Lecture Notes

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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 $3\pi/2$.

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