# Line integral of a vector field defined in polar coordinate

For example, how to find

$$\vec I = \int_0^{\pi/2} \theta \hat{\theta} d\theta$$

where $$\hat{\theta}$$ is the azimuthal unit vector. The integrand is a vector field in azimuthal direction whose magnitude increases with $$\theta$$, and the integral is counter-clockwise along the upper-right quarter of a unit circle.

The trouble is that $$\hat{\theta}$$ itself varies with position. As a result the final answer cannot be something like $$A \hat{\theta}$$, it must also prescribe which $$\hat{\theta}$$ vector.

Of course, we can always resort to the omnipotent Cartesian coordinate. With $$\hat \theta \cdot \hat i = -\sin\theta, \hat \theta \cdot \hat j = \cos\theta$$ We can proceed component-wise $$\vec I = \left(\int_0^{\pi/2} \theta \hat{\theta}\cdot \hat i d\theta \right)\hat i + \left(\int_0^{\pi/2} \theta \hat{\theta}\cdot \hat j d\theta \right)\hat j = -\hat i+\left(\frac{\pi}{2}-1 \right) \hat j$$ It's not really satisfying to jump out of the polar coordinate just to do an integration.

How to find $$\vec I$$ within polar coordinate?

## 1 Answer

You could do integration by parts

$$\int_0^{\frac{\pi}{2}} \theta \hat{\theta}d\theta = \theta \hat{r}\Bigr |_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \hat{r}d\theta = \frac{\pi}{2}\hat{r}\left(\frac{\pi}{2}\right) + \hat{\theta}\left(\frac{\pi}{2}\right) - \hat{\theta}(0)$$

But as you can see, we have stumbled on a fundamental problem of using polar unit vectors: they don't uniquely describe constant vectors and are entirely useless for doing so.

$$\hat{r}\left(\frac{\pi}{2}\right) = \hat{\theta}(0) = \hat{y}$$

Once you plug in numbers for the polar unit vectors, it's game over. There is no point in continuing to use polar coordinates. Polar unit vectors are better used for vector fields, whose values change at different points in space.