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I haven't been able to find an answer to something I've been thinking about.

If you are taking the integral of a circle in polar coordinates you always use the limits for theta as $0$ to $2\pi$. What if I decide to use $7\pi$ to $9\pi$, for example? I get the same solution but I believe this would cause problems on more complicated problems.

A situation I think this would cause problems in double integrals where, for example, the circle from $7\pi$ to $9\pi$ lies on a completely different region than the $0$ to $2\pi$ circle.

Am I thinking about this correctly or am I way off?

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In general, you will be OK integrating from $7\pi$ to $9\pi$. As $\theta$ roams over the interval $[7\pi,9\pi)$, precisely the same circle is swept out as when it roams over the interval $[0,2\pi)$. –  André Nicolas Jan 15 '12 at 21:44
    
But of course you need to be sure that your integrand is actually periodic in $\theta$ with period $2\pi$. (And if it isn't, you should look long and hard on whether integrating from $0$ to $2\pi$ is justified in the first place). –  Henning Makholm Jan 15 '12 at 21:55
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What do you mean by "the integral of a circle"? One usually integrates functions, not geometrical objects. If you mean the integral over a circle, then surely Andre's comment is right - or do you actually have an example where changing the limits of integration gives you a different answer? –  Gerry Myerson Jan 15 '12 at 21:57

1 Answer 1

up vote 2 down vote accepted

Polar coordinates are generally pretty robust. You can mess with $\theta$ all you want and things will still work out ok.

The transformation: $$ \begin{array}{ccc} x & = & r \cos(\theta) \\ y & = & r \sin(\theta) \end{array} $$ satisfies the necessary assumptions so the change of variables formula can be applied as long as $r>0$ and $\theta$ ranges over an interval of length $2\pi$. So one can use $[0,2\pi)$, $[-\pi,\pi)$, $[-\pi/2,3\pi/2)$, $[12345+\sqrt{3},12345+\sqrt{3}+2\pi)$, or $[a,a+2\pi)$ for any $a\in\mathbb{R}$. The only thing to watch out for is $r$. Remember that $r$ is the determinant of the Jacobian of this transformation, so the change of variables formula (for double integrals) tells us to insert the absolute value of the Jacobian -- which is $r$ since $r>0$. On the other hand if (for some really odd unexplained reason) you choose to let $r$ range over all real numbers and $\theta$ range over an interval of length $\pi$, you'll need to use "$|r|$" in the integral.

By the way, the same reasoning applies to triple integrals. When converting to cylindrical coordinates, you can let $\theta$ range over any interval of length $2\pi$ and all is fine (just don't mess with "$r>0$").

For spherical coordinates, if you use the "same" $\theta$ as in cylindrical coordinates you have $$ \begin{array}{ccc} x & = & \rho\cos(\theta)\sin(\phi) \\ y & = & \rho\sin(\theta)\sin(\phi) \\ z & = & \rho\cos(\phi) \end{array} $$ The standard domain for the spherical coordinates is $\rho>0$, $0 \leq \theta < 2\pi$, and $0 \leq \phi < \pi$. The determinant of the Jacobian of this transform is then $\rho^2\sin(\phi)$. Since $\sin(\phi)\geq 0$ when $0 \leq \phi < \pi$, there is no need for an absolute value when using the change of variables formula. This remains true if we change the interval $\theta$ ranges over to something else of the form $[a,a+2\pi)$. However, notice if we wanted to let $\phi$ range over $[0,2\pi)$ and $\theta$ over $[0,\pi)$, we would still have a valid change of coordinates, but our Jacobian determinant $\rho^2\sin(\phi)$ would now need absolute values since sine could take on negative values.

Long story short: Don't worry about changing $\theta$'s interval.

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