In this beginner’s tutorial for double and triple integrals in calculus, we’ll walk you through how to evaluate these types of integrals over rectangular regions and boxes. You’ll learn what double and triple integrals are, how to set them up, and how to evaluate them by integrating one variable at a time. We’ll also work through examples so you can see these concepts used in practice.
Integrating multivariate functions with respect to a single variable is similar to computing partial derivatives, so if you haven’t already, please refer to The Ultimate Beginner’s Guide to Partial Derivatives with Step-by-Step Examples. To understand double and triple integrals, you must be able to calculate definite integrals. For a review of this topic, please check out the article Basic Integration Problems for Beginners.
Double Integrals Over Rectangular Regions
A double integral calculates area or volume over a two-dimensional region.
Double integrals are written as
$$\iint_R f(x, y)\, dA,$$
where \( R \) is a region in the \( xy \)-plane, and \( f(x, y) \) is the function we’re integrating.
Rectangular Regions
In this tutorial, we’ll focus only on rectangular regions, which are defined by constant bounds in the form
$$R = [a, b] \times [c, d].$$
Which means \( x \in [a, b] \), and \( y \in [c, d] \). Rectangles are the simplest to work with because only numbers appear in the limits of integration.
Examples
Let’s look at some examples.
Example 1: Evaluate \( \int_1^3 \int_0^2 (x + 2y) \, dx dy \).
Solution: The inner integral is with respect to x, so we treat \( y \) as a constant and integrate with respect to \( x \) to get
$$\int_1^3 \frac{1}{2}x^2 + 2yx|_0^2 dy = \int_1^3 2 + 4y dy.$$
Now we evaluate the remaining integral to obtain
$$\left[ 2y + 2y^2 \right]_1^3 = 20.$$
We now do a more interesting example. This example uses substitution, so if you haven’t already, please check out The Ultimate Step-by-Step Guide to Solving Integrals Using Substitution.
Example 2: Evaluate \( \int_0^1 \int_0^1 x e^{xy} \, dy \, dx \).
Solution: The Inner Integral is with respect to y, so we treat \( x \) as a constant and integrate with respect to \( y \). We begin with a substitution. Let \( u = xy \), then \( du = x\,dy \). The new limits are \( 0x = 0 \) and \( 1x = x \). Substituting into the integral gives
$$\int_0^1 \int_0^x e^u \, du \, dx = \int_0^1 [e^u]_0^x \, dx = \int_0^1 e^x – 1 dx.$$
Now we evaluate the remaining integral to obtain
$$[e^x – x]_0^1 = e – 2.$$
Triple Integrals Over Boxes
A triple integral calculates volume or mass over a three-dimensional region, just like double integrals sum over an area, triple integrals sum over a volume.
Triple integrals are written as
$$\iiint_E f(x, y, z)\, dV,$$
where \( E \) is a 3D region, and \( f(x, y, z) \) is the function we’re integrating.
Boxes
In this section, we’ll focus on boxes defined by constant bounds, which have the form
$$E = [a, b] \times [c, d] \times [e, f],$$
where \( x \in [a, b] \), \( y \in [c, d] \), and \( z \in [e, f] \).
Examples
Example 3: Evaluate \( \int_0^1 \int_0^1 \int_0^1 xe^{y+z} \, dx\,dy\,dz \).
Solution: The inner integral is with respect to x, so we treat \( y \) and \( z \) as constants and integrate with respect to \( x \) to get
$$\int_0^1 \int_0^1 \frac{1}{2}x^2e^{y+z}|_0^1 \,dy\,dz = \int_0^1 \int_0^1 \frac{1}{2}e^{y+z} \,dy\,dz.$$
We now integrate with respect to y while treating z as a constant. This gives
$$\int_0^1 \frac{1}{2}e^{y+z}|_0^1 \,dz = \int_0^1 \frac{1}{2}e^{z + 1} – \frac{1}{2}e^z \,dz.$$
Now we evaluate the remaining integral to obtain
$$\frac{1}{2}e^{z + 1} – \frac{1}{2}e^z|_0^1 = \frac{1}{2}e^2 – e + \frac{1}{2}.$$
Example 4: Evaluate \( \int_0^2 \int_0^{\pi/2} \int_0^1 xy\cos(z) \, dx\,dz\,dy \).
