Table of Contents
Multivariable Calculus
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Chalkboard VideoDouble integrals can be used to find areas of regions in the $xy$-plane and volumes. If a density is defined for a region in the $xy$-plane, the mass, moments, center of mass, and moment of inertia can be calculated.
Video ChapterDouble integrals are still defined as Riemann sums, but now over a bounded area instead of a line. Here's an example.
\begin{equation} \int_{0}^{1} \int_{0}^{1} xy \text{ d}y\text{d}x = \lim_{m,n \rightarrow \infty} \sum_{i=1}^{n} \sum_{j=1}^{m} x_{i}y_{j} ~\Delta x \Delta y \quad \Delta x = \frac{1}{n} \quad \Delta y = \frac{1}{m} \end{equation}For larger values of $m$ and $n$, the Riemann sum of the volume of many rectangular prisms with base area $\Delta x \Delta y$ 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,
\begin{equation} \sum_{i=k}^{N} k = 1 + 2 + \cdots + N = \frac{N(N+1)}{2} \end{equation}Ignoring the limits for a moment,
\begin{align} \sum_{i=1}^{n} \sum_{j=1}^{m} x_{i}y_{j} ~\Delta x \Delta 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}} \left(\sum_{i=1}^{n} i\right) \left(\sum_{j=1}^{m} j\right) &= \frac{1}{n^{2}}\frac{1}{m^{2}} \frac{n(n+1)}{2} \frac{m(m+1)}{2} \\ &= \frac{n^{2}+n}{2n^{2}} \frac{m^{2}+m}{2m^{2}} \\ &= \left(\frac{1}{2}+\frac{1}{2n}\right)\left(\frac{1}{2}+\frac{1}{2m}\right) \\ \end{align}Now evaluating the limits,
\begin{align} \int_{0}^{1} \int_{0}^{1} xy \text{ d}y\text{d}x &= \lim_{m,n \rightarrow \infty} \sum_{i=1}^{n} \sum_{j=1}^{m} x_{i}y_{j} ~\Delta x \Delta y \\ &= \lim_{m,n \rightarrow \infty} \left(\frac{1}{2}+\frac{1}{2n}\right)\left(\frac{1}{2}+\frac{1}{2m}\right) \\ &= \left(\frac{1}{2}+0\right)\left(\frac{1}{2}+0\right) = \frac{1}{4} \end{align} Video ChapterThere'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{align} \int_{0}^{1} \int_{0}^{1} xy \text{ d}y\text{d}x &= \int_{0}^{1} \left[ x\frac{1}{2}y^{2} \right]_{y=0}^{y=1} \text{d}x \\ &= \int_{0}^{1} \frac{1}{2}x \text{ d}x \\ &= \int_{0}^{1} \left[ \frac{1}{4}x^{2} \right]_{x=0}^{x=1} \text{d}x \\ &= \frac{1}{4} \end{align}The difficulty in double integrals is actually in creating the bounds because the interior integral bounds can now have variables in them.
Solution VideoCreate the integral bounds for the $\text{d}x\text{d}y$ and $\text{d}y\text{d}x$ integral orders for the triangle with vertices $(0,0)$, $(2,0)$, and $(2,3)$.
Solution VideoCreate the integral bounds for the $\text{d}x\text{d}y$ and $\text{d}y\text{d}x$ integral orders for the triangle with vertices $(0,1)$, $(0.5)$, and $(4,3)$.
Solution VideoCreate the integral bounds for the $\text{d}x\text{d}y$ and $\text{d}y\text{d}x$ integral orders for the following region.
\begin{equation} y = x^{2} \qquad y = x \qquad 0 \leq y \leq 4 \end{equation}