mathematical-analysis mathematics real-mathematical-analysis vector-mathematical-analysis Riemann Integrals (Real Scalar Fields) The concept of the Riemann integral as the surface area bounded by the graph of a real function can be extended to multiple dimensions via real scalar fields . For a real scalar field \(f: \mathbb{R}^n \to \mathbb{R}\) , the "multidimensional Riemann integral" gives us the \((n+1)\) -dimensional "volume" bounded by the graph of \(f\) .
Definition: Darboux Sums
Let \(f: \mathcal{D} \subseteq \mathbb{R}^n \to \mathbb{R}\) be a real scalar field which is bounded on a compact rectangular region \([a_1,b_1] \times \cdots \times [a_n, b_n] \subset \mathbb{R}^n\) and let
\[\begin{aligned}Z_1 & = \{x_{1,1}, \dotsc, x_{1,m_1}\} \\ Z_2 & = \{x_{2,1}, \dotsc, x_{2,m_2}\} \\ & \vdots \\ Z_n & = \{x_{n,1}, \dotsc, x_{n,m_n}\}\end{aligned}\]
be partitions of the intervals \([a_1,b_1],\dotsc, [a_n, b_n]\) , respectively:
\[\begin{aligned}a_1 = x_{1,1} < x_{1,2} & < \cdots < x_{1,m_1} = b_1 \\ a_2 = x_{2,1} < x_{2,2} & < \cdots < x_{2,m_2} = b_2 \\ & \vdots & & \\ a_n = x_{n,1} < x_{n,2} & < \cdots < x_{n,m_n} = b_n \end{aligned}\]
The lower Darboux sum of \(f\) with respect to \(Z_1, \dotsc, Z_n\) is defined using the infima of \(f\) as follows:
\[L_{Z_1, \dotsc, Z_n}(f) = \sum_{j_1=1}^{m_1-1} \cdots \sum_{j_n=1}^{m_n-1} \inf_{\substack{x_1 \in [x_{1, j_1}, x_{1, j_1+1}] \\ \vdots \\ x_n \in [x_{n, j_n}, x_{n, j_n+1}]}} f(x_1, \dotsc, x_n) \prod_{i=1}^n (x_{i, j_i+1} - x_{i, j_i})\]
The upper Darboux sum of \(f\) with respect to \(Z_1, \dotsc, Z_n\) is defined using the suprema of \(f\) :
\[U_{Z_1, \dotsc, Z_n}(f) = \sum_{j_1=1}^{m_1-1} \cdots \sum_{j_n=1}^{m_n-1} \sup_{\substack{x_1 \in [x_{1, j_1}, x_{1, j_1+1}] \\ \vdots \\ x_n \in [x_{n, j_n}, x_{n, j_n+1}]}} f(x_1, \dotsc, x_n) \prod_{i=1}^n (x_{i, j_i+1} - x_{i, j_i})\]
Definition: Darboux Integrals
Let \(f: \mathcal{D} \subseteq \mathbb{R}^n \to \mathbb{R}\) be a real scalar field which is bounded on a compact rectangular region \(R = [a_1,b_1] \times \cdots \times [a_n, b_n] \subset \mathbb{R}^n\) .
The lower Darboux integral of \(f\) on \(R\) is the supremum of \(f\) 's lower Darboux sums on \(R\) :
\[L_R(f) = \sup \{L_{Z_1, \dotsc, Z_n} (f) \mid (Z_1, \dotsc, Z_n) \text{ is a partition of } R\}\]
The upper Darboux integral of \(f\) on \(R\) is the infimum of \(f\) 's upper Darboux sums on \(R\) :
\[U_R(f) = \inf \{U_{Z_1, \dotsc, Z_n} (f) \mid (Z_1, \dotsc, Z_n) \text{ is a partition of } R\}\]
Definition: Riemann-Integrability (Rectangular Regions)
Let \(f: \mathcal{D} \subseteq \mathbb{R}^n \to \mathbb{R}\) be a real scalar field .
