Seleccione una de las palabras clave de la izquierda ...

Multivariable CalculusCustom integration

Tiempo de leer: ~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.

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