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Michael Oberguggenberger and Alexander OstermannAnalysis for Computer ScientistsUndergraduate Topics in Computer Sciencehttps://doi.org/10.1007/978-3-319-91155-7_17

17. Integration of Functions of Two Variables

Michael Oberguggenberger¹ and Alexander Ostermann¹

(1)

University of Innsbruck, Innsbruck, Austria

Michael Oberguggenberger (Corresponding author)

Email: michael.oberguggenberger@uibk.ac.at

Alexander Ostermann

Email: alexander.ostermann@uibk.ac.at

In Sect. 11.3 we have shown how to calculate the volume of solids of revolution. If there is no rotational symmetry, however, one needs an extension of integral calculus to functions of two variables. This arises, for example, if one wants to find the volume of a solid that lies between a domain D in the (x, y)-plane and the graph of a non-negative function $$z = f(x, y)$$ . In this section we will extend the notion of Riemann integrals from Chap. 11 to double integrals of functions of two variables. Important tools for the computation of double integrals are their representation as iterated integrals and the transformation formula (change of coordinates). The integration of functions of several variables occurs in numerous applications, a few of which we will discuss.

17.1 Double Integrals

We start with the integration of a real-valued function $$z = f(x, y)$$ which is defined on a rectangle $R = [a,b]\times [c, d]$ . More general domains of integration $D\subset \mathbb R^2$ will be discussed below. Since we know from Sect. 11.1 that Riemann integrable functions are necessarily bounded, we assume in the whole section that f is bounded. If f is non-negative, the integral should be interpretable as the volume of the solid with base R and top surface given by the graph of f (see Fig. 17.2). This motivates the following approach in which the solid is approximated by a sum of cuboids.

We place a rectangular grid G over the domain R by partitioning the intervals [a, b] and [c, d] like in Sect. 11.1:

$\begin{aligned} Z_x : a = x_0< x_1< x_2<\cdots< x_{n-1}< x_n = b,\\ Z_y : c = y_0< y_1< y_2<\cdots< y_{m-1} < y_m = d.\\ \end{aligned}$

The rectangular grid is made up of the small rectangles

$[x_{i-1}, x_i]\times [y_{j-1}, y_j],\quad i = 1,\ldots , n,\ j = 1,\ldots , m.$

The mesh size $\varPhi (G)$ is the length of the largest subinterval involved:

$\varPhi (G) = \max \big (|x_i - x_{i-1}|, |y_j - y_{j-1}|\;;\, i=1,\ldots , n,\, j=1,\ldots , m\big ).$

Finally we choose an arbitrary intermediate point $\mathbf {p}_{ij} = (\xi _{ij},\eta _{ij})$ in each of the rectangles of the grid, see Fig. 17.1.

images/215236_2_En_17_Chapter/215236_2_En_17_Fig1_HTML.gif — Fig. 17.1
Partitioning the rectangle R

The double sum

$S = \sum _{i = 1}^n\sum _{j = 1}^m f(\xi _{ij},\eta _{ij}) (x_i - x_{i-1})(y_j - y_{j-1})$

is again called a Riemann sum . Since the volume of a cuboid with base $[x_{i-1}, x_i]\times [y_{j-1}, y_j]$ and height $f(\xi _{ij},\eta _{ij})$ is

$f(\xi _{ij},\eta _{ij}) (x_i - x_{i-1})(y_j - y_{j-1}),$

the above Riemann sum is an approximation to the volume under the graph of f (Fig. 17.2).

images/215236_2_En_17_Chapter/215236_2_En_17_Fig2_HTML.gif — Fig. 17.2
Volume and approximation by cuboids

Like in Sect. 11.1, the integral is now defined as a limit of Riemann sums. We consider a sequence $G_1, G_2, G_3, \ldots$ of grids whose mesh size $\varPhi (G_N)$ tends to zero as $N \rightarrow \infty$ and the corresponding Riemann sums $$S_N$$ .

