Objectives
After completing this section you should...
Be able to change coordinates of a double integral between Cartesian and polar coordinates
Section11.3Switching Coordinates: Cartesian to Polar
¶We now want to explore how to perform \(u\)-substitution in high dimensions. Let's start with a review from first semester calculus.
Review11.3.1
Consider the integral \(\ds\int_{-1}^4 e^{-3x} dx\text{.}\)
Let \(u=-3x\text{.}\) Solve for \(x\) and then compute \(dx\text{.}\)
Explain why \(\ds\int_{-1}^4 e^{-3x} dx=\int_{3}^{-12}e^u \left(-\frac{1}{3}\right)du\text{.}\)
Explain why \(\ds\int_{-1}^4 e^{-3x} dx=\int_{-12}^{3}e^u \left|-\frac{1}{3}\right| du\text{.}\)
If the \(u\)-values are between \(-3\) and \(2\text{,}\) what would the \(x\)-values be between? How does the length of the \(u\) interval \([-3,2]\) relate to the length of the corresponding \(x\) interval?
In the exercise above, we used a change of coordinates \(u=-3x\text{,}\) or \(x=-1/3 u\text{.}\) By taking derivatives, we found that \(dx=-\frac{1}{3}du\text{.}\) The negative means that the orientation of the interval was reversed. The fraction \(\frac13\) tells us that lengths \(dx\) using \(x\) coordinates will be \(1/3\)rd as long as lengths \(du\) using \(u\) coordinates. When we write \(dx = \frac{dx}{du}du\text{,}\) the number \(\frac{dx}{du}\) is called the Jacobian of \(x\) with respect to \(u\text{.}\) The Jacobian tells us how lengths are altered when we change coordinate systems. We now generalize this to polar coordinates. Before we're done with this section, we'll generalize the Jacobian to any change of coordinates.
Exercise11.3.2
Consider the polar change of coordinates \(x=r\cos\theta\) and \(y=r\sin\theta\text{,}\) which we could just write as
\begin{equation*} \vec T(r,\theta)=(r\cos\theta,r\sin\theta). \end{equation*}If you need a reminder of how to compute determinants, refer to Section 2.3.1
(a)
Compute the derivative \(D\vec T(r,\theta)\text{.}\) You should have a 2 by 2 matrix.
(b)
We need a single number from this matrix that tells us something about area. Determinants are connected to area.
Compute the determinant of \(D\vec T(r,\theta)\) and simplify.
The determinant you found above is called the Jacobian of the polar coordinate transformation. Let's summarize these results in a theorem.
Theorem11.3.1
If we use the polar coordinate transformation \(x=r\cos\theta, y=r\sin\theta\text{,}\) then we can switch from \((x,y)\) coordinates to \((r,\theta)\) coordinates if we use
\begin{equation*} dxdy=|r|drd\theta. \end{equation*}Ask me in class to give you an informal picture approach that explains why \(dxdy=rdrd\theta\text{.}\)
The number \(|r|\) is called the Jacobian of \(x\) and \(y\) with respect to \(r\) and \(\theta\text{.}\) If we require all bounds for \(r\) to be nonnegative, we can ignore the absolute value. If \(R_{xy}\) is a region in the \(xy\) plane that corresponds to the region \(R_{r\theta}\) in the \(r\theta\) plane (where \(r\geq 0\)), then we can write
\begin{equation*} \iint_{R_{xy}} f(x,y) dxdy = \iint_{R_{r\theta}} f(r\cos\theta,r\sin\theta) r\ drd\theta. \end{equation*}Subsection11.3.1Practice Changing Coordinates
We need some practice using this idea. We'll start by describing regions using inequalities on \(r\) and \(\theta\text{.}\)
Exercise11.3.3
For each region \(R\) below, draw the region in the \(xy\)-plane. Then give a set of inequalities of the form \(a\leq r\leq b, \alpha(r)\leq \theta \leq \beta(r)\) or \(\alpha\lt \theta\lt \beta, a(\theta)\leq r\leq b(\theta)\text{.}\) For example, if the region is the inside of the circle \(x^2+y^2=9\text{,}\) then we could write \(0\leq \theta\leq 2\pi\text{,}\) \(0\leq r\leq 3\text{.}\)
(a)
The region \(R\) is the quarter circle in the first quadrant inside the circle \(x^2+y^2=25\text{.}\)
(b)
The region \(R\) is below \(y=\sqrt{9-x^2}\text{,}\) above \(y=x\text{,}\) and to the right of \(x=0\text{.}\)
(c)
The region \(R\) is the triangular region below \(y=\sqrt 3 x\text{,}\) above the \(x\)-axis, and to the left of \(x=1\text{.}\)
Exercise11.3.4
Consider the opening exercise for this unit. We want to find the volume under \(f(x,y)=9-x^2-y^2\) where \(x\geq0\) and \(z\geq 0\text{.}\) We obtained the integral formula
\begin{equation*} \iint_R f dA = \ds\int_{y=-3}^{y=3} \int_{x=0}^{x=\sqrt{9-y^2}}9-x^2-y^2 dx dy. \end{equation*}See Sage.
(a)
Write bounds for the region \(R\) by giving bounds for \(r\) and \(\theta\text{.}\)
(b)
Rewrite the double integral as an iterated integral using the bounds for \(r\) and \(\theta\text{.}\) Don't forget the Jacobian (as \(dxdy=rdrd\theta\)).
(c)
Compute the integral in the previous part by hand. [Suggestion: you'll want to simplify \(9-x^2-y^2\) to \(9-r^2\) before integrating.]
Exercise11.3.5
Find the centroid of a semicircular disc of radius \(a\) (\(y\geq 0\)). Actually compute any integrals.
Exercise11.3.6
Compute the integral \(\ds \int_{0}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{2}{(1+x^2+y^2)^2}dydx\text{.}\)
Hint
try switching coordinate systems to polar coordinates. This will require you to first draw the region of integration, and then then obtain bounds for the region in polar coordinates.
We're now ready to define the Jacobian of any transformation.
Subsection11.3.2Computational Practice
These are provided to help you achieve better skills in basic computational answers.