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Summary of polar coordinates

The polar coordinate system is a coordinate system to describe points \((r,\theta)\) in 2D using \(r\), the radial distance, and \(\theta\), the azimuthal angle.

Here, \(\theta\) is the angle measured from the \(x\)-axis to the \(y\)-axis, so anticlockwise. The point \(P\) in the sketch is the point \((r,\theta)\) with \(r=1\) and \(\theta=\frac{\pi}{3}\).
Now, when solving integrals with polar coordinates, it is crucial to remember these identities: $$\boxed{x=r\cos\theta \qquad y=r\sin\theta}$$ $$\boxed{x^2+y^2=r^2}$$ And whereas we have \(dA=dx\,dy\) in Cartesian coordinates, in polar coordinates we have $$\boxed{dA=\red{r}\,dr\,d\theta}$$ So, never forget to write this factor \(\red{r}\)! (It's called the Jacobian for the change from cartesian coordinates to polar coordinates...)

So, if we have the polar region $$D=\{(r,\theta)\mid \alpha\leq\theta\leq\beta,~ h_1(\theta)\leq r\leq h_2(\theta)\}$$ Then the integral can be written as $$\boxed{\iint_Df(x,y)dA=\int_\alpha^\beta\int_{h_1(\theta)}^{h_2(\theta)}f(r\cos\theta,r\sin\theta)\,\red r\,dr\,d\theta}$$ (Notice that \(f(r\cos\theta,r\sin\theta)\) just means: \(f(x,y)\), but rewritten in polar coordinates.)

Example 1

Question: calculate the volume of the solid body bounded by the function \(z=f(x,y)=x^4+2x^2y^2+y^4\) and the \(xy\)-plane above the circular region in the \(xy\)-plane given in the plot:

Solution: the region in the plot is just half a "donut": $$D=\{(r,\theta)\mid1\leq r\leq 2,~-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\}$$ (The bounds for theta are such because we start "in the bottom" at \(\theta=-\frac{\pi}{2}\) and end "at the top" at \(\theta=\frac{\pi}{2}\). Remember that \(\theta\) increases counterclockwise)

The integrand is \(x^4+2x^2y^2+y^4=(x^2+y^2)^2=(r^2)^2=r^4\), so we can write the integral to calculate the volume: $$V=\iint_D \blue{f(x,y)}\red{dA}=\int_{-\pi/2}^{\pi/2}\int_1^2 \blue{r^4} \red{r\,dr\,d\theta}=\int_{-\pi/2}^{\pi/2}d\theta\int_1^2 r^5 dr=\pi\left[\frac{1}{6}r^6\right]_1^2=\boxed{\frac{21}{2}\pi}$$ Here we used that in polar coordinates, \(dA=r\cdot dr\,d\theta\).

Conclusion: the volume is \(\frac{21}{2}\pi\).

Example 2

Question: calculate the volume of the solid body bounded by the function \(z=f(x,y)=y\sqrt{x^2+y^2}\) and the \(xy\)-plane above the region in the \(xy\)-plane given in the plot:

Solution: we want to calculate \(\displaystyle V=\iint_Dy\sqrt{x^2+y^2}dA\), where \(D\) is the region of integration. But this region of integration happens to be a bit complicated.

First we translate the problem to polar coordinates. The integrand is \(y\sqrt{x^2+y^2}=r\sin(\theta)\sqrt{r^2}=r^2\sin\theta\). Furthermore, the red curve is given by $$\sqrt{x^2+y^2}=3+\frac{2x}{\sqrt{x^2+y^2}}\quad\red{\rightsquigarrow}\quad\sqrt{r^2}=3+\frac{2r\cos\theta}{\sqrt{r^2}}\quad\red{\rightsquigarrow}\quad r=3+2\cos\theta$$ Let's now try to write the region of integration. We see that \(\theta\) goes from \(0\) to \(\pi\).

But how about \(r\)? Its lower bound is \(1\) everywhere (because \(x^2+y^2=1\) is a circle with radius \(1\), and this is where we "start"). But the upper bound is not given by a single function. At the right in the graph, the upper bound of \(r\) is given by the red curve (\(r=3+2\cos\theta\)), but at the left, it's given by the blue circle with \(r=2\).

We could write down the region of integration this way: $$D=\{(r,\theta) \mid 1\leq r\leq \max(2,~3+2\cos\theta),~0\leq\theta\leq\pi\}$$ But this would give us the integral \(\displaystyle\int_0^\pi\int_1^{\max(2,~3+2\cos\theta)} (r^2\sin\theta) r\,dr\,d\theta\) which looks not-so-nice.