If \(f\) is bounded on a compact rectangular region \(R = [a_1,b_1] \times \cdots \times [a_n, b_n] \subset \mathbb{R}^n\) , then we say that \(f\) is Riemann-integrable on \(R\) if its lower Darboux integral and upper Darboux integral are equal:
\[L_R(f) = U_R(f)\]
In this case, this common value is known as \(f\) 's Riemann integral on \(R\) .
Definition: Riemann-Integrability (Jordan-Measurable Sets)
Let \(f: \mathcal{D} \subseteq \mathbb{R}^n \to \mathbb{R}\) be a real scalar field .
If \(f\) is bounded on a Jordan-measurable \(S \subseteq \mathcal{D}\) , then \(f\) is Riemann-integrable on \(S\) if the product \(f \cdot \mathbf{1}_S\) is Riemann-integrable on any compact rectangular region containing \(S\) , where \(\mathbf{1}_S\) is the indicator function of \(S\) :
\[\mathbf{1}_S(\boldsymbol{x}) = \begin{cases}0 & \text{if} & \boldsymbol{x} \notin S \\ 1 & \text{if} & \boldsymbol{x} \in S\end{cases}\]
In this case, the Riemann integral of \(f \cdot \mathbf{1}_S\) is simply called the Riemann integral of \(f\) .
Notation
The Riemann integral of \(f\) on \(S\) is most commonly denoted as follows:
\[\int \cdots \int_{S} f(\boldsymbol{x}) \,\mathrm{d}^n \boldsymbol{x}\]
\[\int_{S} f(\boldsymbol{x}) \,\mathrm{d}^n \boldsymbol{x}\]
If labels \(x_1, \dotsc, x_n\) are introduced for the components, we write:
\[\int \cdots \int_{S} f(x_1, \dotsc, x_n) \,\mathrm{d}(x_1, \dotsc, x_n)\]
For \(n = 2\) , we commonly write \(\mathrm{d}^2 \boldsymbol{x}\) as \(\mathrm{d}S\) or \(\mathrm{d}A\) . For \(n = 3\) , we commonly write \(\mathrm{d}^3 \boldsymbol{x}\) as \(\mathrm{d}V\) .
Sometimes, a single \(\int\) sign is used.
For \(n = 2\) and \(n = 3\) , multidimensional Riemann integrals are also called double integrals and triple integrals , respectively.
Theorem: Linearity of Riemann Integrals
Let \(f: \mathcal{D}_f \subseteq \mathbb{R}^n \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R}^n \to \mathbb{R}\) be real scalar fields and let \(S \subseteq \mathcal{D}_f \cap \mathcal{D}_g\) be Jordan-measurable .
If \(f\) and \(g\) are Riemann-integrable on \(S\) , then so is \(\alpha f + \beta g\) for all \(\alpha, \beta \in \mathbb{R}\) :
\[\int \cdots \int_{S} (\alpha f + \beta g) (\boldsymbol{x}) \,\mathrm{d}^n \boldsymbol{x} = \alpha \int \cdots \int_{S} f (\boldsymbol{x}) \,\mathrm{d}^n \boldsymbol{x} + \beta \int \cdots \int_{S} g (\boldsymbol{x}) \,\mathrm{d}^n \boldsymbol{x}\]
Proof TODO
Theorem: Riemann integrals via Iterated Integrals
Let \(f: \mathcal{D} \subseteq \mathbb{R}^n \to \mathbb{R}\) be a real scalar field and let \(R = [a_1, b_1] \times \cdots \times [a_n, b_n] \subset \mathcal{D}\) be a compact rectangular region .
If \(f\) is continuous on \(R\) , then it is Riemann-integrable on \(R\) and its Riemann-integral is given by any of its iterated integrals :
\[\int \cdots \int_R f(x_1, \dotsc, x_n) \,\mathrm{d}(x_1, \dotsc, x_n) = \int_{a_{\sigma(1)}}^{b_{\sigma(1)}} \cdots \int_{a_{\sigma(n)}}^{b_{\sigma(n)}} f(x_1, \dotsc, x_n) \,\mathrm{d}x_{\sigma(n)} \cdots \mathrm{d}x_{\sigma(1)},\]
where \(\sigma\) is any permutation of \(\{1, \dotsc, n\}\) .