Definition 17.1

A bounded function $$z = f(x, y)$$

is called Riemann integrable on $R = [a,b]\times [c, d]$ if for arbitrary sequences of grids $(G_N)_{N \ge 1}$ with $\varPhi (G_N) \rightarrow 0$ the corresponding Riemann sums $(S_N)_{N \ge 1}$ tend to the same limit I(f), independently of the choice of intermediate points. This limit

$I(f) = \iint _R f(x,y) \,{\mathrm{d}}(x, y)$

is called the double integral of f on R.

Experiment 17.2

Study the M-file mat17_1.m and experiment with different randomly chosen Riemann sums for the function $$z = x^2+y^2$$ on the rectangle $[0,1]\times [0,1]$ . What happens if you choose finer and finer grids?

As in the case of one variable, one may use the definition of the double integral for obtaining a numerical approximation to the integral. However, it is of little use for the analytic evaluation of integrals. In Sect. 11.1 the fundamental theorem of calculus has proven helpful, here the representation as iterated integral does. In this way the computation of double integrals is reduced to the integration of functions in one variable.

Proposition 17.3

(The double integral as iterated integral) If a bounded function f and its partial functions $x \mapsto f(x, y)$ , $y \mapsto f(x, y)$ are Riemann integrable on $R = [a,b]\times [c, d]$ , then the mappings $x \mapsto \int _c^d f(x, y)\,{\mathrm{d}}y$ and $y \mapsto \int _a^b f(x, y)\,{\mathrm{d}}x$ are Riemann integrable as well and

$\iint _R f(x,y) \,{\mathrm{d}}(x,y) = \int _a^b\left( \int _c^d f(x,y)\,{\mathrm{d}}y\right) \mathrm{d}x = \int _c^d\left( \int _a^b f(x, y)\,{\mathrm{d}}x\right) \mathrm{d}y.$

Outline of the proof. If one chooses intermediate points in the Riemann sums of the special form $\mathbf {p}_{ij} = (\xi _i,\eta _j)$ with $\xi _i\in [x_{i-1}, x_i]$ , $\eta _j\in [y_{j-1}, y_j]$ , then

$\begin{aligned} \iint _R f(x,y)\,{\mathrm{d}}(x, y) \approx \sum _{i = 1}^n\left( \sum _{j = 1}^m f(\xi _i,\eta _j) (y_j - y_{j-1}) \right) (x_i - x_{i-1})\\ \approx \sum _{i = 1}^n\left( \int _c^d f(\xi _i, y)\,{\mathrm{d}}y \right) (x_i - x_{i-1}) \approx \int _a^b\left( \int _c^d f(x, y)\,{\mathrm{d}}y \right) \mathrm{d}x \end{aligned}$

and likewise for the second statement by changing the order. For a rigorous proof of this argument, we refer to the literature, for instance [4, Theorem 8.13 and Corollary]. $\square$

Figure 17.3 serves to illustrate Proposition 17.3. The volume is approximated by summation of thin slices parallel to the axis instead of small cuboids. Proposition 17.3 states that the volume of the solid is obtained by integration over the area of the cross sections (perpendicular to the x- or y-axis). In this form Proposition 17.3 is called Cavalieri’s principle.¹ In general integration theory one also speaks of Fubini’s theorem.² Since in the case of integrability the order of integration does not matter, one often omits the brackets and writes

$\iint _R f(x,y)\,{\mathrm{d}}(x,y) = \iint _R f(x,y)\,{\mathrm{d}}x \,{\mathrm{d}}y = \int _a^b\!\!\int _c^d f(x, y)\,{\mathrm{d}}y \,{\mathrm{d}}x.$

images/215236_2_En_17_Chapter/215236_2_En_17_Fig3_HTML.gif — Fig. 17.3
The double integral as iterated integral

images/215236_2_En_17_Chapter/215236_2_En_17_Fig4_HTML.gif — Fig. 17.4
The body B

