Application: Polar Coordinates

Section 5.4 Application: Polar Coordinates

Cartesian coordinates such as discussed in Section 1.1 are not the only way to represent points in the plane. We can also represent them by indicating the distance, \(r\text{,}\) from the origin and the angle, \(\varphi\text{,}\) from the \(x\)-axis as shown in Figure 5.22.

Figure 5.22. Polar coordinates of a point in the plane

If we assume that \(r>0\) and \(\varphi\in[0,2\pi)\) then every point except the origin has the unique representation

\begin{align*} x \amp =r\cos\varphi\\ y \amp =r\sin\varphi\text{.} \end{align*}

Hence \(\mathbb R^2\) is the image of the strip \([0,\infty)\times[0,2\pi)\) under the transformation

\begin{equation*} \vect g(r,\varphi):=(r\cos\varphi,r\sin\varphi)\text{.} \end{equation*}

We want to apply the transformation formula from the previous section to find out how to integrate a function given in polar coordinates. To do so we need to compute the Jacobian determinant of \(\vect g\text{.}\) The Jacobian matrix is given by

\begin{align*} J_{\vect g}(r,\varphi)\amp = \begin{bmatrix} \dfrac{\partial}{\partial r}r\cos\varphi \amp \dfrac{\partial}{\partial\varphi}r\cos\varphi \\ \dfrac{\partial}{\partial r}r\sin\varphi \amp \dfrac{\partial}{\partial\varphi}r\sin\varphi \end{bmatrix}\\ \amp= \begin{bmatrix} \cos\varphi \amp -r\sin\varphi \\ \sin\varphi \amp r\cos\varphi \end{bmatrix}\text{.} \end{align*}

Hence the Jacobian determinant is

\begin{align*} \det J_{\vect g}(r,\varphi) \amp=\det \begin{bmatrix} \cos\varphi \amp -r\sin\varphi \\ \sin\varphi \amp r\cos\varphi \end{bmatrix}\\ \amp=r\det \begin{bmatrix} \cos\varphi \amp -\sin\varphi \\ \sin\varphi \amp \cos\varphi \end{bmatrix}\\ \amp=r(\cos^2\varphi+\sin^2\varphi) =r\geq 0\text{.} \end{align*}

Applying the transformation formula we see that

\begin{equation} \iint_{\vect g(D)}f(x,y)\,dx\,dy =\iint_D f(r\cos\varphi,r\sin\varphi)r\,dr\,d\varphi\text{.}\tag{5.8} \end{equation}

Example 5.23. Are of a disc.

Compute the area of a disc of radius \(R\text{.}\)

Solution.

A disc, \(D\text{,}\) with radius \(R\) is given in polar coordinates by

\begin{equation*} \{(r,\varphi)\colon 0\leq r\leq R, 0\leq\varphi\lt 2\pi\}\text{.} \end{equation*}

Hence by (5.2), (5.8) and Fubini’s theorem

\begin{align*} \area(D) \amp = \iint_D1\,dx\,dy = \int_0^R\int_0^{2\pi} r\,dr\,d\varphi\\ \amp = \left.\int_0^{2\pi}\frac{r^2}{2}\right|_0^R\,d\varphi\\ \amp = \int_0^{2\pi}\frac{R^2}{2}\,d\varphi =\pi R^2\text{.} \end{align*}

Example 5.24. Area of an ellipse.

Compute the area of the ellipse

\begin{equation*} \frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1\text{.} \end{equation*}

Solution.

We aim to find a function mapping a circle onto the given ellipse. If we set \(x=au\) and \(y=bv\) then

\begin{equation*} \frac{x^2}{a^2}+\frac{y^2}{b^2} =\frac{a^2u^2}{a^2}+\frac{b^2v^2}{b^2} =u^2+v^2=1, \end{equation*}

showing that the ellipse, \(E\text{,}\) is the image of the disc, \(D\text{,}\) given by \(u^2+v^2\leq 1\text{,}\) under the (linear) transformation

\begin{equation*} \vect g(u,v):=(au,bv)= \begin{bmatrix} a \amp 0 \\0\amp b \end{bmatrix} \begin{bmatrix} u \\v \end{bmatrix}\text{.} \end{equation*}

According to the transformation formula (5.7) we get

\begin{equation*} \area(E)=\iint_E1\,dx\,dy =\det \begin{bmatrix} a \amp 0 \\0\amp b \end{bmatrix} \iint_D1\,du\,dv =ab\iint_D1\,du\,dv \end{equation*}

Using the result from the previous example to compute the last integral we see that the area of the ellipse is \(ab\pi\text{.}\)

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