5.28 Substitution in two variables
In this lecture we are going to introduce a technique to compute integrals of two variables, which is based in the substitution method in one variable.
So, let us recall first the result for functions in one variable.
Now let us consider the situation for functions on two variables.
Let \(g: S \rightarrow \mathbb{R}\) be an integrable function on a bounded set \(S \subseteq \mathbb{R}^2\) with area.
Suppose we want to compute the Riemann integral \(\int_S g\)
The idea is again to think that is something related to the composition of two functions: one function that we have to integrate and one function that has to do with a transformation of the domain.
Since we know how to integrate over rectangles, it would be great if we could find a function that transforms a rectangle in the set \(S \subseteq \mathbb{R}^2\).
Is this even possible?
To try to guess the formula, let us consider a rectangle \(R=[0,1]\times[0,1]\) and a linear function \(u:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}\) invertible
\(u(x,y)=(ax+cy,bx+dy)\), \(a,b,c,d\in\mathbb{R}\) where \(u(1,0)=(a,b)\) and \(u(0,1)=(c,d)\)
\(a(T_1)=\frac{a \cdot b}{2}\), \(a(T_2)=\frac{c \cdot d}{2}\), \(a(T_3)=\frac{(a-c)(d-b)}{2}\), \(a(R_2)=b(a-c)\)
\(a(T)=\frac{c \cdot d}{2}+\frac{(a-c)(d-b)}{2}\) \(+b(a-c)-\frac{a \cdot b}{2}\) \(=-\frac{b \cdot c}{2}+\frac{a \cdot d}{2}=\frac{a \cdot d-b \cdot c}{2}\)
\(a(T) = \frac{1}{2}(ad-bc)\) and \(a(D) = 2a(T) = ad-bc = \det \begin{pmatrix} a & b \\ c & d \end{pmatrix}\)
Who are the numbers \(a,b,c,d\)?
\(u(x,y) = (ax+cy, bx+dy)\) where \(s(x,y) = ax+cy\), \(s: \mathbb{R}^2 \to \mathbb{R}\) and \(t(x,y) = bx+dy\), \(t: \mathbb{R}^2 \to \mathbb{R}\)
\(\Rightarrow a = \frac{\partial s}{\partial x}\), \(c = \frac{\partial s}{\partial y}\), \(b = \frac{\partial t}{\partial x}\), \(d = \frac{\partial t}{\partial y}\) \(\Rightarrow a(D) = \det \begin{pmatrix} \frac{\partial s}{\partial x} & \frac{\partial t}{\partial x} \\ \frac{\partial s}{\partial y} & \frac{\partial t}{\partial y} \end{pmatrix}\) \(= \det \begin{pmatrix} \frac{\partial s}{\partial x} & \frac{\partial s}{\partial y} \\ \frac{\partial t}{\partial x} & \frac{\partial t}{\partial y} \end{pmatrix}\)
\(u(x,y) = (s(x,y), t(x,y))\)
Idea: \(u\) bijective and \(s\), \(t\) are diff. and \(\frac{\partial s}{\partial x},\frac{\partial s}{\partial y},\frac{\partial t}{\partial x},\frac{\partial t}{\partial y}\) exist, then we approximate \(s\) and \(t\) by the tangent plane.
Point \(p=(x_0, y_0)\)
\(s(x,y)\sim\frac{\partial s}{\partial x}(p)(x-x_{0})+\frac{\partial s}{\partial y}(p)(y-y_{0})+s(x_{0},y_{0})\)
\(t(x,y) \sim \frac{\partial t}{\partial x}(p)(x-x_0) + \frac{\partial t}{\partial y}(p)(y-y_0) + t(x_0, y_0)\)
\(s(x,y)-s(x_{0},y_{0})\sim\frac{\partial s}{\partial x}(p)(x-x_{0})+\frac{\partial s}{\partial y}(p)(y-y_{0})\)
\(t(x,y) - t(x_0,y_0) \sim \frac{\partial t}{\partial x}(p)(x-x_0) + \frac{\partial t}{\partial y}(p)(y-y_0)\)
linear map centered at \((x_0, y_0)\)
Idea: in small neighborhood, one can replace the function by the linear approximation
Therefore, the \(\textit{change of area}\) that we have to take into account is given by the \(\textit{Jacobian}\). By taking limits in the Riemann sum one
In case it is, we must take into account how the \(\text{area}\) changes: for functions of one variable we analyzed the length of an interval. Now, we are dealing with functions in two variables, so we should consider areas.
