Chapter 15 - Multiple Integrals

Iterated integrals

Rectangular domains

Rectangular domains

Suppose we are given a function $f:{\mathbb{R}}^2\to \mathbb{R}$, written in terms of components by $f(x,y)\in \mathbb{R}$ for each $(x,y)\in {\mathbb{R}}^2$. Now let the second coordinate remain some arbitrary fixed value $y=y_0$. This determines a function $f(x,y_0)$ of one variable, $x$, for each possible chosen $y_0$.

We can integrate $f(x,y_0)$ in $x$ just as any other function of $x$:

$$F(y_0)={\int }_a^{b}f(x,y_0)dx.$$

The resulting quantity depends on the chosen $y_0$, so we call it a function of $y_0$, and it also depends on the chosen interval $[a,b]$.

Now we can let $y$ vary again, and integrate the function $F$:

$$G={\int }_c^{d}F(y)dy.$$

It is reasonable to notate the composite process with two integration symbols:

$$G={\int }_c^d{\int }_a^{b}f(x,y)dxdy.$$

There is no reason we couldn’t integrate $x$ first, and this raises an obvious question that is answered by Fubini’s Theorem:

Fubini’s Theorem

Provided $f$ is continuous on $[a,b]\times [c,d]$, the order of integration doesn’t affect the result:

$${\int }_c^d{\int }_a^{b}f(x,y)dxdy={\int }_a^b{\int }_c^{d}f(x,y)dydx.$$

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Example

Double integral calculation

Double integral calculation

Problem: Compute ${\int }_1^2{\int }_0^3{x}^{2}y+x{e}^{y}dxdy$.

Solution: Compute the inner integral first, treating $y$ as a constant:

$$\begin{array}{rl}{\int }_1^2{\int }_0^3{x}^{2}y+x{e}^{y}dxdy& ={\int }_1^2{\left(\frac{x^3}{3}y+\frac{x^2}{2}{e}^y\right)|}_0^{3}dy\\ & ={\int }_1^{2}9y+\frac{9}{2}{e}^{y}dy\\ & ={9\frac{y^2}{2}+\frac{9}{2}{e}^y|}_1^2\\ & =\frac{27}{2}+\frac{9}{2}(e^2-e).\end{array}$$

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Exercise

01

Reversing integration order

Check that Fubini’s Theorem holds: evaluate ${\int }_0^3{\int }_1^2{x}^{2}y+x{e}^{y}dydx$.

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Non-rectangular domains

Non-rectangular domains

Notice that the first integral $F(y_0)={\int }_a^{b}f(x,y_0)dx$ places constants $a$ and $b$ as limits of integration, but $y_0$ is also considered a constant with regard to integration by $x$. So, we can allow the limits to depend upon $y_0$ in a non-trivial fashion: we can let them be functions of the chosen $y_0$ and write: $a(y_0)$, $b(y_0)$.

Now let us incorporate these functions in the double integral notation. Instead of using the symbol $y_0$, we can write $y$, remembering that in the first integral $y$ is treated as a constant, on a par with $a$ and $b$:

$${\int }_c^d{\int }_{a(y)}^{b(y)}f(x,y)dxdy.$$ Link to original

Example

Iterated integral, parabolic boundary

Iterated integral, parabolic boundary

Problem: Let $a(y)=y^2$ and $b(y)=4$, $[c,d]=[0,2]$, and $f(x,y)=xy$. Compute:

$${\int }_0^2{\int }_{y^2}^{4}xydxdy.$$

Solution: We have:

$$\begin{array}{rl}{\int }_0^2{\int }_{y^2}^{4}xydxdy& ={\int }_0^2{\frac{x^2}{2}y|}_{y^2}^{4}dy={\int }_0^{2}8y-\frac{y^5}{2}dy\\ & ={4y^2-\frac{y^6}{12}|}_0^2=16-\frac{16}{3}=\frac{32}{3}.\end{array}$$

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Exercise

02

Iterated integral practice

Evaluate the iterated integrals:

• (a) ${\int }_0^{\pi }{\int }_0^{\sin (x)}(1+\cos (x))dydx$
• (b) ${\int }_0^2{\int }_{3y^2-6y}^{2y-y^2}3ydxdy$
• (c) ${\int }_0^{\pi /4}{\int }_{\sqrt{3}}^{\sqrt{3}\cos (\theta )}rdrd\theta$
• (d) Extending to 3D: ${\int }_0^1{\int }_0^y{\int }_x^{1}12xyzdzdxdy$

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Volume under a surface

Volume under a surface

Consider the iterated integral ${\int }_0^2{\int }_{y^2}^{4}xydxdy$, and the picture:

The inner integral produces a function $F(y)={\int }_{y^2}^{4}xydx$. Given a value of $y$, this function returns the integral over the line segment given by $(x,y)$ with $x\in [y^2,4]$ and $y$ set to the given value.

Imagine we set $z=f(x,y)$, in this case $z=xy$. Then the inner integral provides the area under the curve over this line segment.

Computing the outer integral then gives the volume of the region between the bounds $y\in [0,2]$, $x\in [y^2,4]$, and $z=xy$. It does so by computing the area of the vertical slices as a function of $y$, and then integrating this area as $y$ varies over its own domain $y\in [0,2]$.

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Example

Iterated integral and volume

Iterated integral and volume

Consider the function $f(x,y)=2-x-2y$, and let $z=f(x,y)$ describe a plane.
Problem: Find the volume of the region under $z=f(x,y)$ in the first octant ($x,y,z>0$).

Solution: The region is bounded in the $xy$-plane by the axes and the line $y=(2-x)/2$, since $z=0$ there. So, compute the integral:

$$\begin{array}{rl}{\int }_0^2{\int }_0^{(2-x)/2}(2-x-2y)dydx& ={\int }_0^2{(2-x)y-y^2|}_0^{(2-x)/2}dx\\ & ={\int }_0^2\frac{(2-x{)}^2}{4}dx\\ & ={-\frac{(2-x{)}^3}{12}|}_0^2=\frac{2}{3}.\end{array}$$

The meaning of the two integration steps can be explained by writing the inner integral as the area of a cross section:

$$A(x)=\frac{(2-x{)}^2}{4}.$$

The outer integral then sums these areas multiplied by infinitesimal thicknesses $dx$: ${\int }_0^{2}A(x)dx$.

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Exercise

03

Volume of a solid

Find the volume of the region under $f(x,y)=x^{2}y$ for $x\in [1,3]$ and $y\in [x,2x+1]$.

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Order of integration

Order of integration

When the inner bounds of integration are functions, in order to change the order of integration we must consider the region over which the integral is taken, and define new functions for the new inner bounds.

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Example

Bounds for a triangular region

Bounds for a triangular region

Problem: Write two different integration bounds for the region in the figure:

Solution: On the left, we can use ${\int }_0^4{\int }_{x/2}^{2}dydx$. On the right, use ${\int }_0^2{\int }_0^{2y}dxdy$.

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Example

Changed order of integration, new boundary functions

Changed order of integration, new boundary functions

Problem: Find new bounds to change the order of integration in the following double integral:

$${\int }_{-3}^3{\int }_0^{9-x^2}f(x,y)dydx.$$

Solution: The region can be interpreted as the portion under the top of a parabola $y=9-x^2$, facing down, maximum at $(0,3)$ in the $xy$-plane. To integrate $x$ first we take the range $x\in [-\sqrt{9-y},+\sqrt{9-y}]$ and then $y\in [0,3]$. Thus:

$${\int }_0^9{\int }_{-\sqrt{9-y}}^{+\sqrt{9-y}}f(x,y)dxdy.$$

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Exercise

04

Change order of integration

Change the order of integration in the following double integral:

$${\int }_1^9{\int }_{\sqrt{y}}^{3}x{e}^{y}dxdy.$$

(Hint: first sketch the region in question.)

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Integration over measures of area and volume

Introduction

Another perspective on integration extends the concept of the “unit of length” $dx$ to units of another measure. In calculus, one considers the “unit of area” such as the infinitesimal square $dxdy$ or the “unit of volume” such as the infinitesimal cube $dxdydz$.