Example: \(\iint_{R} x + 3x^2 y^2 \,\mathrm{d}(x,y)\) with \(R = [0,1] \times [0,1]\) Consider the real scalar field \(f: \mathbb{R}^2 \to \mathbb{R}\) defined as
\[f(x, y) = x + 3x^2 y^2\]
and the compact rectangular region \(R = [0,1] \times [0,1] \subset \mathbb{R}^2\) .
We see that \(f\) is continuous on \(R\) . Therefore, we have the following for its double integral over \(R\) :
\[\begin{aligned} \iint_R f(x,y) \, \mathrm{d}(x,y) & = \iint_{R} x + 3x^2 y^2 \,\mathrm{d}(x,y) \\ & = \int_0^1 \int_0^1 x + 3x^2 y^2 \,\mathrm{d}y \,\mathrm{d}x \\ & = \int_0^1 \left.(xy + x^2y^3)\right\vert_{y=0}^{y=1} \,\mathrm{d}x \\ & = \int_0^1 x + x^2 \,\mathrm{d}x \\ & = \left.\left( \frac{1}{2}x^2 + \frac{1}{3}x^3 \right)\right\vert_{x = 0}^{x = 1} \\ & = \frac{5}{6}\end{aligned}\]
We could also use the other iterated integral :
\[\begin{aligned} \iint_R f(x,y) \, \mathrm{d}(x,y) & = \iint_{R} x + 3x^2 y^2 \,\mathrm{d}(x,y) \\ & = \int_0^1 \int_0^1 x + 3x^2 y^2 \,\mathrm{d}x \,\mathrm{d}y \\ & = \int_0^1 \left.\left(\frac{1}{2}x^2 + x^3y^2\right)\right\vert_{x=0}^{x=1} \,\mathrm{d}y \\ & = \int_0^1 \frac{1}{2} + y^2 \,\mathrm{d}y \\ & = \left.\left( \frac{1}{2}y + \frac{1}{3}y^3 \right)\right\vert_{y=0}^{y=1} \\ & = \frac{5}{6}\end{aligned}\]
Example: \(\iiint_{R} xyz \,\mathrm{d}(x,y,z)\) with \(R = [0,1]^3\) Consider the real scalar field \(f: \mathbb{R}^3 \to \mathbb{R}\) defined as
\[f(x, y, z) = xyz\]
and the compact rectangular region \(R = [0,1]^3 = [0,1] \times [0,1] \times [0,1] \subset \mathbb{R}^3\) .
We see that \(f\) is continuous on \(R\) . Therefore, we have the following for its triple integral over \(R\) :
\[\begin{aligned} \iiint_R f(x,y,z) \, \mathrm{d}(x,y,z) & = \iiint_{R} xyz \,\mathrm{d}(x,y,z) \\ & = \int_0^1 \int_0^1 \int_0^1 xyz \,\mathrm{d}z \,\mathrm{d}y \,\mathrm{d}x \\ & = \int_0^1 \int_0^1 \left.\left( \frac{1}{2}xyz^2 \right)\right\vert_{z=0}^{z=1} \,\mathrm{d}y \,\mathrm{d}x \\ & = \int_0^1 \int_0^1 \frac{1}{2}xy \,\mathrm{d}y \,\mathrm{d}x \\ & = \int_0^1 \left.\left( \frac{1}{4}xy^2 \right)\right\vert_{y=0}^{y=1} \,\mathrm{d}x \\ & = \int_0^1 \frac{1}{4}x \,\mathrm{d}x \\ & = \left.\left( \frac{1}{8}x^2 \right)\right\vert_{x=0}^{x=1} \\ & = \frac{1}{8}\end{aligned}\]
We could also use any of the other five possible iterated integrals .
