Area of loop in graph

Question

The graph of

$$r = \frac{3\sec(θ)\tan(θ)}{1 + \tan^3(θ)}$$ is shown below. Find the area of the loop.

I know that the equation for finding the area enclosed under a polar curve is A = (1/2) ∫[a, b] [r(θ)]^2 dθ where r(θ) is the polar equation of the curve and [a, b] represents the interval of θ values that correspond to the desired region, but I am unsure about how I can find the values of a and b in the case of this graph. I think the equation is kind of daunting. And I'm honestly not even sure if that integral can apply to a case like this.

Could I have some help?

Answer 1

First determine where the curve self-intersects. Remember that in polar coordinates, $r$ is an indication of distance from the origin. From the plot, it's evident that the origin is where the self-intersection occurs, hence when $r=0$. Solve for $\theta$: $$\frac{3 \sec\theta \tan\theta}{1 + \tan^3\theta} = 0 \implies \sec\theta \tan \theta = 0 \implies \sin\theta=0 \implies \theta=n\pi,n\in\Bbb Z$$ which initially suggests $(a,b)$ should be an interval of length $\pi$, say $(0,\pi)$. However, we also observe that $$1+\tan^3\theta=0 \implies\tan^3\theta=-1 \implies \tan\theta =-1 \implies \theta=-\frac\pi4+n\pi$$ which means $r(\theta)$ is undefined if $\theta=\dfrac{3\pi}4$. Indeed, one draws the entire curve by letting $\theta\in\left(-\dfrac\pi4,\dfrac{3\pi}4\right)$.

The area must be finite, so we look for a removable discontinuity $r=\theta^*$ in our interval for which $\displaystyle\lim_{\theta\to\theta^*}r(\theta)=0$. Given that $\tan\dfrac\pi2$ is undefined, we can start our search there and luck out immediately, since

$$\lim_{\theta\to\frac\pi2} \frac{3\sec\theta\tan\theta}{1+\tan^3\theta} = \lim_{\theta\to\frac\pi2} \frac{3\sin\theta\cos\theta}{\sin^3\theta+\cos^3\theta} = 0$$

and hence the integral for the loop's area is

$$\frac12\int_0^{\tfrac\pi2} \frac{9\sec^2\theta\tan^2\theta}{\left(1+\tan^3\theta\right)^2}\,\mathrm d\theta=\frac12\int_0^{\tfrac\pi2} \frac{3\,\mathrm d\!\left(1+\tan^3\theta\right)}{\left(1+\tan^3\theta\right)^2}=\\[3pt]=\frac12\left[\frac{-3}{1+\tan^3\theta}\right]_0^{\tfrac\pi2}=\frac32\,.$$

Answer 2

Another way of calculation.

Changing to cartesian coordinates we have $x^3+y^3=3xy$ which gives a nodal cubic symmetric respect to the diagonal $y=x$. In order of calculate easier the integral we change of cartesian rotation of an angle of $-45^{\circ}$ so we have $$x=\frac{\sqrt2}{2}(x'+y')\\y=\frac{\sqrt2}{2}(x'-y'))$$ so we get the new equation $$\sqrt2(2x^3+6xy^2)=6(x^2-y^2)\iff y^2=\dfrac{3x^3-\sqrt2 x^3}{3\sqrt2x+3}$$ symmetric respect to the x-axis. Thus the area is $$2\int_0^{\dfrac{3}{\sqrt2}}\left(\frac{\sqrt{3x^2-\sqrt2 x^3}}{\sqrt{3\sqrt2x+3}}\right)dx=\frac32$$