Solution: The inner integral is with respect to x, so we treat \( y \) and \( z \) as constants and integrate with respect to \( x \) to get
$$\int_0^2 \int_0^{\pi/2} \frac{1}{2}x^2y\cos(z)|_0^1 \,dz\,dy = \int_0^2 \int_0^{\pi/2} \frac{1}{2}y\cos(z) \,dz\,dy.$$
We now integrate with respect to z while treating y as a constant. This gives
$$\int_0^2 \frac{1}{2}y\sin(z)_0^{\frac{\pi}{2}} dy = \int_0^2 \frac{1}{2}y dy.$$
Now we evaluate the remaining integral to obtain
$$\frac{1}{4}y^2|_0^2 = 1.$$
Example 5: Evaluate \( \int_0^1 \int_0^{\pi} \int_0^1 y\cos(x)\, dy\,dx\,dz \).
Solution: The inner integral is with respect to y, so we treat \( x \) and \( z \) as constants and integrate with respect to \( y \) to get
$$\int_0^1 \int_0^{\pi} \frac{1}{2}y^2\cos(x)|_0^1 \,dx\,dz = \int_0^1 \int_0^{\pi} \frac{1}{2}\cos(x) \,dx\,dz.$$
We now integrate with respect to x while treating z as a constant. This gives
$$\int_0^1 \frac{1}{2}\sin(x)|_0^{\pi} dz = \int_0^1 0 dz = 0.$$
Example 6: Evaluate \( \int_0^1 \int_0^1 \int_0^2 xze^y \, dy\,dz\,dx \).
Solution: The inner integral is with respect to y, so we treat \( x \) and \( z \) as constants and integrate with respect to \( y \) to get
$$\int_0^1 \int_0^1 xze^y|_0^2 \,dz\,dx = \int_0^1 \int_0^1 e^2xz – xz \,dz\,dx.$$
We now integrate with respect to z while treating x as a constant. This gives
$$\int_0^1 \frac{e^2}{2}xz^2 – \frac{1}{2}xz^2|_0^1 \,dx = \int_0^1 \frac{e^2}{2}x – \frac{1}{2}x \,dx.$$
Now we evaluate the remaining integral to obtain
$$\frac{e^2}{4}x^2 – \frac{1}{4}x^2|_0^1 = \frac{e^2}{4} – \frac{1}{4}.$$
Example 7: Evaluate \( \int_0^1 \int_0^\pi \int_0^1 e^z\cos(x) \, dz\,dx\,dy \).
Solution: The inner integral is with respect to z, so we treat \( x \) and \( y \) as constants and integrate with respect to \( z \) to get
$$\int_0^1 \int_0^\pi e^z\cos(x)|_0^1 \,dx\,dy = \int_0^1 \int_0^\pi e\cos(x) – \cos(x) \,dx\,dy.$$
We now integrate with respect to x while treating y as a constant. This gives
$$\int_0^1 e\sin(x) – \sin(x)_0^{\pi} \,dy = \int_0^1 0 \,dy = 0.$$
Example 8: Evaluate \( \int_0^1 \int_0^2 \int_0^3 xz \, dz\,dy\,dx \).
Solution: The inner integral is with respect to z, so we treat \( x \) and \( y \) as constants and integrate with respect to \( z \) to get
$$\int_0^1 \int_0^2 \frac{1}{2}xz^2|_0^3 \,dy\,dx = \int_0^1 \int_0^2 \frac{9}{2}x \,dy\,dx.$$
We now integrate with respect to y while treating x as a constant. This gives
$$\int_0^1 \frac{9}{2}xy|_0^2 \,dx = \int_0^1 9x \,dx.$$
Now we evaluate the remaining integral to obtain
$$\frac{9}{2}x^2|_0^1 = \frac{9}{2}.$$
Conclusion
In this tutorial, we learned how to evaluate double and triple integrals over rectangular regions and boxes, building the foundation for more advanced topics. You learned how to compute double and triple integrals and how to apply integration techniques like basic integration and substitution to solve them.
