Zmień język

English

Zaloguj się do Mathigon

lub

Nowe konto     Zresetuj hasło     

Multivariable Calculus
by Samuel S. Watson
Introduction
Taylor Series
Differentiation
Optimization
Matrix differentiation
The chain rule
Multivariable integration
Custom integration

Dzielić

Resetuj postęp

Spowoduje to usunięcie Twojego postępu i danych czatu dla wszystkich rozdziałów w tym kursie i nie będzie można tego cofnąć!

Słownik

Wybierz jedno ze słów kluczowych po lewej stronie…

Multivariable CalculusCustom integration

Czas czytania: ~10 min

If we want to integrate over a region which doesn't split nicely along lines parallel to the coordinate axes, we can split the region up along other lines or curves.

For example, consider integrating f(x,y) = y^2 over the region bounded by the hyperbolas xy = 1 and xy = 3 and the lines y = \frac{1}{2}x and y = 2x. This region naturally splits along hyperbolas of the form xy = v where v ranges from 1 to 3 and lines of the form y/x = u where u ranges from \frac{1}{2} to 2. One convenient way of describing this family of curves and lines we're using to slice up the region is to write the region as the image of the rectangle [\frac{1}{2},2] \times [1,3] under the inverse \mathbf{T} of the transformation (x,y) \mapsto (y/x,xy).

To integrate over the region in the xy-plane using this subdivision, we need to find the area of each small piece. Since the areas of the rectangles in the uv-plane are easier to deal with, we find the areas of the piece in the xy-plane by the area of the corresponding rectangle by the area distortion factor of the transformation \mathbf{T}. This local area distortion factor, or Jacobian determinant is the absolute value of the determinant of the Jacobian matrix \frac{\partial \mathbf{T}(\mathbf{x})}{\partial \mathbf{x}}. This is because the area distortion factor of a linear transformation is the absolute value of its determinant, and a differentiable function is essentially linear when zoomed far enough in. Thus we arrive at the formula

\begin{align*}\iint_D f(x,y) \, \mathrm{d}x\, \mathrm{d}y = \iint_R f(\mathbf{T}(u,v)) \left| \frac{\partial \mathbf{T}(u,v)}{\partial (u,v)} \right| \, \mathrm{d}u \, \mathrm{d}v,\end{align*}

where \left|\frac{\partial \mathbf{T}(u,v)}{\partial (u,v)} \right| is notation for the absolute value of the determinant of the transformation \frac{\partial \mathbf{T}(u,v)}{\partial (u,v)}.

Since the area distortion factor of \mathbf{T} is the of the area distortion factor of \mathbf{T}^{-1}, we can work out that \left|\frac{\partial \mathbf{T}(u,v)}{\partial (u,v)} \right| = \left|\begin{array}{cc} x & y \\\ 1/x & -y/x^2 \end{array} \right|^{-1} = (2y/x)^{-1} = \frac{1}{2v}, which implies

\begin{align*}\iint_D f(x,y) \, \mathrm{d}x\, \mathrm{d}y = \int_{1}^{3}\int_{1/2}^{2}uv \left(\frac{1}{2v}\right) \mathrm{d} u \, \mathrm{d}v = \boxed{3}.\end{align*}

Let's apply the Jacobian transformation idea to a familiar problem: the area enclosed by a circle.

Exercise
Use the map (u,v) \mapsto (u \cos v, u \sin v) from the rectangle [0,1] \times [0,2\pi] to the unit disk, and calculate the Jacobian for this transformation. Use your result to integrate 1 over the unit disk and confirm that the result is equal to the area of the unit disk.

Solution. The Jacobian of the given transformation is

\begin{align*}\left|\begin{array}{cc} \cos v & \sin v \\ -u\sin v & u \cos v \end{array} \right| = u(\cos^2 v + \sin^2 v) = u.\end{align*}

So the area of the disk is equal to

\begin{align*}\int_0^1 \int_0^{2\pi} u \, \mathrm{d}u \mathrm{d}v = \boxed{\pi}.\end{align*}

Congratulations! You have finished the Data Gymnasia multivariable calculus course.

Bruno
Bruno Bruno