\[\begin{aligned} \iiint_R f(x,y,z) \, \mathrm{d}(x,y,z) & = \int_0^1 \int_0^1 \int_0^1 xyz \,\mathrm{d}z \,\mathrm{d}x \,\mathrm{d}y \\ & = \int_0^1 \int_0^1 \left.\left( \frac{1}{2}xyz^2 \right)\right\vert_{z=0}^{z=1} \,\mathrm{d}x \,\mathrm{d}y \\ & = \int_0^1 \int_0^1 \frac{1}{2}xy \,\mathrm{d}x \,\mathrm{d}y \\ & = \int_0^1 \left.\left( \frac{1}{4}x^2y \right)\right\vert_{x=0}^{x=1} \,\mathrm{d}y \\ & = \int_0^1 \frac{1}{4}y \,\mathrm{d}y \\ & = \left.\left( \frac{1}{8}y^2 \right)\right\vert_{y=0}^{y=1} \\ & = \frac{1}{8}\end{aligned}\]
\[\begin{aligned} \iiint_R f(x,y,z) \, \mathrm{d}(x,y,z) & = \int_0^1 \int_0^1 \int_0^1 xyz \,\mathrm{d}y \,\mathrm{d}z \,\mathrm{d}x \\ & = \int_0^1 \int_0^1 \left.\left( \frac{1}{2}xy^2z \right)\right\vert_{y=0}^{y=1} \,\mathrm{d}z \,\mathrm{d}x \\ & = \int_0^1 \int_0^1 \frac{1}{2}xz \,\mathrm{d}z \,\mathrm{d}x \\ & = \int_0^1 \left.\left( \frac{1}{4}xz^2 \right)\right\vert_{z=0}^{z=1} \,\mathrm{d}x \\ & = \int_0^1 \frac{1}{4}x \,\mathrm{d}x \\ & = \left.\left( \frac{1}{8}x^2 \right)\right\vert_{x=0}^{x=1} \\ & = \frac{1}{8}\end{aligned}\]
\[\begin{aligned} \iiint_R f(x,y,z) \, \mathrm{d}(x,y,z) & = \int_0^1 \int_0^1 \int_0^1 xyz \,\mathrm{d}y \,\mathrm{d}x \,\mathrm{d}z \\ & = \int_0^1 \int_0^1 \left.\left( \frac{1}{2}xy^2z \right)\right\vert_{y=0}^{y=1} \,\mathrm{d}x \,\mathrm{d}z \\ & = \int_0^1 \int_0^1 \frac{1}{2}xz \,\mathrm{d}x \,\mathrm{d}z \\ & = \int_0^1 \left.\left( \frac{1}{4}x^2z \right)\right\vert_{x=0}^{x=1} \,\mathrm{d}z \\ & = \int_0^1 \frac{1}{4}z \,\mathrm{d}z \\ & = \left.\left( \frac{1}{8}z^2 \right)\right\vert_{z=0}^{z=1} \\ & = \frac{1}{8}\end{aligned}\]
\[\begin{aligned} \iiint_R f(x,y,z) \, \mathrm{d}(x,y,z) & = \int_0^1 \int_0^1 \int_0^1 xyz \,\mathrm{d}x \,\mathrm{d}z \,\mathrm{d}y \\ & = \int_0^1 \int_0^1 \left.\left( \frac{1}{2}x^2yz \right)\right\vert_{x=0}^{x=1} \,\mathrm{d}z \,\mathrm{d}y \\ & = \int_0^1 \int_0^1 \frac{1}{2}yz \,\mathrm{d}z \,\mathrm{d}y \\ & = \int_0^1 \left.\left( \frac{1}{4}yz^2 \right)\right\vert_{z=0}^{z=1} \,\mathrm{d}y \\ & = \int_0^1 \frac{1}{4}y \,\mathrm{d}y \\ & = \left.\left( \frac{1}{8}y^2 \right)\right\vert_{y=0}^{y=1} \\ & = \frac{1}{8}\end{aligned}\]
\[\begin{aligned} \iiint_R f(x,y,z) \, \mathrm{d}(x,y,z) & = \int_0^1 \int_0^1 \int_0^1 xyz \,\mathrm{d}x \,\mathrm{d}y \,\mathrm{d}z \\ & = \int_0^1 \int_0^1 \left.\left( \frac{1}{2}x^2yz \right)\right\vert_{x=0}^{x=1} \,\mathrm{d}y \,\mathrm{d}z \\ & = \int_0^1 \int_0^1 \frac{1}{2}yz \,\mathrm{d}y \,\mathrm{d}z \\ & = \int_0^1 \left.\left( \frac{1}{4}y^2z \right)\right\vert_{y=0}^{y=1} \,\mathrm{d}z \\ & = \int_0^1 \frac{1}{4}z \,\mathrm{d}z \\ & = \left.\left( \frac{1}{8}z^2 \right)\right\vert_{z=0}^{z=1} \\ & = \frac{1}{8}\end{aligned}\]
Proof TODO
Theorem: Domain Additivity of Riemann Integrals
Let \(S_1, \dotsc, S_p \subseteq \mathbb{R}^n\) be Jordan-measurable such that for all \(i \ne j\) , the intersection \(S_i \cap S_j\) is a null set of the Peano-Jordan measure .
