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Computing Volume with Triple Integrals

Represent volume as a triple integral, choose Cartesian bounds that describe the solid, and evaluate and check the resulting integrals.

Intermediate 9 min read

What You'll Learn

  • Express the volume of a bounded solid as the triple integral of 1 over that solid
  • Connect Cartesian integration bounds to the solid’s projection and vertical extent
  • Evaluate volume integrals over a rectangular box and a tetrahedron
  • Set up and compute the volume of a solid with a triangular footprint and sloping top
  • Check bounds for geometric validity and verify computed volumes

Volume as a Triple Integral

A triple integral can add up a quantity throughout a three-dimensional region. When that quantity is the constant \(1\), each small piece contributes its volume alone. The integral therefore gives the volume of the solid.

The main work in many volume problems is translating the solid’s geometry into bounds that describe every point in it exactly once. With suitable Cartesian bounds, the volume becomes an iterated integral of \(1\).

Definition: If \(E\) is a bounded solid with a sufficiently regular boundary, its volume is \(V(E)=\iiint_E 1\,dV\). In Cartesian coordinates, \(dV=dx\,dy\,dz\); an iterated integral uses the differentials in the order of integration.

Here, “sufficiently regular” includes the standard solids bounded by piecewise smooth surfaces considered in this course. The condition ensures that the boundary does not interfere with the usual definition and evaluation of the triple integral.

Why Integrating 1 Gives Volume

Imagine dividing a solid into many small boxes. For a box lying inside the solid, the value of the function \(1\) is \(1\), so its contribution to a Riemann sum is approximately \(1\) times the box’s volume. Adding those contributions approximates the solid’s volume. Refining the division gives the triple integral.

1
Partition the solid.
Divide the region into small pieces whose volumes are \(\Delta V_i\).
2
Evaluate the integrand.
For the volume integrand, every sample value is \(1\), so each term in the sum is \(1\cdot\Delta V_i=\Delta V_i\).
3
Pass to the limit.
The sums of the piece volumes approach the volume of \(E\), and their limit is \(\iiint_E 1\,dV\).

This also explains why the bounds matter. They specify which small pieces belong to \(E\). If the bounds include points outside the solid or exclude points inside it, the integral measures a different region.

Cartesian Bounds and Volume

For a \(z\)-simple solid, the projection onto the \(xy\)-plane is a region \(D\), and each vertical line above a point in \(D\) runs from a lower surface \(z=g_1(x,y)\) to an upper surface \(z=g_2(x,y)\). The bounds describe the solid as

$$ E=\{(x,y,z):(x,y)\in D,\ g_1(x,y)\le z\le g_2(x,y)\}. $$

When \(D\) can be described by \(a\le x\le b\) and \(h_1(x)\le y\le h_2(x)\), the corresponding volume integral is

$$ V(E)=\int_a^b\int_{h_1(x)}^{h_2(x)} \int_{g_1(x,y)}^{g_2(x,y)}1\,dz\,dy\,dx. $$

Integrating the constant \(1\) first with respect to \(z\) gives \(g_2(x,y)-g_1(x,y)\), the vertical height of the solid above \((x,y)\). The remaining integrals add these heights over the projection. This is an interpretation of the same triple integral, not a different definition of volume.

The order can be changed when the region is described with valid bounds in that order. The geometric requirement remains the same: each point of the solid must be included once, and no point outside it may be included.

Worked Example: Volume of a Rectangular Box

Worked Example: Volume of a Rectangular Box

Find the volume of the box \(E\) given by \(0\le x\le 2\), \(0\le y\le 3\), and \(0\le z\le 4\).

The bounds independently describe the three side lengths. Integrate \(1\) over these bounds:

$$ \begin{aligned} V(E) &=\int_0^2\int_0^3\int_0^4 1\,dz\,dy\,dx\\ &=\int_0^2\int_0^3 [z]_0^4\,dy\,dx\\ &=\int_0^2\int_0^3 4\,dy\,dx\\ &=\int_0^2 [4y]_0^3\,dx\\ &=\int_0^2 12\,dx\\ &=[12x]_0^2\\ &=24. \end{aligned} $$

Check: the side lengths are \(2\), \(3\), and \(4\), so their product is \(2\cdot3\cdot4=24\), agreeing with the integral. If the coordinates measure lengths in units, the volume is \(24\) cubic units.

Worked Example: A Tetrahedron in the First Octant

Worked Example: A Tetrahedron in the First Octant

Find the volume of the solid \(E\) bounded by the coordinate planes and the plane \(x+y+z=2\).

The coordinate planes require \(x,y,z\ge0\). For fixed \(x\) and \(y\), the plane gives \(0\le z\le 2-x-y\). The condition that this interval is nonempty is \(x+y\le2\). Thus \(0\le x\le2\), and for each \(x\), \(0\le y\le2-x\). These bounds describe every point of the tetrahedron:

$$ \begin{aligned} V(E) &=\int_0^2\int_0^{2-x}\int_0^{2-x-y}1\,dz\,dy\,dx\\ &=\int_0^2\int_0^{2-x}(2-x-y)\,dy\,dx\\ &=\int_0^2\left[(2-x)y-\frac{y^2}{2}\right]_0^{2-x}\,dx\\ &=\int_0^2\left((2-x)^2-\frac{(2-x)^2}{2}\right)\,dx\\ &=\frac12\int_0^2(2-x)^2\,dx\\ &=\frac12\left[-\frac{(2-x)^3}{3}\right]_0^2\\ &=\frac12\left(0-\left(-\frac83\right)\right)\\ &=\frac43. \end{aligned} $$

Check the bounds at the endpoints: when \(x=0\), \(y\) ranges from \(0\) to \(2\); when \(x=2\), the only allowed value is \(y=0\), and then \(z=0\). The plane and coordinate planes meet at the intercepts \((2,0,0)\), \((0,2,0)\), and \((0,0,2)\), as the bounds encode. The volume is \(4/3\) cubic units.