So, let's use a new idea, which is to split up the region into two regions:

We see that in the region \(D_1\), we have \(1\leq r\leq3+2\cos\theta\), and in region \(D_2\), we have \(1\leq r\leq2\). The split-angle \(\theta_{\text{split}}\) as in the picture can be calculated by setting equal the radii of the curves that we want to find the intersection of: $$\red{3+2\cos\theta_{\text{split}}}=\blue2\quad\Rightarrow\quad\cos\theta_{\text{split}}=-\frac{1}{2}\quad\Rightarrow\quad \theta_{\text{split}}=\frac{2\pi}{3}$$ So for formality, let's write the regions of integration... $$D_1=\{(r,\theta)\mid0\leq\theta\leq\frac{2\pi}{3},~1\leq r\leq3+2\cos\theta\}$$ $$D_2=\{(r,\theta)\mid\frac{2\pi}{3}\leq\theta\leq\pi,~1\leq r\leq2\}$$ $$D=D_1\cup D_2$$ But look! Now we can just split up our integral and write it like this: \begin{align} V&=\iint_D f(x,y)dA=\iint_{D_1}f(x,y)dA+\iint_{D_2}f(x,y)dA\nonumber\\ &\qquad\textcolor{#de8231}{\text{\scriptsize ($*$ The integrand was $r^2\sin\theta$, also do not forget that $dA=\red r\,dr\,d\theta$ $*$)}}\nonumber\\ &=\int_0^{2\pi/3}\int_1^{3+2\cos\theta}(r^2\sin\theta)r\,dr\,d\theta+\int_{2\pi/3}^{\pi}\int_{1}^2(r^2\sin\theta) r\,dr\,d\theta\nonumber\\ &=\int_0^{2\pi/3}\sin\theta\int_1^{3+2\cos\theta}r^3dr\,d\theta+\int_{2\pi/3}^{\pi}\sin\theta d\theta\int_{1}^2r^3 dr\nonumber\\ &=\int_0^{2\pi/3}\sin\theta\left[\frac{r^4}{4}\right]_1^{3+2\cos\theta}d\theta+\left(\left[-\cos\theta\right]_{2\pi/3}^{\pi} \left[\frac{r^4}{4}\right]_1^2\right)\nonumber\\ &=\frac{1}{4}\int_0^{2\pi/3}\sin\theta\left((3+2\cos\theta)^4-1\right)d\theta+\left(\left[-\cos\theta\right]_{2\pi/3}^{\pi} \left[\frac{r^4}{4}\right]_1^2\right)\nonumber\\ &=\frac{1}{4}\left[-\frac{1}{10}(3+2\cos\theta)^5+\cos\theta\right]_0^{2\pi/3}+\frac{15}{8}\nonumber\\ &=\frac{1}{4}\left(-\frac{37}{10}+\frac{3115}{10}\right)+\frac{15}{8}=\boxed{\frac{3153}{40}}\nonumber\\ \end{align} Conclusion: the volume is \(\boxed{\frac{3153}{40}}\)

Example 3

Question: evaluate \(\displaystyle\int_{-\infty}^{\infty}e^{-x^2}dx\) by first evaluating \(\displaystyle\iint_{\mathbb{R}^2}e^{-x^2-y^2}dA\).

Solution: as we know, there is no "nice" antiderivative to \(e^{-x^2}\). So it seems like we have to follow the hint... let's first evaluate \(\displaystyle\iint_{\mathbb{R}^2}e^{-x^2-y^2}dA\).
The region of integration is \(\mathbb{R}\times\mathbb{R}=\mathbb{R}^2\), i.e., all points \((x,y)\). In polar coordinates, we cover all points by letting \(\theta\) run from \(0\) to \(2\pi\) and setting \(0\leq r <\infty\). The integrand can be rewritten in polar coordinates; \(e^{-x^2-y^2}=e^{-r^2}\). So we get: $$\iint_{\mathbb{R}^2}e^{-x^2-y^2}dA=\int_0^{2\pi}\int_0^\infty e^{-r^2}r\,dr\,d\theta=\int_0^{2\pi}d\theta\int_0^\infty re^{-r^2}dr$$ $$=2\pi\left[-\frac{1}{2}e^{-r^2}\right]_0^{\infty}=2\pi\left(0+\frac{1}{2}\right)=\pi$$ So we found that \(\displaystyle\int_{-\infty}^\infty\int_{-\infty}^{\infty}e^{-x^2-y^2}dydx=\pi\). Now, we split the integral, using that \(e^{-x^2-y^2}=e^{-x^2}\cdot e^{-y^2}\): $$\pi=\int_{-\infty}^\infty\int_{-\infty}^{\infty}e^{-x^2-y^2}dydx=\left(\int_{-\infty}^{\infty}e^{-x^2}dx\right)\left(\int_{-\infty}^{\infty}e^{-y^2}dy\right)$$ But look: those two integrals on the right MUST be equal. After all, they are not different, except for the fact that the variable of integration has a different name. In other words, we can just rename the \(y\) in the rightmost integral to \(x\) and nothing will change, so we get: $$\pi=\left(\int_{-\infty}^\infty e^{-x^2}dx\right)^2$$ Now the answer to the question becomes obvious: $$\boxed{\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}}$$ Realize this: there is no (nice) antiderivative for \(e^{-x^2}\), meaning we cannot evaluate \(\displaystyle\int_a^b e^{-x^2}dx\) for any \(a\) and \(b\) (except with advanced mathematics). But when \(a=-\infty\) and \(b=\infty\), then we suddenly can; and the magical answer is \(\sqrt\pi\). (Wow!)