Change of variables

Oftentimes we are faced with either a difficult to integrate function (single, but sometimes multivariable integration) or difficult to handle integral bounds (usually in double or triple integration). So we’re motivated to find a change of variables that will map our inputs to a new coordinate system (usually from rectangular).

For the single integral case, we usually discuss:

• U-substitution
• Trigonometric substitution

For the multi-integral case, our important cases are a change to:

• Polar coordinates (double)
• Cylindrical coordinates (triple)
• Spherical coordinates (triple)

General case

A key theorem is as follows. To transform from the $xy$-plane to the $uv$-plane, we can:

$${\iint }_{R}f(x,y)dA={\iint }_{S}f(g(u,v),h(u,v))|J(u,v)|dA$$

where $|J|$ is the Jacobian determinant:

$$J(u,v)=\left[\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right]$$

We sometimes aren’t able to change directly to a common coordinate system listed above. This forms a pretty large class of problems.

Strategies

We can aim for simple regions of integration in the $uv$-plane. Ideally we should be able to transform to something rectangular in $uv$.

We should also consider whether it’s easier to express $u$ and $v$ as functions of $x,y$ or $x$ and $y$ as functions of $u,v$. In the first case, it’s easiest to draw the region. In the second, it’s easiest to compute the Jacobian.

In one case, we can let the region in $uv$ suggest what new variables we can use. Ideally, we have a region bounded by two pairs of parallel curves, which we can map easily to rectangles.

Example problem

For a region bounded by the curves $y=\frac{1}{\sqrt{x}}$, $y=\frac{2}{\sqrt{x}}$, $y=\frac{1}{x}$, $y=\frac{2}{x}$, then:

• Let $u=y\sqrt{x}$
• Let $v=yx$

So our bounds are: $v\in [1,2],u\in [1,2]$. Do not forget to compute the Jacobian.

In the other, we can let the integrand determine what new variables. For example, if the integrand is:

$$f=\sqrt{\frac{x-y}{x+y+1}}$$

then $u=x-y$ and $v=x+y$ is a valid transformation.

Problems

Problems like the above are easily error-prone. We can use cylindrical coordinates where the top bound is something like $z=\sqrt{2=r^2}$ and lower $z=r^2$. $r$ and $\theta$ aren’t too hard to find.