A real scalar field \(f: \mathcal{D} \subseteq \mathbb{R}^n \to \mathbb{R}\) is Riemann-integrable on the union \(S = S_1 \cup \cdots \cup S_p \subseteq \mathcal{D}\) if and only if it is Riemann-integrable on each of \(S_1, \dotsc, S_p\) . In this case:
\[\int_{S} f(\boldsymbol{x}) \, \mathrm{d}^n \boldsymbol{x} = \sum_{i = 1}^p\int_{S_i} f(\boldsymbol{x}) \, \mathrm{d}^n \boldsymbol{x}\]
Proof TODO
Theorem: Riemann Integrals under Alterations
Let \(f: \mathcal{D}_f \subseteq \mathbb{R}^n \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R}^n \to \mathbb{R}\) be real scalar fields , let \(S \subseteq \mathcal{D}_f \cap \mathcal{D}_g\) be Jordan-measurable and let \(N = \{\boldsymbol{x} \in S \mid f(\boldsymbol{x}) \ne g(\boldsymbol{x})\}\) .
If \(f\) is Riemann-integrable on \(S\) , \(g\) is bounded on \(S\) and \(N\) is a null set of the Peano-Jordan measure , then \(g\) is also Riemann-integrable on \(S\) and its Riemann integral is the same:
\[\int_{S} f(\boldsymbol{x}) \, \mathrm{d}^n\boldsymbol{x} = \int_{S} g(\boldsymbol{x}) \, \mathrm{d}^n\boldsymbol{x}\]
Proof TODO
Theorem: Integration by Substitution
Let \(f: \mathcal{D}_f \subseteq \mathbb{R}^n \to \mathbb{R}\) be a real scalar field and let \(\Omega \subseteq \mathbb{R}^n\) be Jordan-measurable . Let \(\Phi: \mathcal{D}_{\Phi} \subseteq \mathbb{R}^n \to \mathcal{D}_f\) be a real vector field with the following properties:
Then \((f \circ \Phi)|\det J_{\Phi}|\) is Riemann-integrable on \(\Omega\) if and only if \(f\) is Riemann-integrable on \(\Phi(\Omega)\) . In this case:
\[\int_{\Omega} f(\Phi(\boldsymbol{x})) |\det J_{\Phi}(\boldsymbol{x})| \,\mathrm{d}^n \boldsymbol{x} = \int_{\Phi(\Omega)} f(\boldsymbol{u}) \, \mathrm{d}^n \boldsymbol{u}\]
Example Consider the real scalar field \(f: \mathbb{R}^2 \to \mathbb{R}\) defined as follows:
\[f(x,y) = x\]
Let \(B = \{\begin{bmatrix}x & y\end{bmatrix}^{\mathsf{T}} \in \mathbb{R}^2 \mid x^2 + y^2 \le 1, x \ge 0, y \ge 0\}\) .
