Parametrization Question

Question

When computing a line integral, or any integral that requires parametrization, what are you integrating with respect to?

For example, if parametrizing in polar coordinates, with $x=r\cos\theta$ and $y=r\sin\theta$, would you use

$\ dx=-r\sin \theta \ d\theta $ and $\ dy=r\cos\theta \ d\theta$,

or $\ dx \ dy = r \ dr \ d\theta$?

Or, would these be equivalent?

Thanks.

Answer 1

It depends on whether you are computing a line integral or a double integral over a region.

When computing a line integral, you parameterize the path with one variable. You are integrating w.r.t. that variable.

However, if you are doing a double (area) integral, then you parameterize the region in two variables, and you integrate w.r.t. these variables.

If you are computing a line integral around the circle $x^2+y^2 = R^2$ (where the radius of the circle, $R$, is constant), then we have $x = R\cos\theta$, $y = R\sin\theta$ where $0 \le \theta \le 2\pi$, and $dx = -R\sin\theta\,d\theta$, $dy = R\cos\theta\,d\theta$.

If you are computing a double integral over the disk $x^2+y^2 \le R^2$, then we have $x = r\cos\theta$, $y = r\sin\theta$, where $0 \le r \le R$, $0 \le \theta \le 2\pi$, and $\,dx\,dy = r\,dr\,d\theta$.

In either case, the resulting integral is integrating w.r.t. the parameter/parameters which you used to parameterize the curve/region.

Answer 2

You always integrate with respect to the parameters of the parametrization. For line integrals, we have one parameter, and for surfaces we have two parameters. In general, if we have a curve $\gamma(t) = (x(t), y(t), z(t))$, you can write: $$\mathrm{d}x = x'(t)~\mathrm{d}t \qquad \mathrm{d}y = y'(t)~\mathrm{d}t \qquad \mathrm{d}z = z'(t)~\mathrm{d}t$$ If you have a surface $\mathbf{x}(u,v) = (x(u,v), y(u,v), z(u,v))$, you can write: $$\mathrm{d}x~\mathrm{d}y = \frac{\partial(x,y)}{\partial(u,v)}~\mathrm{d}u~\mathrm{d}v \qquad \mathrm{d}x~\mathrm{d}z = \frac{\partial(x,z)}{\partial(u,v)}~\mathrm{d}u~\mathrm{d}v \qquad \mathrm{d}y~\mathrm{d}z = \frac{\partial(y,z)}{\partial(u,v)}~\mathrm{d}u~\mathrm{d}v $$ and solve the simple or double integral.