# Lecture Notes

Below are the lecture notes from my course on multivariable calculus, as well as some handouts I give to students for some of the experiments in the introductory physics lab courses. Note that I update my notes each term, but this website may not link to the latest editions.

### Notes on MTH3015: Calculus III

Visualizing 3D Curves with WolframAlpha

Week 3: Vector-valued Functions; Curves

Week 4: Functions of Several Variables; Limits and Continuity

Week 5: Partial Derivatives; The Chain Rule

Week 7: Directional Derivative and Gradient

Week 8: Taylor Series and Extrema

Week 9: Double Integrals; Polar Integrals

**Week 9: Double Integrals; Polar Integrals**

Sujeet Akula

Simple Domains

In evaluated multiple integrals, we will be restricting ourselves to 'simple domains'. These are sets that have specific forms. In 2D, there are two kinds of simple domains.

**Definition** Let $\D$ be a subset of $\Rtwo$. Let $a,b\in\R$. Let $g,h:\R\to\R$ be continuous
functions on $[a,b]$, and $g(x)\le h(x)$ for every $x\in[a,b]$. $\D$ is said to be a simple domain
of the first kind if it has the form
\begin{equation}
\D = \left\{(x,y) : a \le x \le b, g(x) \le y \le h(x)\right\} .
\end{equation}

**Definition** Let $\D$ be a subset of $\Rtwo$. Let $c,d\in\R$. Let $G,H:\R\to\R$ be continuous
functions on $[c,d]$, and $G(y) \le H(y)$ for every $y\in[c,d]$. $\D$ is said to be a simple domain
of the second kind if it has the form
\begin{equation}
\D = \left\{(x,y) : c \le y \le d, G(y) \le x \le H(y)\right\} .
\end{equation}
Note that we can of course have a set that is simultaneously a simple domain of the first and second
kinds, if $g$ and $h$ are constants. A rectangular domain is an example of this. Rectangular domains
may be specified with the notation $R=[a,b]\times[c,d]$. This signifies that $x$ runs from $a$ to $b$,
and $y$ runs from $c$ to $d$. The rectangle would then have corners at $(a,c)$, $(a,d)$, $(b,d)$, and
$(a,d)$.

Jordan Measure

The 'Jordan Measure' of a set is defined in terms of a double-integral. It is a way to define the 'area' of a set. A set is said to be 'Jordan Measurable' if it has a Jordan Measure.

**Theorem** Let $\D$ be a bounded subset of $\Rtwo$. Then, $\D$ is Jordan Measurable if and
only if the constant function $f(x,y)=1$ is integrable over $\D$. In this case, the area of $\D$ is
given by
\begin{equation}
\area(\D) = \iint_\D 1 .
\end{equation}
So we obtain the area of a set by evaluating the double integral of $f(x,y)=1$. What if instead, we
use some arbitrary function $f$? What would that represent? In fact, if $f$ is non-negative over $\D$,
then we would obtain the volume between the surface $z=f(x,y)$ and the set $\D$.

Double Integrals

Double integrals arise when the Jordan Measure of a set is desired, as mentioned in the previous section, and also in determining volumes. Double integrals are evaluated as iterated integrals, i.e. they are broken up into two integrals, where the result of one integral becomes the integrand of the next.

Iterated Integrals

We begin by discussing the simplest case: a double integral over a rectangular domain.

Integrating over Rectangular Domains

The main ideas behind iterated integrals begin with Fubini's theorem. We present this theorem now.

**Theorem (Fubini's Theorem)** Let $f:\D\to\R$ be a continuous function defined on a
rectangular subset $\D$ of $\Rtwo$, given by $\D=[a,b]\times[c,d]$. Then $f$ is integrable and its
integral can be computed as an iterated integral:
\begin{equation}
\iint_\D f = \int_a^b\left(\int_c^d f(x,y)\,dy\right)dx = \int_c^d\left(\int_a^b f(x,y)\,dx\right)dy .
\end{equation}
So far, we have learned that integrating $f(x,y)=1$ over a 2D set gives us the area, and we are
able to integrate any continuous function on a rectangular domain using Fubini's theorem. This means
that we can accomplish a fairly trivial task: compute the area of a rectangle. Just as a sanity
check, let's proceed to do this, and then consider something more interesting.

