-x-y\).
\[V = \int_0^1\int_0^{1-x}\int_0^{1-x-y} dz\, dy\, dx\]Innermost integral:
\[\int_0^{1-x-y} dz = 1 - x - y\]Middle integral:
\[\int_0^{1-x}(1-x-y)\, dy = \left[(1-x)y - \frac{y^2}{2}\right]_0^{1-x} = \frac{(1-x)^2}{2}\]Outer integral:
\[\int_0^1 \frac{(1-x)^2}{2}\, dx = \frac{1}{2}\cdot\frac{(1-x)^3}{-3}\Bigg|_0^1 \cdot(-1)= \frac{1}{6}\]The volume of a tetrahedron with unit side is \(\boxed{\dfrac{1}{6}}\).
6.4 Example: Triple Integral of a Function
Compute \(\displaystyle\iiint_E z\, dV\) where \(E = \{(x,y,z) \mid 0 \leq x \leq 1,\; 0 \leq y \leq x,\; 0 \leq z \leq x+y\}\).
Inner (over \(z\)):
\[\int_0^{x+y} z\, dz = \frac{(x+y)^2}{2}\]Middle (over \(y\), \(0 \leq y \leq x\)):
\[\int_0^x \frac{(x+y)^2}{2}\, dy = \frac{1}{2}\cdot\frac{(x+y)^3}{3}\Bigg|_0^x = \frac{(2x)^3 - x^3}{6} = \frac{7x^3}{6}\]Outer (over \(x\)):
\[\int_0^1 \frac{7x^3}{6}\, dx = \frac{7}{6}\cdot\frac{1}{4} = \frac{7}{24}\]7. Triple Integrals in Cylindrical Coordinates
When the solid \(E\) has circular symmetry about the \(z\)-axis, cylindrical coordinates are the natural choice [Stewart §15.8]:
\[x = r\cos\theta, \quad y = r\sin\theta, \quad z = z\]The volume element becomes:
\[dV = r\, dr\, d\theta\, dz\]Again, the Jacobian factor \(r\) is mandatory.
7.1 Example: Volume Inside a Cylinder Below a Paraboloid
Find the volume of the solid bounded above by \(z = 9 - r^2\) and below by \(z = 0\), inside the cylinder \(r = 2\).
\[V = \int_0^{2\pi}\int_0^2\int_0^{9-r^2} r\, dz\, dr\, d\theta\]Inner:
\[\int_0^{9-r^2} dz = 9 - r^2\]Middle:
\[\int_0^2 r(9-r^2)\, dr = \int_0^2 (9r - r^3)\, dr = \left[\frac{9r^2}{2} - \frac{r^4}{4}\right]_0^2 = 18 - 4 = 14\]Outer:
\[\int_0^{2\pi} 14\, d\theta = 28\pi\]8. Triple Integrals in Spherical Coordinates
When the solid is a sphere, cone, or spherical shell, spherical coordinates are ideal [Stewart §15.9]:
\[x = \rho\sin\phi\cos\theta, \quad y = \rho\sin\phi\sin\theta, \quad z = \rho\cos\phi\]where \(\rho \geq 0\) is the distance from the origin, \(0 \leq \phi \leq \pi\) is the polar angle from the \(z\)-axis, and \(0 \leq \theta \leq 2\pi\) is the azimuthal angle. Note that \(\rho^2 = x^2 + y^2 + z^2\).
The volume element is:
\[dV = \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta\]8.1 Example: Volume of a Sphere
Find the volume of the ball \(\rho \leq a\).
\[V = \int_0^{2\pi}\int_0^{\pi}\int_0^a \rho^2\sin\phi\, d\rho\, d\phi\, d\theta\] \[= \int_0^{2\pi}d\theta \cdot \int_0^{\pi}\sin\phi\, d\phi \cdot \int_0^a \rho^2\, d\rho = 2\pi \cdot 2 \cdot \frac{a^3}{3} = \frac{4}{3}\pi a^3\]8.2 Example: Integral Over a Ball
Compute \(\displaystyle\iiint_B e^{(x^2+y^2+z^2)^{3/2}}\, dV\) where \(B\) is the unit ball \(x^2+y^2+z^2 \leq 1\).