Worked Example: A Sloping Roof over a Triangular Footprint

Worked Example: A Sloping Roof over a Triangular Footprint

Find the volume of the solid above \(z=0\), below \(z=2+x+y\), and over the triangular region \(0\le x\le1\), \(0\le y\le1-x\).

The footprint condition \(y\le1-x\) is equivalent to \(x+y\le1\). Over this triangle, the roof height \(2+x+y\) is positive, so the lower surface is \(z=0\) and the upper surface is \(z=2+x+y\). The Cartesian bounds give

$$ \begin{aligned} V(E) &=\int_0^1\int_0^{1-x}\int_0^{2+x+y}1\,dz\,dy\,dx\\ &=\int_0^1\int_0^{1-x}(2+x+y)\,dy\,dx\\ &=\int_0^1\left[(2+x)y+\frac{y^2}{2}\right]_0^{1-x}\,dx\\ &=\int_0^1\left((2+x)(1-x)+\frac{(1-x)^2}{2}\right)\,dx\\ &=\int_0^1\left(2-2x+x-x^2+\frac12-x+\frac{x^2}{2}\right)\,dx\\ &=\int_0^1\left(\frac52-2x-\frac{x^2}{2}\right)\,dx\\ &=\left[\frac52x-x^2-\frac{x^3}{6}\right]_0^1\\ &=\frac52-1-\frac16\\ &=\frac43. \end{aligned} $$

Check by reversing the order in the triangular footprint. Its bounds become \(0\le y\le1\) and \(0\le x\le1-y\). Then

$$ \begin{aligned} V(E) &=\int_0^1\int_0^{1-y}(2+x+y)\,dx\,dy\\ &=\int_0^1\left[(2+y)x+\frac{x^2}{2}\right]_0^{1-y}\,dy\\ &=\int_0^1\left((2+y)(1-y)+\frac{(1-y)^2}{2}\right)\,dy\\ &=\int_0^1\left(\frac52-2y-\frac{y^2}{2}\right)\,dy\\ &=\left[\frac52y-y^2-\frac{y^3}{6}\right]_0^1\\ &=\frac43. \end{aligned} $$

The second evaluation agrees with the first. The volume is \(4/3\) cubic units.

A General Justification for the Bounds

For the \(z\)-simple description above, fix a point \((x,y)\) in the projection \(D\). The allowed \(z\)-values form the interval from \(g_1(x,y)\) to \(g_2(x,y)\). Therefore, integrating \(1\) over that interval gives

$$ \int_{g_1(x,y)}^{g_2(x,y)}1\,dz =g_2(x,y)-g_1(x,y). $$

This difference is exactly the vertical length of the solid above that point. The remaining double integral adds these vertical lengths over \(D\), with each small piece of the projection contributing its height times its area. Those products are volume contributions. Thus the iterated integral agrees with the volume represented by the triple integral. The argument requires valid bounds for the solid and the usual integrability conditions; it does not justify bounds that fail to describe the region.

Theorem: If \(E\) is described by \(a\le x\le b\), \(h_1(x)\le y\le h_2(x)\), and \(g_1(x,y)\le z\le g_2(x,y)\), with integrable boundary functions and \(g_1\le g_2\), then \(V(E)=\int_a^b\int_{h_1(x)}^{h_2(x)}\int_{g_1(x,y)}^{g_2(x,y)}1\,dz\,dy\,dx\).

A Common Bounds Error

A frequent error is to treat dependent bounds as independent. For the tetrahedron \(x,y,z\ge0\), \(x+y+z\le2\), one might incorrectly write

$$ \int_0^2\int_0^2\int_0^{2-x-y}1\,dz\,dy\,dx. $$

These bounds are invalid: at \(x=1.5\) and \(y=1\), the proposed upper limit is \(2-1.5-1=-0.5\), below the lower limit \(z=0\). Worse, the integral includes negative contributions where \(x+y>2\), even though there are no points of the tetrahedron there.

The correction is to make the range of \(y\) depend on \(x\), so that \(x+y\le2\) throughout the projection:

$$ 0\le x\le2,\qquad 0\le y\le2-x,\qquad 0\le z\le2-x-y. $$

Before evaluating, check that each upper bound is at least its lower bound throughout the preceding ranges. Also check that the bounds trace the intended projection and vertical extent.

Why Volume Integrals Are Useful

Writing volume as \(\iiint_E1\,dV\) gives one consistent method for solids whose boundaries may be slanted or curved, rather than only boxes with familiar side lengths. The integrand remains simple; the geometry is carried by the bounds. This makes volume calculations a direct test of whether a solid has been translated correctly into Cartesian inequalities.

Units provide a useful final check. If \(x\), \(y\), and \(z\) measure lengths, then \(dV\) has units of cubic length, so the answer must be a volume. A negative result or a result with units of length usually signals an error in the bounds or in the integration.

Check Your Understanding

Use the geometry and the meaning of the volume integral to answer these questions.

  1. What quantity does \(\iiint_E1\,dV\) represent, and why does the integrand equal \(1\)?
  2. For a \(z\)-simple solid, what geometric quantity results from integrating \(1\) with respect to \(z\) first?
  3. Write Cartesian bounds for the solid \(x,y,z\ge0\), \(x+y+z\le1\), using the order \(dz\,dy\,dx\).
  4. Why are independent bounds \(0\le x\le2\), \(0\le y\le2\) invalid for the projection of the tetrahedron \(x+y+z\le2\) in the first octant?
  5. What units should a volume integral have if each coordinate measures length?

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