Example 17.4

Let $R = [0,1]\times [0,1]$ . The volume of the body

$B=\{(x,y, z)\in \mathbb R^3 : (x, y)\in R,\, 0\le z \le x^2+y^2\}$

is obtained using Proposition 17.3 as follows, see also Fig. 17.4:

$\begin{aligned}&\iint _R\left( x^2+y^2\right) \,{\mathrm{d}}(x, y) = \int _0^1\left( \int _0^1\big (x^2+y^2\big ) \,{\mathrm{d}}y\right) \mathrm{d}x\\&\qquad = \int _0^1\left( x^2y + \frac{y^3}{3}\right) \Big |^{y=1}_{y=0} \,{\mathrm{d}}x = \int _0^1 \left( x^2+\frac{1}{3}\right) \,{\mathrm{d}}x = \left( \frac{x^3}{3} + \frac{x}{3}\right) \Big |^{x=1}_{x=0} = \frac{2}{3}. \end{aligned}$

We now turn to the integration over more general (bounded) domains $D \subset \mathbb R^2$ . The indicator function of the domain D is

$\begin{aligned} {\small 1}\!\!1_D(x, y)= \left\{ \begin{array} {rl} 1, &{} \quad (x, y) \in D, \\ 0, &{} \quad (x, y) \notin D. \end{array} \right. \end{aligned}$

We can enclose the bounded domain D in a rectangle R ( $D\subset R$ ). If the Riemann integral of the indicator function of D exists, then it represents the volume of the cylinder of height one and base D and thus the area of D (Fig. 17.5). The result obviously does not depend on the size of the surrounding rectangle since the indicator function assumes the value zero outside the domain D.

images/215236_2_En_17_Chapter/215236_2_En_17_Fig5_HTML.gif — Fig. 17.5
Area as volume of the cylinder of height one

Definition 17.5

Let D be a bounded domain and R an enclosing rectangle.

(a)

If the indicator function of D is Riemann integrable then the domain D is called measurable and one sets

$\iint _D \,{\mathrm{d}}(x, y) = \iint _R {\small 1}\!\!1_D(x,y)= \,{\mathrm{d}}(x, y).$
(b)

A subset $N\subset \mathbb R^2$ is called set of measure zero , if $\iint _N \,{\mathrm{d}}(x, y) = 0$ .
(c)

For a bounded function , its integral over a measurable domain D is defined as

$\iint _D f(x,y) \,{\mathrm{d}}(x,y) = \iint _R f(x, y){\small 1}\!\!1_D(x,y)= \,{\mathrm{d}}(x, y),$

if $f(x, y){\small 1}\!\!1_D(x, y)$ is Riemann integrable.

Sets of measure zero are, for example, single points, straight line segments or segments of differentiable curves in the plane. Item (c) of the definition states that the integral of a function f over a domain D is determined by continuing f to a larger rectangle R and assigning the value zero outside D.

Remark 17.6

(a) If D is a measurable domain, N a set of measure zero and f is integrable over the respective domains then

$\iint _D f(x,y) \,{\mathrm{d}}(x,y) = \iint _{D\setminus N} f(x,y) \,{\mathrm{d}}(x, y).$

(b) Let $D = D_1\cup D_2$ . If $D_1\cap D_2$ is a set of measure zero then

$\iint _D f(x,y) \,{\mathrm{d}}(x, y) = \iint _{D_1} f(x,y) \,{\mathrm{d}}(x, y) + \iint _{D_2} f(x,y) \,{\mathrm{d}}(x, y).$

The integral over the entire domain D is thus obtained as sum of the integrals over subdomains. The proof of this statement can easily be obtained by working with Riemann sums.

An important class of domains D on which integration is simple are the so-called normal domains .

Definition 17.7

(a) A subset $D\subset \mathbb R^2$ is called normal domain of type I if

$D = \{(x, y) \in \mathbb R^2\;;\; a\le x \le b,\ v(x) \le y \le w(x)\}$

with certain continuously differentiable lower and upper bounding functions $x \mapsto v(x)$ , $x \mapsto w(x)$ .