From this point of view, the double integral ${\int }_c^d{\int }_a^{b}f(x,y)dxdy$ is the integral of the function $f(x,y)$ over the rectangular box $[a,b]\times [c,d]$, summing the infinitesimal quantities of area times height, $f(x,y)dxdy$. This sum can be written as the limit of a Riemann sum:

$${\iint }_{\mathcal{R}}f(x,y)dxdy=\underset{n,m\to \infty }{\lim }\sum _{i=1}^n\sum _{j=1}^{m}f(p_{ij})\Delta x_i\Delta y_j,$$

where $\mathcal{R}$ is the box $[a,b]\times [c,d]$, and $p_{ij}$ is a point that lives somewhere in the square $[x_i,x_{i+1}]\times [y_j,y_{j+1}]$ inside a grid over $\mathcal{R}$.

In line with this perspective on integration, sometimes the unit of area $dxdy$ is written suggestively as $dA$ with $A$ for “area”, and similarly $dxdydz$ can be written $dV$ with $V$ for “volume”.

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Example

Integration over measure of area

Integration over measure of area

Problem: Compute the integral ${\iint }_{\mathcal{R}}{x}^{2}ydA$ where $\mathcal{R}$ is the region in the figure:

Solution: We rewrite the integral as an iterated integral and then compute. The region can be written as $x\in [1,3]$ and $y\in [1/x,\sqrt{x}]$. So we have:

$$\begin{array}{rl}{\iint }_{\mathcal{R}}{x}^{2}ydA& ={\int }_1^3{\int }_{1/x}^{\sqrt{x}}{x}^{2}ydydx\\ & ={\int }_1^3{\frac{1}{2}{x}^2{y}^2|}_{1/x}^{\sqrt{x}}dx={\int }_1^3\frac{1}{2}{x}^3-\frac{1}{2}dx\\ & ={\frac{1}{8}{x}^4-\frac{1}{2}x|}_1^3=9.\end{array}$$

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Introduction (part 2)

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Area of a region

Area of a region

By integrating the function $f(x,y)=1$ over a region, we obtain the area of the region:

$$A(\mathcal{R})={\iint }_{\mathcal{R}}dA={\iint }_{\mathcal{R}}dxdy.$$Link to original

Example

Area under curve as double integral

Area under curve as double integral

Problem: Find the area under the curve $y=x^2$ for $x\in [0,1]$ using a double integral.

Solution: Compute:

$$\begin{array}{rl}{\iint }_{\mathcal{R}}dydx& ={\int }_0^1{\int }_0^{x^2}dydx\\ & ={\int }_0^{1}y{|}_0^{x^2}dx={\int }_0^1{x}^{2}dx\\ & ={\frac{x^3}{3}|}_0^1=\frac{1}{3}.\end{array}$$

Note that this agrees with the usual method: ${\int }_0^1{x}^{2}dx=\frac{1}{3}$.

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Double integration in polar coordinates

Introduction

Iterated integrals are defined in other coordinate systems in exactly the same manner. However, they do not give the expected geometric quantity unless extra factors are added to create an infinitesimal unit of area. For polar coordinates, the unit of area is $dA=rdrd\theta$. For spherical coordinates, the unit of volume is $dV=r^2\sin (\phi )drd\theta d\phi$.

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Example

Double integral in polar

Double integral in polar

Problem: Compute ${\iint }_{\mathcal{R}}(3x+4y^2)dA$ for $\mathcal{R}$ the region with $y\ge 0$ that lies between the circles of radii 1 and 2.

Solution: We convert the integral into polar coordinates using $dA=rdrd\theta$. Obtain:

$$\begin{array}{rl}{\iint }_{\mathcal{R}}(3x+4y^2)dA& ={\int }_0^{\pi }{\int }_1^2(3r\cos (\theta )+4(r\sin (\theta ){)}^2)rdrd\theta \\ & ={\int }_0^{\pi }(7\cos (\theta )+15{\sin }^2(\theta ))d\theta =\frac{15}{2}\pi .\end{array}$$

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Introduction (part 2)

Area measure requires correction factors

What happens if we neglect the factor ‘$r$’ in $dA=rdrd\theta$ and simply compute ${\int }_c^d{\int }_a^{b}f(r,\theta )drd\theta$? We can still perform this iterated integral. But the resulting quantity corresponds to the integral of the function $f(r,\theta )$ over the box $(r,\theta )\in [a,b]\times [c,d]$, as if $r$ and $\theta$ were Cartesian coordinates.