We want to evaluate the Riemann integral of \(f\) over \(B\) :
\[\iint_B f(x,y) \, \mathrm{d}(x,y)\]
We notice that \(B\) is a quarter-circle. Specifically, it is the image of \(\Omega = [0,1] \times \left[0, \frac{\uppi}{2}\right]\) under the coordinate transformation \(\Phi: [0, +\infty] \times [0, 2\uppi] \to \mathbb{R}^2\) from polar coordinates :
\[\Phi(\rho, \varphi) = \begin{bmatrix} \rho \cos \varphi \\ \rho \sin \varphi\end{bmatrix}\]
\[B = \Phi(\Omega)\]
We see that the conditions in the theorem are met. Therefore:
\[\iint_B f(x,y) \, \mathrm{d}(x,y) = \iint_{\Omega} f(\Phi(\rho, \varphi)) |\det J_{\Phi}(\rho, \varphi)| \, \mathrm{d}(\rho,\varphi)\]
For the Jacobian matrix of \(J_{\Phi}\) , we have:
\[J_{\Phi}(\rho, \varphi) = \begin{bmatrix} \cos \varphi & -\rho \sin \varphi \\ \sin \varphi & \rho \cos \varphi \end{bmatrix}\]
Therefore:
\[|\det J_{\Phi}(\rho, \varphi)| = |\rho| = \rho\]
We have:
\[\begin{aligned}\iint_B f(x,y) \, \mathrm{d}(x,y) & = \iint_{\Omega} f(\Phi(\rho, \varphi)) |\det J_{\Phi}(\rho, \varphi)| \, \mathrm{d}(\rho,\varphi) \\ & = \iint_{\Omega} \rho \cos (\varphi) \rho \, \mathrm{d}(\rho,\varphi) \\ & = \iint_{\Omega} \rho^2 \cos \varphi \, \mathrm{d}(\rho,\varphi) \end{aligned}\]
Since \(\rho^2 \cos \varphi\) is continuous , we have:
\[\begin{aligned}\iint_{\Omega} \rho^2 \cos \varphi \, \mathrm{d}(\rho,\varphi) & = \int_0^{\frac{\uppi}{2}} \int_0^1 \rho^2 \cos \varphi \, \mathrm{d}\rho \, \mathrm{d}\varphi \\ & = \left( \int_0^{\frac{\uppi}{2}} \cos \varphi \, \mathrm{d}\varphi \right) \left( \int_0^1 \rho^2 \, \mathrm{d}\rho \right) \\ & = 1 \cdot \frac{1}{3} \\ & = \frac{1}{3}\end{aligned}\]
Proof TODO
Theorem: Riemann Integrals over Elementary Regions
Definition: Elementary Regions
An elementary region of \(\mathbb{R}^1\) is a compact interval .
An elementary region of \(\mathbb{R}^n\) (\(n \gt 1\) ) is a subset \(E \subset \mathbb{R}^n\) for which there exists a permutation \(\sigma\) of \(\{1, \dotsc, n\}\) such that \(E\) can be specified as
\[E = \left\{\begin{bmatrix}x_1 \\ \vdots \\ x_n \end{bmatrix} \in \mathbb{R}^n : \begin{bmatrix}x_{\sigma(1)} \\ \vdots \\ x_{\sigma(n-1)} \end{bmatrix}\in E' \text{ and } l\left( \begin{bmatrix}x_{\sigma(1)} \\ \vdots \\ x_{\sigma(n - 1)}\end{bmatrix}\right) \le x_{\sigma(n)} \le u\left( \begin{bmatrix}x_{\sigma(1)} \\ \vdots \\ x_{\sigma(n - 1)}\end{bmatrix}\right)\right\}\]
for some elementary region \(E'\) of \(\mathbb{R}^{n-1}\) and some continuous real scalar fields \(l, u: E' \to \mathbb{R}\) .
Let \(f: \mathcal{D} \subseteq \mathbb{R}^n \to \mathbb{R}\) be a real scalar field and let \(E \subseteq \mathcal{D}\) be an elementary region as defined above.
