## Area and Perimeter of Astroids

tags: Calculus

It is about time I revisited my first adventure Envelopes and Astroids. In the end of the post, I named the envelope of the line segments an Astroid. I will now find two basic properties of this shape: area and perimeter.

Definition 1. For a given real number $$l$$, the curve determined by $x^{\frac{2}{3}} + y^{\frac{2}{3}} = l^{\frac{2}{3}}$ is called an $$l$$-astroid.

Theorem 2. The area of an $$l$$-astroid is $$\dfrac{3\pi l^2}{8}$$.

Note: The proof relies on a fact on the Jacobian Determinant, $$J$$, (somehow) represents the change in size when variables are changed. Something I learnt in Calculus III (multivariable calculus) but it was never proven in class. I'll look into it.

Proof: The equation has the following parametric form: \begin{aligned} x = l \sin^3(\alpha) \\ y = l \cos^3(\alpha) \end{aligned} This suggests a polar-like coordinate switch: \begin{aligned} x = r \sin^3(\alpha) \\ y = r \cos^3(\alpha) \end{aligned} Let $$A$$ denote the region bounded by the astroid. So we have $\text{Area} = \iint \limits_{A} \,dx dy = \int \limits_0^{2\pi} \int \limits_0^l |J| \,dr d\alpha$ where $J = \left|\begin{matrix} \frac{dx}{dr} & \frac{dx}{d\alpha} \\ \frac{dy}{dr} & \frac{dy}{d\alpha}\end{matrix}\right| = \frac{3}{4} r \sin^2(2\alpha)$

Therefore, \begin{aligned} \text{Area} &= \int \limits_0^{\frac{\pi}{2}} \int \limits_0^l 3 r \sin^2(2\alpha) \,dr d\alpha \\ &= \frac{3l^2}{2} \int_0^{\frac{\pi}{2}} \sin^2(2\alpha) \,d\alpha \\ &= \frac{3\pi l^2}{8} \end{aligned}

$$\Box$$

Theorem 3. The perimeter of an $$l$$-astroid is $$6l$$.

Proof: For this I want to use the integral of a curve's length: $\text{length} = \int \sqrt{\frac{dy}{dx}^2+1} \, dx$

However, this integral assumes a proper function, i.e., a unique $$y$$-value for every $$x$$. So, I'll restrict myself to the first quadrant then multiply the result by $$4$$ since one quadrant contains exactly one quarter of the astroid. Also, recall from the last post that the astroid satisfies the following equation: $\frac{dy}{dx} = - \left(\frac{y}{x}\right)^{\frac{1}{3}}$ So, \begin{aligned} \text{Perimeter} &= 4 \int_0^l \sqrt{\left(\frac{y}{x}\right)^{\frac{2}{3}}+1} \, dx \\ &= 4 \int_0^l x^{-\frac{1}{3}} \sqrt{ y^{\frac{2}{3}}+x^{\frac{2}{3}}} \, dx \\ &= 4l^{\frac{1}{3}} \int_0^l x^{-\frac{1}{3}} \, dx \\ &= 6l \end{aligned}

$$\Box$$

Note: Informal justification for the integral of a curve's length.. $P = \sum \sqrt{\Delta y^2+ \Delta x^2} = \sum \sqrt{\left(\frac{\Delta y}{\Delta x}\right)^2+1}\, \Delta x = \int \sqrt{y'^2+1} \, dx$

I wanted to wrap up the post by giving a proof on the Envelope Theorem I used in the last post but I'm afraid I would have to introduce the basics of multivariable calculus first. I'll postpone it for later (again). But for now, I'll leave you with a problem to play with.

From a fixed point in on earth, projectiles are thrown at various angles. Assuming gravity is constant and no air resistance, show that the envelope of these projectiles is an Elliptic Paraboloid. Then use this to solve Problem 317 on Project Euler without writing a program.