In spherical coordinates, \(x^2+y^2+z^2 = \rho^2\), so \((x^2+y^2+z^2)^{3/2} = \rho^3\).
\[\int_0^{2\pi}\int_0^{\pi}\int_0^1 e^{\rho^3}\rho^2\sin\phi\, d\rho\, d\phi\, d\theta = 2\pi \cdot 2 \cdot \int_0^1 \rho^2 e^{\rho^3}\, d\rho\]Let \(u = \rho^3\), \(du = 3\rho^2\, d\rho\):
\[4\pi \cdot \frac{1}{3}\int_0^1 e^u\, du = \frac{4\pi}{3}(e - 1)\]9. A Quick Guide: Which Coordinates to Use
| Region shape | Best coordinates |
|---|---|
| Rectangle or general polygon | Cartesian \((x, y)\) or \((x, y, z)\) |
| Disk, ring, sector | Polar \((r, \theta)\) |
| Cylinder, cone, solid with circular cross-sections | Cylindrical \((r, \theta, z)\) |
| Sphere, spherical shell, cone from origin | Spherical \((\rho, \phi, \theta)\) |
The rule of thumb: match the coordinate system to the symmetry of the region. A sphere described in Cartesian coordinates requires painful algebra; in spherical coordinates it is two lines.
10. Common Mistakes to Avoid
1. Forgetting the Jacobian. The area element in polar is \(r\, dr\, d\theta\), not \(dr\, d\theta\). The volume element in cylindrical is \(r\, dr\, d\theta\, dz\), and in spherical it is \(\rho^2\sin\phi\, d\rho\, d\phi\, d\theta\). Omitting these factors is the single most common error in this chapter.
2. Wrong limits on the inner integral. The inner limits generally depend on the outer variables. For a Type I region, the \(y\)-limits are functions of \(x\). Always draw and label the region before writing the integral.
3. Integrating in an impossible order. If the inner integral has no closed form, try switching the order. Always check whether the integrand suggests a preferred order.
4. Confusing \(\phi\) and \(\theta\) in spherical. Stewart uses \(\phi\) for the polar angle (from the \(z\)-axis) and \(\theta\) for the azimuthal angle. Some textbooks reverse this convention. Be consistent.
5. Setting up limits for the wrong region. Sketch the region in 3D (or at least its projection onto the \(xy\)-plane). A solid bounded above by a paraboloid and below by a plane requires \(z\) to run from the plane up to the paraboloid — not the other way around.
11. Summary
| Concept | Key Formula |
|---|---|
| Double integral (rectangle) | \(\iint_R f\, dA = \int_a^b\int_c^d f\, dy\, dx\) |
| Double integral (Type I) | \(\int_a^b\int_{g_1(x)}^{g_2(x)} f\, dy\, dx\) |
| Polar area element | \(dA = r\, dr\, d\theta\) |
| Triple integral (general solid) | \(\iint_D\int_{u_1}^{u_2} f\, dz\, dA\) |
| Cylindrical volume element | \(dV = r\, dr\, d\theta\, dz\) |
| Spherical volume element | \(dV = \rho^2\sin\phi\, d\rho\, d\phi\, d\theta\) |
The single most important skill in this chapter is setting up the limits correctly. The integration itself is usually routine single-variable calculus — it is the geometry of the region that requires care. Draw every region. Label every boundary. Then write the integral.
Good luck — and remember that every triple integral in spherical coordinates is just three single integrals waiting to be separated.
References
All section numbers refer to:
Stewart, J. (2021). Calculus, 9th edition. Cengage Learning.
- §15.1 — Double Integrals over Rectangles
- §15.2 — Double Integrals over General Regions
- §15.3 — Double Integrals in Polar Coordinates
- §15.4 — Applications of Double Integrals
- §15.7 — Triple Integrals
- §15.8 — Triple Integrals in Cylindrical Coordinates
- §15.9 — Triple Integrals in Spherical Coordinates