If \(f\) is continuous on \(E\) , then its Riemann integral over \(E\) exists and is given by the following Riemann integral of a parametric integral :
\[\int_E f(x_1, \dotsc, x_n) \, \mathrm{d}(x_1, \dotsc, x_n) = \int_{E'} \left( \int\limits_{\displaystyle l(x_{\sigma(1)}, \dotsc, x_{\sigma(n-1)})}^{\displaystyle u(x_{\sigma(1)}, \dotsc, x_{\sigma(n-1)})} f(x_1, \dotsc, x_n) \, \mathrm{d}x_{\sigma(n)}\right)\, \mathrm{d}(x_{\sigma(1)}, \dotsc, x_{\sigma(n-1)})\]
Example: \(\iint_{E} xy^2 \, \mathrm{d}(x,y)\) with \(E = \{(x,y) \in \mathbb{R}^2 \mid 0 \le x \le 1 + y^2, -1 \le y \le 1\}\) Consider the real scalar field \(f: \mathbb{R}^2 \to \mathbb{R}\) defined as follows:
\[f(x, y) = xy^2\]
We want to evaluate the Riemann integral
\[\iint_{E} f(x,y) \, \mathrm{d}(x,y) = \iint_{E} xy^2 \, \mathrm{d}(x,y),\]
where \(E\) is defined as follows:
\[E = \{(x,y) \in \mathbb{R}^2 \mid 0 \le x \le 1 + y^2, -1 \le y \le 1\}\]
We see that \(E\) is an elementary region of \(\mathbb{R}^2\) because
\[E = \{(x, y) \in \mathbb{R}^2 \mid y \in E' \text{ and } l(y) \le x \le u(y)\}\]
with \(E' = [-1, 1]\) (a compact interval and thus an elementary region of \(\mathbb{R}\) ) and with the continuous \(l,u: E' \to \mathbb{R}\) defined as \(l(y) = 0\) and \(u(y) = 1 + y^2\) .
Since \(f\) is continuous on \(E\) , we can apply the theorem:
\[\begin{aligned} \iint_{E} f(x,y) \, \mathrm{d}(x,y) & = \iint_{E} xy^2 \, \mathrm{d}(x,y) \\ & = \int_{E'}\left( \int_{l(y)}^{u(y)} f(x,y) \, \mathrm{d}x \right) \, \mathrm{d}y \\ & = \int_{-1}^{1} \left( \int_{0}^{1+y^2} xy^2 \, \mathrm{d}x \right) \, \mathrm{d}y \\ & = \int_{-1}^{1} \left( \left.\frac{1}{2}x^2 y^2\right\vert_{x=0}^{x=1+y^2} \right) \, \mathrm{d}y \\ & = \int_{-1}^{1} \frac{1}{2}(1+y^2)^2 y^2 \, \mathrm{d}y \\ & = \frac{1}{2} \int_{-1}^{1} (1 + 2y^2 + y^4) y^2 \, \mathrm{d}y \\ & = \frac{1}{2} \int_{-1}^{1} (y^2 + 2y^4 + y^6) \, \mathrm{d}y \\ & = \frac{1}{2} \left[ \frac{y^3}{3} + \frac{2y^5}{5} + \frac{y^7}{7} \right]_{-1}^{1} \\ & = \frac{1}{2} \left( \left( \frac{1}{3} + \frac{2}{5} + \frac{1}{7} \right) - \left( -\frac{1}{3} - \frac{2}{5} - \frac{1}{7} \right) \right) \\ & = \frac{1}{3} + \frac{2}{5} + \frac{1}{7} \\ & = \frac{35}{105} + \frac{42}{105} + \frac{15}{105} \\ & = \frac{92}{105} \end{aligned}\]
Example: \(\iint_S x^2 y \,\mathrm{d}(x,y)\) Consider the real scalar field \(f: \mathbb{R}^2 \to \mathbb{R}\) defined as follows:
\[f(x, y) = xy^2\]
We want to evaluate the Riemann integral
\[\iint_{S} f(x,y) \, \mathrm{d}(x,y) = \iint_{S} x^2y \, \mathrm{d}(x,y),\]
where \(S\) is defined as the subset bounded by the graphs of the functions \(y = 2\) , \(y = x\) and \(y = \frac{1}{2}\) :
