Question

Determine the area of the shaded region bounded by the implicitly defined curve
x^2 = y^4( 1 - y^3 )

Answer 1

Suppose x ≥ 0. Then

x² = y⁴ (1 - y³)   ⇒   x = √(y⁴ (1 - y³)) = y² √(1 - y³)

On the other hand, if x < 0, then

x² = y⁴ (1 - y³)   ⇒   x = - y² √(1 - y³)

Either expression for x in terms of y describes half of the curve, to either side of the y-axis (i.e. the line x = 0).

Find where these two curves intersect:

y² √(1 - y³) = - y² √(1 - y³)

2y² √(1 - y³) = 0

2y² = 0   or   √(1 - y³) = 0

y = 0   or   1 - y³ = 0

y = 0   or   y³ = 1

y = 0   or   y = 1

So, the two halves meet at the points (0, 0) and (0, 1).

The area between the halves is then given by the integral

$\displaystyle \int_{y=0}^{y=1} \left(y^2\sqrt{1-y^3} - \left(-y^2\sqrt{1-y^3}\right)\right) \, dy$

Let's calculate it:

$\displaystyle \int_{y=0}^{y=1} 2y^2 \sqrt{1-y^3} \, dy$

Substitute z = 1 - y³ and dz = -3y² dy.

$\displaystyle \int_{z = 1 - 0^3}^{z = 1 - 1^3} -\frac23 \sqrt{z} \, dz$

$\displaystyle -\frac23 \int_{z = 1}^{z = 0} \sqrt{z} \, dz$

$\displaystyle \frac23 \int_{z = 0}^{z = 1} z^{1/2} \, dz$

$\displaystyle \frac23 \left(\frac23 z^{3/2}\right) \bigg|_{z=0}^{z=1}$

$\displaystyle \frac49 z^{3/2} \bigg|_{z=0}^{z=1}$

$\displaystyle \frac49 \cdot 1^{3/2} - 0 = \boxed{\frac49}$