(b) A subset $D\subset \mathbb R^2$ is called normal domain of type II

$D = \{(x, y) \in \mathbb R^2\;;\; c\le y \le d,\ l(y) \le x \le r(y)\}$

with certain continuously differentiable left and right bounding functions $x \mapsto l(x)$ , $x \mapsto r(x)$ .

Figure 17.6 shows examples of normal domains.

images/215236_2_En_17_Chapter/215236_2_En_17_Fig6_HTML.gif — Fig. 17.6
Normal domains of type I and II

Proposition 17.8

(Integration over normal domains) Let D be a normal domain and $f:D\rightarrow \mathbb R$ continuous. For normal domains of type I, one has

$\iint _D f(x,y) \,{\mathrm{d}}(x,y) = \int _a^b\left( \int _{v(x)}^{w(x)} f(x, y)\,{\mathrm{d}}y\right) \mathrm{d}x$

and for normal domains of type II

$\iint _D f(x,y) \,{\mathrm{d}}(x,y) = \int _c^d\left( \int _{l(y)}^{r(y)} f(x, y)\,{\mathrm{d}}x\right) \mathrm{d}y.$

Proof

The statements follow from Proposition 17.3. We recall that f is extended by zero outside of D. For details we refer to the remark at the end of [4, Chap. 8.3]. $\square$

Example 17.9

For the calculation of the volume of the body lying between the triangle $D = \{(x, y)\;;\; 0 \le x \le 1, 0 \le y \le 1-x\}$ and the graph of $$z = x^2+y^2$$

, we interpret D as normal domain of type I with the boundaries $$v(x) = 0$$

. Consequently

$\begin{aligned}&\iint _D\left( x^2+y^2\right) \,{\mathrm{d}}(x, y) = \int _0^1\left( \int _0^{1-x}\big (x^2+y^2\big ) \,{\mathrm{d}}y\right) \mathrm{d}x\nonumber \\&\qquad \quad = \int _0^1\left( x^2y + \frac{y^3}{3}\right) \Big |^{y=1-x}_{y=0} \,{\mathrm{d}}x = \int _0^1 \left( x^2(1-x)+\frac{(1-x)^3}{3}\right) \,{\mathrm{d}}x = \frac{1}{6}, \end{aligned}$

as can be seen by multiplying out and integrating term by term.

17.2 Applications of the Double Integral

For modelling purposes it is useful to introduce a simplified notation for Riemann sums. In the case of equidistant partitions $$Z_x, Z_y$$

where all subintervals have the same lengths, one writes

$\varDelta x = x_i - x_{i-1},\quad \varDelta y = y_j - y_{j-1}$

and calls

$\varDelta A = \varDelta x\varDelta y$

the area element of the grid G. If one then takes the right upper corner $\mathbf {p}_{ij} = (x_i, y_j)$ of the subrectangle $[x_{i-1}, x_i]\times [y_{j-1}, y_j]$ as an intermediate point, the corresponding Riemann sum reads

$S = \sum _{i = 1}^n\sum _{j = 1}^m f(x_i, y_j)\varDelta A = \sum _{i = 1}^n\sum _{j = 1}^m f(x_i, y_j)\varDelta x\varDelta y.$

Application 17.10

(Mass as integral of the density) A thin plane object D has density $\rho (x, y)$ [mass/unit area] at the point (x, y). If the density $\rho$ is constant everywhere then its total mass is simply the product of density and area. In the case of variable density (e.g. due to a change of the material properties from point to point), we partition D in smaller rectangles with sides $\varDelta x$ , $\varDelta y$ . The mass contained in such a small rectangle around (x, y) is approximately equal to $\rho (x, y)\varDelta x\varDelta y$ . The total mass is thus approximately equal to

$\sum _{i = 1}^n\sum _{j = 1}^m \rho (x_i, y_j)\varDelta x\varDelta y.$

However, this is just a Riemann sum for

$M = \iint _D\rho (x, y)\,{\mathrm{d}}x\,{\mathrm{d}}y.$

This consideration shows that the integral of the density function is a feasible model for representing the total mass of a two-dimensional object.

