Area under the curve without finding relations in $x$ and $y$

Question

Suppose that we are given a point $P$ whose $x$ and $y$ coordinates are given by $$x=f(\theta)$$$$y=g(\theta)$$where $\theta$ is the angle made with the positive $x$ axis in anti-clockwise direction. So basically, the point $P$ moves in the $xy$ plane as $\theta$ varies. I am intersected in finding the area under the locus of point $P$ in some interval.

One way of doing so is to find a function $h$ such that $y=h(x)$ and then integrating this function in that interval. However, is there any other way to find the area without finding $h$?

Any help will be highly appreciated!

Answer 1

I'm assuming that $\theta$ is the polar angle of the running point $(x,y)$. Given two $\theta$-values $\theta_1<\theta_2$ the integral $${1\over2}\int_{\theta_1}^{\theta_2} r^2(\theta)\>d\theta={1\over2}\int_{\theta_1}^{\theta_2} \bigl(f^2(\theta)+g^2(\theta)\bigr)\>d\theta$$ gives the area of the sector centered at $(0,0)$ created by this part of the curve. The area you want differs by some triangle areas from this sector area.

Answer 2

Consider the triangle with a verices at the origin, $(x,y)$ and $(x+\Delta x,y+\Delta y)$, in that orientation. Its signed area is equal to $$\frac12\begin{vmatrix}x&x+\Delta x\\y&y+\Delta y\end{vmatrix} = \frac12(x\Delta y-y\Delta x).$$ Passing from the Riemann sum to an integral, we then have for the area swept out along the curve $C$ the line integral $\frac12\int_Cx\,dy-y\,dx$. (I should really show that the Riemann sum converges, but I’m going to skip that in the interest of brevity.) If you’re familiar with differential forms, we have $d\left(\frac12(x\,dy-y\,dx)\right)=dx\wedge dy$, so being able to use this expression to compute area is a consequence of Stokes’ theorem. You can in fact use any primitive of the area element $dx\wedge dy$: alternatives that might be more convenient are $x\,dy$ and $-y\,dx$. Applying your parameterization to this, we obtain via the chain rule $$\frac12\int_Cx\,dy-y\,dx = \frac12\int_{\theta_1}^{\theta_2}f(\theta)g'(\theta)-f'(\theta)g(\theta)\,d\theta.$$ To take a simple example, let $x=\cos \theta$ and $y=\sin\theta$. The area of the sector from $\theta_1$ to $\theta_2$ is then $$\frac12\int_{\theta_1}^{\theta_2}\cos^2\theta+\sin^2\theta\,d\theta = \frac12(\theta_2-\theta_1),$$ as expected.