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Exercise

01

Deriving polar area formula

Use the formula $rdrd\theta$ for infinitesimal area in polar double integrals to derive the formula $\frac{1}{2}\int r(\theta {)}^{2}d\theta$ for the area enclosed under the loop of a polar graph $r=r(\theta )$.

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Applications of multiple integrals

Density and total quantity

Density and total quantity

When the function of several variables is a density $\rho (x,y)$ or $\rho (x,y,z)$, then the integral gives the total quantity:

$$\mathrm{Total}={\iint }_{\mathcal{R}}\rho (x,y)dA,\mathrm{Total}={\iiint }_{\mathcal{V}}\rho (x,y,z)dV.$$

If $\rho$ measures the ordinary density, then the integral gives total mass. If $\rho$ measures a charge density, then the integral gives total charge. If $\rho$ measures the probability density, then the integral gives total probability.

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Average value

Average value

The average value of a distributed quantity is the total amount of that quantity divided by the total measure of the region over which it is spread.

When $f(x,y)$ is defined on a region $\mathcal{R}\subset {\mathbb{R}}^2$, then the average value of $f$ over that region is given by:

$$\bar{f}=\frac{1}{A}{\iint }_{\mathcal{R}}fdA,$$

where $A={\iint }_{\mathcal{R}}dA$. Similarly, if $g(x,y,z)$ is defined on a region $\mathcal{V}\subset {\mathbb{R}}^3$, then the average value of $g$ over that region is given by:

$$\bar{g}=\frac{1}{V}{\iiint }_{\mathcal{V}}gdV,$$

where $V={\iiint }_{\mathcal{V}}dV$.

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Center of mass

Center of mass

The center of mass of a planar region or a solid body is the average position of an element of material in the region or body. If the density of the body varies, it is the average position weighted by the density.

Position in a planar region is given by the vector $(x,y)$. Let $A$ be the area of a planar region $\mathcal{R}$. Since integration is linear and respects vector components, we can compute the average position componentwise:

$$\mathbf{CoM}=\frac{1}{A}{\iint }_{\mathcal{R}}(x,y)dxdy=\left(\frac{1}{A}{\iint }_{\mathcal{R}}xdxdy,\frac{1}{A}{\iint }_{\mathcal{R}}ydxdy\right)$$

(and analogously for the $\mathbf{CoM}$ of a solid body, using three components). If there is a varying density, then we need the average weighted by density:

$$\mathbf{CoM}=\frac{1}{M}{\iint }_{\mathcal{R}}\rho (x,y)(x,y)dxdy=\left(\frac{1}{M}{\iint }_{\mathcal{R}}x\rho (x,y)dxdy,\frac{1}{M}{\iint }_{\mathcal{R}}y\rho (x,y)dxdy\right),$$

where $M={\iint }_{\mathcal{R}}\rho (x,y)dxdy$ is now the mass, which will not be proportional to area (or volume) if the density is not uniform.

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Moments

Moments

Given a quantity $f(x,y,z)$ that depends on points in space, the higher moments of this quantity are given by the integral of the powers of the quantity over the region. These moments are often computed for the case where $f$ returns one of the coordinates, or a radius. For example, when $f=x$ then the first moment is the $x$-coordinate of the center of mass.

An important moment in physics is the moment of inertia, which is the second moment of the distance to the axis of rotation:

$$I={\iiint }_{\mathcal{V}}{d}_{\ell }(x,y,z{)}^2\cdot \rho (x,y,z)dxdydz$$

where $d_{\ell }(x,y,z)$ gives the distance from $(x,y,z)$ to the line $\ell$ that forms the axis of rotation.

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