Application 17.11

(Centre of gravity) We consider a two-dimensional flat object D as in Application 17.10. The two statical moments of a small rectangle close to (x, y) with respect to a point $$(x^*, y^*)$$

are

$(x-x^*)\rho (x,y)\varDelta x\varDelta y,\quad (y-y^*)\rho (x, y)\varDelta x\varDelta y,$

see Fig. 17.7.

images/215236_2_En_17_Chapter/215236_2_En_17_Fig7_HTML.gif — Fig. 17.7
The statical moments

The relevance of the statical moments can be seen if one considers the object under the influence of gravity. Multiplied by the gravitational acceleration g one obtains the moments of force with respect to the axes through $$(x^*, y^*)$$

in direction of the coordinates (force times lever arm). The centre of gravity of the two-dimensional object D is the point $(x_\mathsf{S}, y_\mathsf{S})$ with respect to which the total statical moments vanish:

$\sum _{i = 1}^n\sum _{j = 1}^m (x_i - x_\mathsf{S}) \rho (x_i, y_j)\varDelta x\varDelta y \approx 0,\quad \sum _{i = 1}^n\sum _{j = 1}^m (y_j - y_\mathsf{S}) \rho (x_i, y_j)\varDelta x\varDelta y \approx 0.$

In the limit, as the mesh size of the grid tends to zero, one obtains

$\iint _D (x-x_\mathsf{S})\rho (x, y)\,{\mathrm{d}}x\,{\mathrm{d}}y = 0, \quad \iint _D (y-y_\mathsf{S})\rho (x, y)\,{\mathrm{d}}x\,{\mathrm{d}}y = 0$

as defining equations for the centre of gravity; i.e.,

$x_\mathsf{S} = \frac{1}{M}\iint _D x\rho (x, y)\,{\mathrm{d}}x\,{\mathrm{d}}y, \quad y_\mathsf{S} = \frac{1}{M}\iint _D y\rho (x, y)\,{\mathrm{d}}x\,{\mathrm{d}}y,$

where M denotes the total mass as in Application 17.10.

For the special case of a constant density $\rho (x, y)\equiv 1$ one obtains the geometric centre of gravity of the domain D.

Example 17.12

(Geometric centre of gravity of a quarter circle) Let D be the quarter circle of radius r about (0, 0) in the first quadrant; i.e., $D = \{(x, y)\;;\; 0\le x \le r,\, 0 \le y \le \sqrt{r^2-x^2}\}$ (Fig. 17.8). With density $\rho (x, y)\equiv 1$ one obtains the area M as $r^2\pi /4$ . The first statical moment is

$\begin{aligned} \iint _D x\,{\mathrm{d}}x\,{\mathrm{d}}y&= \int _0^r\left( \int _0^{\sqrt{r^2-x^2}} x\,{\mathrm{d}}y\right) \mathrm{d}x = \int _0^r\left( xy\Big |_{y=0}^{y = \sqrt{r^2-x^2}}\,\right) \mathrm{d}x\\&= \int _0^r x\sqrt{r^2-x^2}\,{\mathrm{d}}x = -\frac{1}{3}\big (r^2-x^2\big )^{3/2} \Big |_{x=0}^{x=r} = \frac{1}{3}r^3. \end{aligned}$

The x-coordinate of the centre of gravity is thus given by $x_\mathsf{S} = \frac{4}{r^2\pi }\cdot \frac{1}{3}r^3 = \frac{4r}{3\pi }$ . For reasons of symmetry, one has $y_\mathsf{S} = x_\mathsf{S}$ .

images/215236_2_En_17_Chapter/215236_2_En_17_Fig8_HTML.gif — Fig. 17.8
Centre of gravity of the quarter circle

17.3 The Transformation Formula

Similar to the substitution rule for one-dimensional integrals (Sect. 10.2), the transformation formula for double integrals makes it possible to change coordinates on the domain D of integration. For the purpose of this section it is convenient to assume that D is an open subset of $\mathbb R^2$ (see Definition 9.1).