**Example** Find the area of the set $\D_1=[0,4]\times[0,2]$. Note that this is explicitly
\[ \D_1 = \left\{(x,y): 0 \le x \le 4, 0\le y \le 2 \right\} . \]
From basic geometry, we already know that $\area(\D_1)=4\times2 = 8$. Let's evaluate the area using a
double integral.
\begin{align}
\area(\D_1) &= \iint_{\D_1} 1 \\
&= \int_0^4\left(\int_0^2 1\,dy\right)dx \\
&= 2\int_0^4\,dx \\
&= 2\times4 = 8 .
\end{align}
OK, now let's do something more interesting. In the previous section, I noted that if instead of
evaluating the double-integral of $f(x,y)=1$ over a rectangular domain, we evaluate the double-integral
of an arbitrary non-negative function $f$, we would obtain the volume between the surface $z=f(x,y)$ and
the domain "below'' it.
\pagebreak

**Example** Let the set $\D_1$ be as in the previous example. ($\D_1=[0,4]\times[0,2]$.) Find
the volume $V$ of the solid defined by the set $\D_2$, where $\D_2$ is given by
\[ \D_2 = \left\{(x,y,z) : (x,y)\in\D_1, 0 \le z \le x+y^2 \right\} .\]
The volume of $\D_2$ is given by
\begin{align}
V &= \iint_{\D_1}f \text{ , where $f(x,y) = x+y^2$.} \\
&= \int_0^4\left(\int_0^2 f(x,y)\,dy\right)dx \\
&= \int_0^4\left(\int_0^2 \left(x+y^2\right)dy\right)dx \\
&= \int_0^4\left[xy+\frac{1}{3}y^3\right]_{y=0}^{y=2}\,dx \\
&= \int_0^4\left(2x+\frac{8}{3} - 0\right)dx \\
&= \left[x^2+\frac{8}{3}x\right]_{x=0}^{x=4} \\
&= 16+\frac{32}{3} = \frac{80}{3}
\end{align}
Note that here we have chosen to evaluate the integral in $y$ first, but if we chose to evaluate the $x$
integral first, we would arrive at the same result. (You should check this yourself.)

Integrating over Simple Domains

We present a theorem that allows us to break up double integrals over simple domains into iterated integrals.

**Theorem** Let $f_1:\D_1\to\R$ be a continuous function defined on a simple domain of the first
kind, $\D_1$. Let $f_2:\D_2\to\R$ be a continuous function defined on a simple domain of the second kind,
$\D_2$. Then $f_1$ is integrable and $f_2$ is integrable and their integrals are equal to iterated integrals:
\begin{align}
\iint_{\D_1}f_1 &= \int_a^b\left(\int_{g(x)}^{h(x)}f_1(x,y)\,dy\right)dx , \\
\iint_{\D_2}f_2 &= \int_c^d\left(\int_{G(y)}^{H(y)}f_2(x,y)\,dx\right)dy .
\end{align}
Note that unlike in the case of a rectangular domain in Fubini's theorem, we are not free to choose the
order of integration. The kind of simple domain given to us dictates in which order the integrals must be
evaluated.

**Example** Find the volume under the surface $z=f(x,y)=(x+y)e^x$ over the domain $\D$, where
$\D$ is the triangle with vertices $(0,0)$, $(2,0)$, and $(2,1)$. \\
Our first challenge is to write $\D$ as a simple domain. We can write that
\[ \D = \left\{(x,y) : 0 \le x \le 2, 0 \le y \le \frac{1}{2}x \right\} . \]
The volume $V$ is given by
\begin{align}
V &= \iint_\D f \text{ , where $f(x,y) = (x+y)e^x$.} \\
&= \int_0^2\left(\int_0^{x/2}(x+y)e^x\,dy\right)dx \\
\end{align}We must evaluate the inner integral first:\begin{align}
\int_0^{x/2}(x+y)e^x\,dy &= \left[(xy+\frac{1}{2}y^2)e^x\right]_{y=0}^{y=x/2} = \frac{5}{8}x^2e^x \\
\Rightarrow V &= \int_0^2 \frac{5}{8}x^2e^x\,dx = \left[\frac{5}{8}(x^2-2x+2)e^x\right]_{x=0}^{x=2} = \frac{5}{4}\left(e^2-1\right) .
\end{align}

Moment of Inertia and Center of Mass

We now consider a simple application of double integrals that is common in Physics. We will be discussing the moment of inertia and center of mass of a system, which is given by a mass density function. Many of these ideas may be readily generalized to problems involving densities of a different sort, say a probability density function, or a charge density function. Suppose that we are given the mass density of an object, that occupies the 2D domain $\D$, $f(x,y)$, and that $f$ is continuously defined on $\D$. Then, we may compute the total mass of the object and the moments of inertia about the $x$ and $y$ axes, each as double integrals. These then give the center of mass of the object. \begin{align} \end{align}We begin with the mass, $m$:\begin{align} m &= \iint_\D f . \end{align}The moment of inertia about the $x$--axis, $I_x$:\begin{align} I_x &= \iint_\D yf . \end{align}The moment of inertia about the $y$--axis, $I_y$:\begin{align} I_y &= \iint_\D xf . \end{align}Finally, we compute the co-ordinates of the center of mass of the object $(\xcm,\ycm)$:\begin{align} \xcm = \frac{I_y}{m} &, \ycm = \frac{I_x}{m} . \end{align}

