Multiple integrals · Definition
Polar coordinates
Before you start you need to know
They turn any region with circular symmetry into a simple integral, where in \(x,y\) coordinates it forces square roots and awkward limits.
Definition
Polar coordinates locate a point of the plane by its distance \(r\) from the origin and the angle \(\theta\) its radius makes with the positive \(x\)-axis.
The pair \((r,\theta)\) is related to \((x,y)\) by \(x=r\cos\theta\), \(y=r\sin\theta\), and conversely \(r^2=x^2+y^2\), \(\tan\theta=y/x\). The representation is not unique: \(\theta\) and \(\theta+2\pi\) give the same point, and the origin has \(r=0\) with any angle. When recovering \(\theta\) you must watch the quadrant, because \(\arctan(y/x)\) only returns angles between \(-\tfrac{\pi}{2}\) and \(\tfrac{\pi}{2}\).
A «polar rectangle» is \(\{a\le r\le b,\ \alpha\le\theta\le\beta\}\): a ring or a sector. There \(\iint_D f\,dA=\int_\alpha^\beta\!\int_a^b f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta\). The factor \(r\) is the Jacobian of the change of variables: a cell at distance \(r\) measures \(dr\) in the radial direction and \(r\,d\theta\) along the arc, so its area is not \(dr\,d\theta\).
Guiding example: \(\iint_D (x^2+y^2)\,dA\) over the unit disc is \(\int_0^{2\pi}\!\int_0^1 r^3\,dr\,d\theta=\tfrac{\pi}{2}\), whereas in Cartesian form it requires integrating \(\sqrt{1-x^2}\) and powers of it. With the same method, \(\iint_{\mathbb{R}^2}e^{-(x^2+y^2)}dA=\pi\), hence \(\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt\pi\). If the region is \(g_1(\theta)\le r\le g_2(\theta)\), the inner limits depend on \(\theta\), just as in a general region.
Typical mistakes: forgetting the factor \(r\), leaving \(x\) or \(y\) unsubstituted in the integrand, and choosing polar coordinates when neither the region nor the integrand is circular. This change of variables is the planar case of cylindrical coordinates, and both are examples of the general change of variables with the Jacobian.
Think about it
Describe the upper half-disc of radius \(R\) centered at the origin in polar coordinates.
\(0\le r\le R\), \(0\le\theta\le\pi\).
Why does \(\iint_D 1\,dA\) over the disc of radius \(R\) give \(\pi R^2\) and not \(2\pi R\)?
Because \(\int_0^R r\,dr=\tfrac{R^2}{2}\) includes the factor \(r\); without it you would get \(2\pi R\), the circumference, not the area.
Exercises 10
- Find the \(x\)-coordinate of the point with \(r=4\) and \(\theta=\tfrac{\pi}{3}\).
- The Cartesian point \((-1,1)\) written in polar form with \(r>0\) and \(0\le\theta<2\pi\) is:
- The circle \(x^2+y^2=9\) is written in polar coordinates as:
- Using \(dA=r\,dr\,d\theta\), compute the area of the disc of radius \(2\): \(\int_0^{2\pi}\!\int_0^2 r\,dr\,d\theta\).
- Compute \(\iint_D (x^2+y^2)\,dA\), where \(D\) is the unit disc \(x^2+y^2\le1\).
- Find the area of the half-disc \(0\le r\le3\), \(0\le\theta\le\pi\).
- The polar curve \(r=2\cos\theta\) is:
- Compute \(\iint_D x\,dA\), with \(D\) the quarter disc of radius \(2\) in the first quadrant.
- Integrating \(e^{-(x^2+y^2)}\) over the disc of radius \(a\) and letting \(a\to\infty\), the value of \(\iint_{\mathbb{R}^2}e^{-(x^2+y^2)}\,dA\) is:
- In \(dA=r\,dr\,d\theta\), why does the factor \(r\) appear?
References
- James Stewart, Daniel K. Clegg, Saleem Watson, Calculus: Early Transcendentals, 9th edition, Cengage, 2021. §10.3 y §15.3
- Joel Hass, Christopher Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr., Thomas' Calculus, 15th edition, Pearson, 2023. §15.4
- Jerrold E. Marsden, Anthony Tromba, Vector Calculus, 6th edition, W. H. Freeman, 2012. §6.2