Definition 17.13

A bijective, differentiable mapping $\mathbf {F}: D \rightarrow B=\mathbf {F}(D)$ between two open subsets $D, B \subset \mathbb R^2$ is called a diffeomorphism if the inverse mapping $\mathbf {F}^{-1}$ is also differentiable.

We use the following notation for the variables:

$\mathbf {F}:D\rightarrow B: \begin{bmatrix}u\\v\end{bmatrix} \mapsto \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}x(u,v)\\y(u, v)\end{bmatrix}.$

Figure 17.9 shows the image B of the domain $D = (0,1)\times (0,1)$ under the transformation

$\mathbf {F}: \begin{bmatrix}u\\v\end{bmatrix} \mapsto \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}u+v/4\\u/4 + v + u^2v^2\end{bmatrix}.$

The aim is to transform the integral of a real-valued function f over the domain B to one over D.

images/215236_2_En_17_Chapter/215236_2_En_17_Fig9_HTML.gif — Fig. 17.9
Transformation of a planar domain

For this purpose we lay a grid G over the domain D in the (u, v)-plane and select a rectangle, for instance with the left lower corner (u, v) and sides spanned by the vectors

$\begin{bmatrix}\varDelta u\\0\end{bmatrix},\ \begin{bmatrix}0\\\varDelta v\end{bmatrix}.$

The image of this rectangle under the transformation $\mathbf {F}$ will in general have a curvilinear boundary. In a first approximation we replace it by a parallelogram. In linear approximation (see Sect. 15.4) we have the following:

$\begin{aligned} \mathbf {F}(u+\varDelta u,v)\approx & {} \mathbf {F}(u,v) + \mathbf {F}'(u, v)\begin{bmatrix}\varDelta u\\0\end{bmatrix},\\ \mathbf {F}(u,v+\varDelta v)\approx & {} \mathbf {F}(u,v) + \mathbf {F}'(u, v)\begin{bmatrix}0\\\varDelta v\end{bmatrix}. \end{aligned}$

The approximating parallelogram is thus spanned by the vectors

$\begin{bmatrix}\dfrac{\partial x}{\partial u}(u,v)\\ \dfrac{\partial y}{\partial u}(u,v)\end{bmatrix}\varDelta u,\quad \begin{bmatrix}\dfrac{\partial x}{\partial v}(u,v)\\ \dfrac{\partial y}{\partial v}(u, v)\end{bmatrix}\varDelta v$

and has the area (see Appendix A.5)

$\Bigl |\det \begin{bmatrix}\dfrac{\partial x}{\partial u}(u,v)&\ \dfrac{\partial x}{\partial v}(u,v) \\ \dfrac{\partial y}{\partial u}(u,v)&\ \dfrac{\partial y}{\partial v}(u,v)\end{bmatrix}\varDelta u\varDelta v\Bigr | \ = \ \bigl |\det \mathbf {F}'(u, v)\bigr |\varDelta u\varDelta v.$

In short, the area element $\varDelta A = \varDelta u\varDelta v$ is changed by the transformation $\mathbf {F}$ to the area element $\varDelta \mathbf {F}(A) = \bigl |\det \mathbf {F}'(u, v)\bigr |\varDelta u\varDelta v$ (see Fig. 17.10).

images/215236_2_En_17_Chapter/215236_2_En_17_Fig10_HTML.gif — Fig. 17.10
Transformation of an area element

Proposition 17.14

(Transformation formula for double integrals) Let D, B be open, bounded subsets of $\mathbb R^2$ , $\mathbf {F}: D \rightarrow B$ a diffeomorphism and $f: B \rightarrow \mathbb R$ a bounded mapping. Then

$\iint _B f(x,y)\,{\mathrm{d}}x\,{\mathrm{d}}y = \iint _D f\big (\mathbf {F}(u,v)\big ) \bigl |\det \mathbf {F}'(u, v)\bigr |\,{\mathrm{d}}u\,{\mathrm{d}}v,$

as long as the functions f and $f(\mathbf {F})\left| \det \mathbf {F}'\right|$ are Riemann integrable.

