Integral of $x^2$ over $x^2+y^2 ≤ a^2$

Question

This is in my lecture notes, I understand that $x=rcos\theta$ so therefore its the integral of $x^2=(rcos\theta)^2$ but why is there an r in $rdrd\theta$ in there?

Any help would be appreciated.

Answer 1

When we convert co-ordinates $(x,y)$ to $(r,\theta)$,

$dx\ dy = |J|\ dr \ d\theta$

where $J$ is the Jacobian of transformation.

$x = r \cos\theta \ , \ y = r \sin\theta$

Taking partial derivatives,

$x_r = \cos\theta \ , y_r = \sin\theta$

$x_{\theta} = - r \sin\theta \ , \ y_{\theta} = r \cos\theta $

$J = \begin{vmatrix} \cos\theta & \sin\theta \\ - r \sin\theta & r \cos\theta \end{vmatrix}$

$J = r(\cos\theta)(\cos\theta) + r(\sin\theta)(\sin\theta) = r(\cos^2\theta + \sin^2\theta) = r$

$$|J| = r$$ Thus, $$dx \ dy = r \ dr \ d\theta$$

Answer 2

Here's the underlying geometric intuition (not a rigorous argument).

In Cartesian coordinates the size of an infinitesimal rectangle with sides $dx$ and $dy$ is just the product $dxdy$, independent of where it is in the plane.

In polar coordinates, changing the angle changes the area more when you're farther from the origin. The area of a small not quite rectangular patch determined by $d\theta$ and $dr$ is $ dr \times r d\theta = r dr d\theta$.

There's a picture here: http://citadel.sjfc.edu/faculty/kgreen/vector/block3/jacob/node4.html

Answer 3

That is the formula for changing variables from cartesian coordinates to polar cordinates in double integrals $\mathrm d x\,\mathrm d y$ is replaced with $r\,\mathrm d r\,\mathrm d\theta$.

Intuitively, this come from the fact that the area of a small circular sector corresponding to a small increment $\mathrm d r$ of the radius and a small increment of the angle $\mathrm d\theta$ is approximately $\;\mathrm dr\cdot r\,\mathrm d\theta$