Definition
Let \(D,S\) be two domains (open and connected) sets in \(\mathbb{R}^2\).
Let \(u: D \rightarrow S\) be a function \(u(x,y) = (s(x,y), t(x,y))\) with \(s: D \rightarrow \mathbb{R}\) and \(t: D \rightarrow \mathbb{R}\).
Assume that the partial derivatives of \(s\) and \(t\) exist.
We define the Jacobian matrix of \(u\) at the point \(p\) by
The Jacobian of \(u\) at the point \(p\) is the determinant of the Jacobian matrix
Remarks
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Since by assumption the partial derivatives of the functions \(s: D \to \mathbb{R}\) and \(t: D \to \mathbb{R}\) exist, the Jacobian defines a function \(J_u: D \to \mathbb{R}\)
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To compute upper and lower sums of functions over a rectangle \(R\) we take into account partitions on smaller rectangles \(R_{ij}\)
Theorem
Let \(D, S \subseteq \mathbb{R}^2\) be two sets with area. Let \(u: D \to S\) be a bijective function and assume that \(u(x,y) = (s(x,y), t(x,y))\) with \(s: D \to \mathbb{R}\) and \(t: D \to \mathbb{R}\) differentiable.
Then for any function \(f: D \to \mathbb{R}\) continuous we have
Remark
This change the way we measure areas by Jacobian
Usually we want to compute \(\int_S f\) using the right hand side. So, in order to apply the theorem we need:
- Find a good bijective function \(u: D \to S\).
- Determine who is exactly \(D = u^{-1}(S)\).
- Compute the Jacobian \(J_u\).
- Compute the integral over \(D\).
Example
Let's compute the area of the circle of radius \(R>0\)
\(a(\overline{B(0,0),R}) = \pi R^2\), Our definition \(a(S) = \int_{S} 1\) where \(S = \left\{ (s,t) \in \mathbb{R}^2 \middle| s^2+t^2 \le R^2 \right\}\)
Consider the function \(u: (0, R] \times [0, 2\pi) \rightarrow \mathbb{R}^2\) where \(u(x, y) = (x \cos(y), x \sin(y))\)
\(s(x,y) = x \cos(y)\) and \(t(x,y) = x \sin(y)\)
\(\frac{\partial s}{\partial x} = \cos(y)\), \(\frac{\partial s}{\partial y} = -x \sin(y)\), \(\frac{\partial t}{\partial x} = \sin(y)\), \(\frac{\partial t}{\partial y} = x \cos(y)\), \(JM_{u}=\begin{pmatrix} \cos(y) & -x\sin(y) \\ \sin(y) & x\cos(y)\end{pmatrix}\)
Jacobian of \(x\), \(J_n(x,y) = \text{det} \begin{pmatrix} \frac{\partial s}{\partial x} & \frac{\partial s}{\partial y} \\ \frac{\partial t}{\partial x} & \frac{\partial t}{\partial y} \end{pmatrix}\) \(= \text{det} \begin{pmatrix} \text{cos}(y) & -x\text{sin}(y) \\ \text{sin}(y) & x\text{cos}(y) \end{pmatrix}=x\)
By our formula: \(\displaystyle \int_{S} 1 = \iint_{R} 1 \times dxdy = \int_{R} x\)
\(f(s,t)=1\) and \(f(u(x,y))=1\)
Now we use iterated integrals \(\displaystyle \int_{R} x = \int_{0}^{R} \int_{0}^{2\pi}x \, dy \, dx\)
Hypothesis: \(\displaystyle F(x) = \int_{0}^{2\pi}x \, dy = xy \big|_{0}^{2\pi}= 2\pi x\) exists for all x and is integrable.
By theorem \(\displaystyle = \int_{0}^{R} \left( \int_{0}^{2\pi}x \, dy \right) dx = \int_{0}^{R} 2\pi x \, dx = 2\pi \int_{0}^{R} x \, dx = 2\pi \frac{x^{2}}{2}\bigg|_{0}^{R} = \pi R^{2} \checkmark\)
In general, if \(S\) is a set with area and \(S = u(D)\) then \(\displaystyle a(S) = \int_{S} 1 = \iint_{D} |J_{u}(x, y)| \, dx \, dy\)
That is, we compute the area by integrating the absolute value of the Jacobian of \(u\).
This implies in general that \(a(S) \neq a(D)\), because \(\displaystyle a(D) = \int_{D} 1 = \iint_{D} 1 \, dx \, dy\)