Outline of the proof. We use Riemann sums on the transformed grid and obtain

$\begin{aligned} \iint _B f(x, y)\,{\mathrm{d}}x\,{\mathrm{d}}y\approx & {} \sum _{i = 1}^n\sum _{j = 1}^m f(x_i, y_j)\varDelta \mathbf {F}(A)\\\approx & {} \sum _{i = 1}^n\sum _{j = 1}^m f\big (x(u_i,v_j),y(u_i,v_j)\big ) \bigl |\det \mathbf {F}'(u_i,v_j)\bigr | \varDelta u\varDelta v\\\approx & {} \iint _D f\big (x(u,v),y(u,v)\big )\bigl |\det \mathbf {F}'(u, v)\bigr |\,{\mathrm{d}}u\,{\mathrm{d}}v. \end{aligned}$

A rigorous proof is tedious and requires a careful study of the boundary of the domain D and the behaviour of the transformation $\mathbf {F}$ near the boundary (see for instance [3, Chap. 19, Theorem 4.7]). $\square$

Example 17.15

The area of the domain B from Fig. 17.9 can be calculated using the transformation formula with $$f(x, y)= 1$$

as follows. We have

$\begin{aligned} \mathbf {F}'(u, v) = \begin{bmatrix} 1&1/4\\ 1/4 + 2uv^2&1 + 2u^2v \end{bmatrix},\\ \bigl |\det \mathbf {F}'(u, v)\bigr | = \Bigl |\frac{15}{16} + 2u^2v - \frac{1}{2}uv^2\Bigr | {,} \end{aligned}$

and thus

$\begin{aligned} \iint _B \,{\mathrm{d}}x\,{\mathrm{d}}y= & {} \iint _D \bigl |\det \mathbf {F}'(u, v)\bigr |\,{\mathrm{d}}u\,{\mathrm{d}}v\\= & {} \int _0^1\left( \int _0^1 \left( \frac{15}{16} + 2u^2v - \frac{1}{2}uv^2\right) \,{\mathrm{d}}v\right) \mathrm{d}u\\= & {} \int _0^1 \left( \frac{15}{16} + u^2 - \frac{1}{6}u\right) \,{\mathrm{d}}u = \frac{15}{16} + \frac{1}{3} - \frac{1}{12} = \frac{19}{16}. \end{aligned}$

Example 17.16

(Volume of a hemisphere in polar coordinates) We represent a hemisphere of radius R by the three-dimensional domain

$\{(x,y, z)\;;\; 0 \le x^2+y^2\le R^2, 0 \le z \le \sqrt{R^2-x^2-y^2}\}.$

Its volume is obtained by integration of the function $f(x, y) = \sqrt{R^2-x^2-y^2}$ over the base $B = \{(x, y)\;;\; 0 \le x^2+y^2\le R^2\}$ . In polar coordinates

$\mathbf {F}:\mathbb R^2\rightarrow \mathbb R^2: \begin{bmatrix}r\\\varphi \end{bmatrix} \mapsto \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}r\cos \varphi \\r\sin \varphi \end{bmatrix}$

the area B can be represented as the image $\mathbf {F}(D)$ of the rectangle $D = [0,R]\times [0,2\pi ]$ . However, in order to fulfil the assumptions of Proposition 17.14 we have to switch to open domains on which $\mathbf {F}$ is a diffeomorphism. We can obtain this, for instance, by removing the boundary and the half ray $\{(x, y)\;;\; 0 \le x \le R,\, y = 0\}$ of the circle B and the boundary of the rectangle D. On the smaller domains $$D', B'$$

