Polar coordinates to find area

Question

Is there a way to use polar coordinates to find the area enclosed by the loop of the curve $y^2=(x+1)^2(3-x)$ instead of using cartesian? Wondering if there is a neater method.

Thanks!

Answer 1

I solved the equation

$$y = (x+1) \sqrt{3-x} \Rightarrow r(\theta) \cos (\theta) = (r(\theta) \sin (\theta) + 1) \sqrt{3 - r(\theta) \sin (\theta)}$$

to find $r(\theta)$... which was horrendously complicated. Nevertheless, when plotted in a radial plot gave:

which matched the direct rectilinear plot.

Don't try this by hand!!

And finding the area in polar coordinates is very difficult.

It is trivial in rectilinear:

$\int\limits_{x=-1}^3 \int\limits_{-(x+1)\sqrt{3-x}}^{(x+1)\sqrt{3-x}} 1 dy\ dx = {256 \over 15}$

---

@MOMO (below) suggests solving the cubic for $r$ (equivalent to what I did, above). Just for the record, here's a single branch of the three-branch solution:

$$r = \frac{1}{3} \cos (2 t) \sec ^3(t)+\frac{\sec ^3(t) \sqrt[3]{162 \cos ^6(t)+90 \cos (2 t) \cos ^4(t)+4 \cos ^3(2 t)+3 \sqrt{\frac{3}{2}} \sqrt{-976 \sin ^2(t) \cos (2 t) \cos ^6(t)-279 \sin ^2(t) \cos (4 t) \cos ^6(t)-793 \sin ^2(t) \cos ^6(t)}}}{3\ 2^{2/3}}-\frac{2^{2/3} \left(-15 \cos ^4(t)-\cos ^2(2 t)\right) \sec ^3(t)}{3 \sqrt[3]{162 \cos ^6(t)+90 \cos (2 t) \cos ^4(t)+4 \cos ^3(2 t)+3 \sqrt{\frac{3}{2}} \sqrt{-976 \sin ^2(t) \cos (2 t) \cos ^6(t)-279 \sin ^2(t) \cos (4 t) \cos ^6(t)-793 \sin ^2(t) \cos ^6(t)}}}$$

Answer 2

Put $x=r\cos t,y=r\sin t$ in your equation to get (after a little work and a trigonometric identity): $$r^3\cos ^3 t-r^2 \cos (2t)-5r\cos t = 3$$ From here you can find $r=f(t)$ using the solutions of a cubic equation (Notice that $f$ will not be single-valued for every $t$). Then you can find the area by solving the integral $\frac{1}{2}\int_0^{2\pi}g(t)^2\text{d}t\ $ where $g(t)$ is the smaller value between the 2 (at most) possible values of $f(t)$.