**Example** A thin plate occupies the domain between the parabola $y=x^2$ and the line $y=4$. Its
mass density is given by the function $f(x,y)=y$. Find the center of mass of this object.\\
The first thing to do is to write the domain of the object as a simple domain. We can write that
\[ \D = \left\{(x,y) : -2 \le x \le 2, x^2 \le y \le 4\right\} . \]
Now we must compute the quantities that give us the center of mass co-ordinates. We begin with the mass:
\begin{align}
m &= \int_{-2}^2\left(\int_{x^2}^4 y\,dy\right)dx \\
&= \int_{-2}^2\left(8-\frac{1}{2}x^4\right)dx \\
&= \frac{128}{5}
\end{align}The moment of inertia about the $y$--axis:\begin{align}
I_y &= \int_{-2}^2\left(\int_{x^2}^4 xy\,dy\right)dx \\
&= \int_{-2}^2\frac{1}{2}x\left(16-x^4\right)dx \\
&= 0
\end{align}The moment of inertia about the $x$--axis:\begin{align}
I_x &= \int_{-2}^2\left(\int_{x^2}^4 y^2\,dy\right)dx \\
&= \int_{-2}^2\frac{1}{3}\left(64-x^6\right)dx \\
&= \frac{512}{7}
\end{align}Finally, the center of mass co-ordinates:\begin{align}
\therefore \left(\xcm,\ycm\right) &= \left(0,\frac{20}{7}\right) .
\end{align}
Notice that domain has a symmetry about the $y$--axis; this is why $I_y=0$. (And so, $\ycm=0$.)

Multiple Integrals in Polar Co-ordinates

So far in our study of multiple integrals, we are only able to evaluate multiple integrals as iterated integrals, if we are able to write the domain of integration as a simple domain. This imposes significant constraints on the possible geometry of the domain. We can open up to a larger set of possible geometries by constructing simple domains in polar co-ordinates.

Simple Domain in Polar Co-ordinates

We define the form of a simple domain in polar co-ordinates, in a way analogous to simple domains of the first kind in rectangular co-ordinates.

**Definition** Let $\D$ be a subset of $\Rtwo$. Let $\alpha,\beta\in\R$. Let $g,h:\R\to\R$ be
continuous functions on $[\alpha,\beta]$, with $0\le g(\theta) \le h(\theta)$. $\D$ is said to be a
simple domain in polar co-ordinates if it has the form
\begin{equation}
\D = \left\{(x,y) : x=r\cos\theta, y=r\sin\theta, \alpha \le \theta \le \beta, g(\theta) \le r \le h(\theta)\right\} .
\label{polar_domain}
\end{equation}

Iterated Integral in Polar Co-ordinates

We now give the theorem that allows us to write double integrals over simple domains in polar co-ordinates as iterated integrals. This is a very simple generalization of the theorem presented in the previous section.

**Theorem** Let $f:\D\to\R$ be a continuous function defined on $\D$. Let $\D$ be a simple domain
in polar co-ordinates, of the form Eq. \eqref{polar_domain}. Then $f$ is integrable, and its integral is given by
\begin{equation}
\iint_\D f = \int_\alpha^\beta\left(\int_{g(\theta)}^{h(\theta)}f(r\cos\theta,r\sin\theta)r\,dr\right)d\theta .
\end{equation}
Notice that there is an extra factor of $r$ in the integrand. This is a special case of the transformation theorem
which we will discuss towards at the end of the quarter. Specifically, the transformation theorem gives us that
when we switch to polar co-ordinates, we have that $dxdy \to rdrd\theta$. (It's important to not forget the extra $r$.)

**Example** Let $\D$ be the domain of the half of the annulus between the circles of radius 1 and 2
centered at the origin in the half-plane $y+x\ge0$. (An annulus is the donut-shape between two circles; here
we are cutting the annulus in half by restricting $y+x\ge0$.) Find the area of $\D$.\\
We write that
\[ \D=\left\{(x,y) : x=r\cos\theta, y=r\sin\theta, -\pi/4 \le \theta \le 3\pi/4, 1 \le r \le 2\right\} . \]
We know that the area of $\D$ is given by
\begin{align}
\area(\D) &= \iint_\D f \text{ , where we set $f(x,y)=1$.} \\
&= \int_{-\pi/4}^{3\pi/4}\left(\int_1^2 r\,dr\right)d\theta \\
&= \int_{-\pi/4}^{3\pi/4}\frac{1}{2}\left[r^2\right]_{r=1}^{r=2}\,d\theta \\
&= \int_{-\pi/4}^{3\pi/4}\frac{1}{2}\left(4-1\right)d\theta \\
&= \frac{3}{2}\int_{-\pi/4}^{3\pi/4}\,d\theta \\
&= \frac{3\pi}{2} .
\end{align}
This is the result that we would obtain by simply considering the geometry: the area of the entire annulus is the
difference between the area of the circle of radius 2, $4\pi$, and the circle of radius 1, $\pi$. So, the area of
the annulus would be $3\pi$. However, we restricted ourselves to half, so the area is $3\pi/2$.

Week 10: Triple Integrals; Cylindrical and Spherical Co-ordinates

Week 11: The Transformation Theorem

### Notes on IPL/CPS Physics Labs

Experiment 12: The Simple Pendulum