The area of the area bounded by curve: $ \{ (x,y) \in \mathbb{R}^2 | x > 0, y < 0, x^4 -xy^2 - y^3 = 0 \} \cup { (0,0) }$, bad parametrization?

Show that the set:$$ \{ (x,y) \in \mathbb{R}^2 | x > 0, y < 0, x^4 -xy^2 - y^3 = 0 \} \cup { (0,0) }$$

is a closed curve. Calculate the area of the area bounded by this curve.

Hint: substitute: $y = -tx$ and try to apply Green's theorem. The result should be equal to $\frac{1}{120}$.

Showing that the curve is closed

Given the hint, we substitute $y=−tx$ into the equation $x^4 − xy^2 − y^3 = 0$ and get:$$x^4 − x(−tx)^2 − (−tx)^3 = 0 \iff x^3 (x − t^2 + t^3) = 0$$

Since $x \neq 0$ when we take the initial condition $x>0$, we get:$$x − t^2 + t^3 = 0 \iff x = t^2 − t^3$$

The parametric equations derived are:

• $x = t^2 − t^3$
• $y = −tx= −t(t^2 − t^3) = −t^3 + t^4$

Hence, we can express $(x,y)$ as: $(x,y) = (t^2 − t^3, −t^3 + t^4)$

Since $x = t^2 − t^3$ must be positive, we get: $t^2 − t^3 > 0 \iff t^2 (1−t) > 0$

Thus, $0<t<1$. When $t=0$ or $t=1$, $x=0$, which corresponds to the point $(0,0)$.

As $t$ approaches $0$ or $1$, both coordinates of $(x,y)$ approach $(0,0)$. Therefore, including the point $(0,0)$ makes the curve closed.

Calculating the area bounded by the curve

The parametric equations we derived for the curve are: (x,y)=(t2−t3,−t3+t4),0≤t≤1

To apply Green's theorem correctly, we set up our integrand. Green's theorem relates the area of a closed curve to a line integral around the curve:$A = \iint_C x \ dy$.

We will use Green's theorem in the form:$A = \frac{1}{2}​ \iint_C ​(x \ dy − y \ dx)$

First, we need to compute $dx$ and $dy$ from the parameterization - differentiating with respect to $t$:

• $dx = \frac{d​}{dt}(t^2 − t^3) = 2t − 3t^2 \ dt$
• $dy = \frac{d}{dt}(−t^3 + t^4) = −3t^2 + 4t^3 \ dt$

Now, we set up the line integral using the parameter $t$ ranging from $0$ to $1$:$$A = \frac{1}{2}​ \int_0^1​ \left[ (t^2 − t^3)(−3t^2 + 4t^3) − (−t^3 + t^4)(2t − 3t^2) \right] \ dt = $$$$ = \frac{1}{2}​ \int_0^1​ \left[ t^2(−3t^2 + 4t^3) − t^3(−3t^2 + 4t^3) + t^3(2t − 3t^2) - t^4(2t − 3t^2) \right] \ dt = $$$$ = \frac{1}{2}​ \int_0^1​ \left[ −3t^4 + 4t^5 + 3t^5 - 4t^6 + 2t^4 − 3t^5 -2t^5 + 3t^6 \right] \ dt = $$$$ = \frac{1}{2}​ \int_0^1​ \left[ −t^4 + 2t^5 - t^6 \right] \ dt $$

Now we can integrate each term and get:

• $\int_0^1 ​−t^4 \ dt = \left[−5 t^5 ​\right]_0^1 = -\frac{1}{5}$​
• $\int_0^1​ 2t^5 \ dt = \left[ \frac{2}{6}t^6​​\right]_0^1 = \frac{1}{3}$
• $\int_0^1 −t^6 \ dt = \left[ -\frac{1}{7} t^7​​\right]_0^1​ = -\frac{1}{7}$

From that we get:$$A = \frac{1}{2}​ \left( −\frac{1}{5} ​+ \frac{1}{3} ​− \frac{1}{7}​ \right) = -\frac{1}{210}$$

But why my area is $-\frac{1}{210}$ and not $\frac{1}{210}$ ?