While \(S\) is not itself an elementary region , it can be expressed as the union of two elementary regions \(E_1\) and \(E_2\) :
\[E_1 = \left\{(x,y) \in \mathbb{R}^2 \mid \frac{1}{2} \le x \le 1, \frac{1}{x} \le y \le 2 \right\}\]
\[E_2 = \left\{(x, y) \in \mathbb{R}^2 \mid 1 \le x \le 2, x \le y \le 2\right\}\]
Since \(S\) is Jordan-measurable and since \(f\) is continuous on \(S\) , the Riemann integral of \(f\) over \(S\) exists. Moreover, since
\[E_1 \cap E_2 = \{(x, y) \in \mathbb{R}^2 \mid x = 1, 1 \le y \le 2\}\]
is a null set of the Peano-Jordan measure , we can split the Riemann integral :
\[\iint_S f(x, y) \,\mathrm{d}(x, y) = \iint_{E_1} f(x, y) \,\mathrm{d}(x, y) + \iint_{E_2} f(x, y) \,\mathrm{d}(x, y)\]
We now use the fact that \(E_1\) is an elementary region
\[E_1 = \left\{(x,y) \in \mathbb{R}^2 \mid \frac{1}{2} \le x \le 1, \frac{1}{x} \le y \le 2 \right\}\]
with \(E_1' = \left[\frac{1}{2}, 1\right]\) , \(l_1(x) = \frac{1}{x}\) and \(u_1(x) = 2\) and the fact that \(E_2\) is an elementary region
\[E_2 = \left\{(x, y) \in \mathbb{R}^2 \mid 1 \le x \le 2, x \le y \le 2\right\}\]
with \(E_2' = \left[1, 2\right]\) , \(l_2(x) = x\) and \(u_2(x) = 2\) . Therefore:
\[\begin{aligned}\iint_{E_1} f(x,y) \,\mathrm{d}(x, y) + \iint_{E_2} f(x,y) \,\mathrm{d}(x, y) & = \int_{E_1'} \left( \int_{l_1(x)}^{u_1(x)} f(x, y) \, \mathrm{d}y \right) \,\mathrm{d}x + \int_{E_2'} \left( \int_{l_2(x)}^{u_2(x)} f(x, y) \, \mathrm{d}y \right) \,\mathrm{d}x \\ & = \int_{\frac{1}{2}}^{1} \left( \int_{\frac{1}{x}}^{2} x^2 y \, \mathrm{d}y \right) \,\mathrm{d}x + \int_{1}^{2} \left( \int_{x}^{2} x^2 y \, \mathrm{d}y \right) \,\mathrm{d}x \\ & = \int_{\frac{1}{2}}^{1} \left(\left.\frac{1}{2}x^2 y^2 \right\vert_{y = \frac{1}{x}}^{2} \right)\,\mathrm{d}x + \int_{1}^{2} \left(\left.\frac{1}{2}x^2 y^2 \right\vert_{y = x}^{2} \right) \,\mathrm{d}x \\ & = \int_{\frac{1}{2}}^{1} \left( 2x^2 - \frac{1}{2} \right) \,\mathrm{d}x + \int_{1}^{2} \left( 2x^2 - \frac{1}{2}x^4 \right) \,\mathrm{d}x \\ & = \left( \frac{2}{3}x^3 - \frac{1}{2}x \right)\Bigg|_{\frac{1}{2}}^{1} + \left( \frac{2}{3}x^3 - \frac{1}{10}x^5 \right)\Bigg|_{1}^{2} \\ & = \left[ \left(\frac{2}{3} - \frac{1}{2}\right) - \left(\frac{1}{12} - \frac{1}{4}\right) \right] + \left[ \left(\frac{16}{3} - \frac{32}{10}\right) - \left(\frac{2}{3} - \frac{1}{10}\right) \right] \\ & = \left[ \frac{1}{6} - \left(-\frac{1}{6}\right) \right] + \left[ \frac{32}{15} - \frac{17}{30} \right] \\ & = \frac{1}{3} + \frac{47}{30} \\ & = \frac{10}{30} + \frac{47}{30} \\ & = \frac{57}{30} \\ & = \frac{19}{10} \end{aligned}\]
Proof TODO
April 3, 2026 April 3, 2026 
