In this section you will learn how to...
switch coordinate systems for integration (\(u\)-substitution for higher dimensions)
To find the Jacobian of the transformation \(\vec T(u,v)=(x(u,v), y(u,v))\text{,}\) we first find the derivative of \(\vec T\text{.}\) This is a square matrix, so it has a determinant, which should give us information about area. As the determinant may be positive or negative, we then take the absolute value to obtain the Jacobian. Formally:
Suppose \(\vec T(u,v)=(x(u,v),y(u,v))\) is a differentiable coordinate transformation. The Jacobian of the transformation \(\vec T\) is the absolute value of the determinant of the derivative. Notationally we write
\begin{equation*} J(u,v) = \frac{\partial (x,y)}{\partial (u,v)} = |\det(D\vec T(u,v))|. \end{equation*}For a tongue twister, say “the absolute value of the determinant of the derivative” ten times really fast.
Consider the transformation \(u=x+2y\) and \(v=2x-y\text{.}\)
Solve for \(x\) and \(y\) in terms of \(u\) and \(v\text{.}\)
Next compute the Jacobian \(\frac{\partial (x,y)}{\partial (u,v)}\text{.}\)
We were give \(u\) and \(v\) in terms of \(x\) and \(y\text{,}\) so we could have directly computed \(\frac{\partial (u,v)}{\partial (x,y)}\text{.}\) Do so now.
Make a conjecture about the relationship between \(\frac{\partial (x,y)}{\partial (u,v)}\) and \(\frac{\partial (u,v)}{\partial (x,y)}\text{.}\)
Suppose that \(f\) is integrable over a region \(R_{xy}\) in the \(xy\) plane. Suppose that \(\vec T(u,v)=(x(u,v),y(u,v))\) is a coordinate transformation that has the Jacobian \(\ds \frac{\partial (x,y)}{\partial (u,v)}\text{.}\) Suppose the region \(R_{uv}\) in the \(uv\)-plane corresponds to the region \(R_{xy}\) in the \(xy\)-plane. Provided the Jacobian is nonzero except possibly on regions with zero area, we can then write
\begin{equation*} \iint_{R_{xy}} f(x,y) dxdy = \iint_{R_{uv}} f(x(u,v),y(u,v)) \frac{\partial (x,y)}{\partial (u,v)} dudv. \end{equation*}We can remember this in differential form as
\begin{equation*} dxdy = \frac{\partial (x,y)}{\partial (u,v)} dudv. \end{equation*}Let's use this to rapidly find the area inside of an ellipse.
Consider the region \(R\) inside the ellipse \(\left(\dfrac{x}{a}\right)^2+\left(\dfrac{y}{b}\right)^2=1\text{.}\) We'll consider the change of coordinates given by \(u=(x/a)\) and \(v=(y/b)\text{.}\)
Draw the region \(R\) in the \(xy\)-plane.
After substituting \(u=x/a\) and \(v=y/b\text{,}\) draw the region \(R_{uv}\) in the \(uv\)-plane. You should have a circle.
What is the area inside this circle in the \(uv\)-plane?
Solve for \(x\) and \(y\text{,}\) and then compute the Jacobian \(\dfrac{\partial (x,y)}{\partial (u,v)}\text{.}\)
Show how to get the same result from directly computing \(\dfrac{\partial (u,v)}{\partial (x,y)}\text{.}\)
We know the area in the \(xy\)-plane of the ellipse is \(\iint_{R_{xy}} dxdy\text{.}\) Use the previous theorem to switch to an integral over the region \(R_{uv}\text{.}\) Then evaluate this integral by using facts about area to prove that the area in the \(xy\) plane is \(\pi a b\text{.}\)
HintLet \(R\) be the region in the plane bounded by the curves:
\begin{align*} x+2y\amp =1\\ x+2y\amp =4\\ 2x-y\amp =0\\ 2x-y\amp =8 \end{align*}We want to compute the integral \(\iint_R xdxdy\) using a change of coordinates. [Challenge: Try doing this without looking at the steps in the next paragraph]
Draw the region \(R\) in the \(xy\)-plane.
Use the change of coordinates \(u=x+2y\) and \(v=2x-y\) to evaluate this integral.
What are the bounds for \(u\) and \(v\text{?}\) You'll want to solve for \(x\) and \(y\) in terms of \(u\) and \(v\text{,}\) and then you'll need a Jacobian.
Make sure you provide a sketch of the region \(R_{uv}\) in the \(uv\)-plane (it should be a rectangle).
This is problem 7 in section 15.8.
Use the transformation \(u=3x+2y\) and \(v=x+4y\) to evaluate the integral
\begin{equation*} \iint_R (3x^2+14xy+8y^2)dxdy =\iint_R (3x+2y)(x+4y)dxdy \end{equation*}for the region \(R\) that is bounded by the lines \(y=-(3/2)x+1\text{,}\) \(y=-(3/2)x+3\text{,}\) \(y=-(1/4)x\text{,}\) and \(y=(-1/4)x+1\text{.}\)