obtained in this way, $\mathbf {F}$ is a diffeomorphism. However, since B differs from $$B'$$

and D differs from $$D'$$

by sets of measure zero, the value of the integral is not changed if one replaces B by $$B'$$

and D by

, see Remark 17.6. We have

$\mathbf {F}'(r,\varphi ) = \begin{bmatrix}\cos \varphi&\;&-r\sin \varphi \\ \sin \varphi&\;&r\cos \varphi \end{bmatrix},\quad \bigl |\det \mathbf {F}'(r,\varphi )\bigr | = r.$

Substituting $x = r\cos \varphi$ , $y = r\sin \varphi$ results in $$x^2+y^2 = r^2$$

and we obtain the volume from the transformation formula as

$\begin{aligned} \iint _B \sqrt{R^2-x^2-y^2}\,{\mathrm{d}}x\,{\mathrm{d}}y= & {} \int _0^R\int _0^{2\pi }\sqrt{R^2-r^2}\;r\,{\mathrm{d}}\varphi \,{\mathrm{d}}r\\= & {} \int _0^R 2\pi r \sqrt{R^2-r^2}\,{\mathrm{d}}r\\= & {} -\frac{2\pi }{3}\big (R^2-r^2\big )^{3/2}\Big |_{r=0}^{r=R} = \frac{2\pi }{3}R^3, \end{aligned}$

which coincides with the known result from elementary geometry.

17.4 Exercises

1.

Compute the volume of the parabolic dome $$z = 2 - x^2-y^2$$ above the quadratic domain $D: -1\le x \le 1$ , $-1\le y \le 1$ .

2.

(From statics) Compute the axial moment of inertia $\iint _D y^2\,{\mathrm{d}}x\,{\mathrm{d}}y$ of a rectangular cross section $D: 0 \le x \le b$ , $-h/2 \le y \le h/2$ , where $$b> 0, h >0$$ .

3.

Compute the volume of the body bounded by the plane $$z = x+y$$ above the domain $D: 0 \le x \le 1$ , $0 \le y \le \sqrt{1-x^2}$ .

4.

Compute the volume of the body bounded by the plane $$z = 6 - x - y$$ above the domain D, which is bounded by the y-axis and the straight lines $$x+y = 6$$ , $$x+3y = 6$$ ( $x \ge 0, y \ge 0$ ).

5.

Compute the geometric centre of gravity of the domain $D: 0 \le x \le 1$ , $0 \le y \le 1-x^2$ .

6.

Compute the area and the geometric centre of gravity of the semi-ellipse

$\frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1,\quad y \ge 0.$

Hint. Introduce elliptic coordinates $x = ar\cos \varphi$ , $y = br\sin \varphi$ , $0 \le r \le 1$ , $0 \le \varphi \le \pi$ , compute the Jacobian and use the transformation formula.

7.

(From statics) Compute the axial moment of inertia of a ring with inner radius $$R_1$$ and outer radius $$R_2$$ with respect to the central axis, i.e. the integral $\iint _D (x^2+y^2)\,{\mathrm{d}}x\,{\mathrm{d}}y$ over the domain $D : R_1 \le \sqrt{x^2+y^2} \le R_2$ .

8.

Modify the M-file mat17_1.m so that it can evaluate Riemann sums over equidistant partitions with $\varDelta x \ne \varDelta y$ .

9.

Let the domain D be bounded by the curves

$y = x \quad \text{ and }\quad y = x^2, \qquad 0\le x\le 1.$

(a): Sketch D.
(b): Compute the area of D by means of the double integral $F = \iint _D\,{\mathrm{d}}(x, y)$ .
(c): Compute the statical moments $\iint _Dx\,\,{\mathrm{d}}(x, y)$ und $\iint _Dy\,\,{\mathrm{d}}(x, y)$ .

10.