Courses

Multivariable Calculus

Double Integrals
1. Cartesian $d x d y$
2. Polar $r d r d θ$
Vectors
Divergence
Curl

Additional Resources

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Double Integrals in Cartesian Coordinates | Chalkboard Video

Double integrals can be used to find areas of regions in the $x y$ -plane and volumes. If a density is defined for a region in the $x y$ -plane, the mass, moments, center of mass, and moment of inertia can be calculated.

Riemann Sum | Video Chapter

Double integrals are still defined as Riemann sums, but now over a bounded area instead of a line. Here's an example.

\int_{0}^{1} \int_{0}^{1} x y d y d x = lim_{m, n \to \infty} \sum_{i = 1}^{n} \sum_{j = 1}^{m} x_{i} y_{j} Δ x Δ y Δ x = \frac{1}{n} Δ y = \frac{1}{m}

For larger values of $m$ and $n$ , the Riemann sum of the volume of many rectangular prisms with base area $Δ x Δ y$ and height specified by the integrand becomes a better approximation to the volume under the surface.

This one can be evaluated fairly easily using the formula for,

\sum_{i = k}^{N} k = 1 + 2 + \dots + N = \frac{N (N + 1)}{2}

Ignoring the limits for a moment,

\begin{aligned} \sum_{i = 1}^{n} \sum_{j = 1}^{m} x_{i} y_{j} Δ x Δ y & = \sum_{i = 1}^{n} \sum_{j = 1}^{m} \frac{i}{n} \frac{j}{m} \frac{1}{n} \frac{1}{m} \\ = \frac{1}{n^{2}} \frac{1}{m^{2}} (\sum_{i = 1}^{n} i) (\sum_{j = 1}^{m} j) & = \frac{1}{n^{2}} \frac{1}{m^{2}} \frac{n (n + 1)}{2} \frac{m (m + 1)}{2} \\ = \frac{n^{2} + n}{2 n^{2}} \frac{m^{2} + m}{2 m^{2}} \\ = (\frac{1}{2} + \frac{1}{2 n}) (\frac{1}{2} + \frac{1}{2 m}) \end{aligned}

Now evaluating the limits,

\begin{aligned} \int_{0}^{1} \int_{0}^{1} x y d y d x & = lim_{m, n \to \infty} \sum_{i = 1}^{n} \sum_{j = 1}^{m} x_{i} y_{j} Δ x Δ y \\ = lim_{m, n \to \infty} (\frac{1}{2} + \frac{1}{2 n}) (\frac{1}{2} + \frac{1}{2 m}) \\ = (\frac{1}{2} + 0) (\frac{1}{2} + 0) = \frac{1}{4} \end{aligned}

Evaluating Double Integrals | Video Chapter

There's nothing particularly special about evaluating a double integral vs. evaluating a single integral. The interior integral is evaluated first, which creates a new integrand. Then the exterior integral is evaluated.

\begin{aligned} \int_{0}^{1} \int_{0}^{1} x y d y d x & = \int_{0}^{1} {[x \frac{1}{2} y^{2}]}_{y = 0}^{y = 1} d x \\ = \int_{0}^{1} \frac{1}{2} x d x \\ = \int_{0}^{1} {[\frac{1}{4} x^{2}]}_{x = 0}^{x = 1} d x \\ = \frac{1}{4} \end{aligned}

The difficulty in double integrals is actually in creating the bounds because the interior integral bounds can now have variables in them.

Example 1 | Solution Video

Create the integral bounds for the $d x d y$ and $d y d x$ integral orders for the triangle with vertices $(0, 0)$ , $(2, 0)$ , and $(2, 3)$ .

Example 2 | Solution Video

Create the integral bounds for the $d x d y$ and $d y d x$ integral orders for the triangle with vertices $(0, 1)$ , $(0.5)$ , and $(4, 3)$ .

Example 3 | Solution Video

Create the integral bounds for the $d x d y$ and $d y d x$ integral orders for the following region.

y = x^{2} y = x 0 \leq y \leq 4