Question

what is the area enclosed by the curve $x^{2}+(y-x^{2/3})^{2}=a^{2}$?
$\frac{a^{2}}{2}$
$\frac{pi a^{2}}{2}$
$a^{2}$
$pi a^{2}$
$\frac{3}{2}a^{2}$
$\frac{3}{2}pi a^{2}$
none of these options

Explanation:

Step1: Recognize curve type

The equation $x^2 + (y - x^{2/3})^2 = a^2$ describes a family of circles with radius $a$, where each circle is centered at $(x, x^{2/3})$ for all valid $x$. To find the enclosed area, we use Cavalieri's principle (or apply the concept of the area swept by a moving circle).

Step2: Determine bounds for x

For real $y$, the equation requires $x^2 \leq a^2$, so $x \in [-a, a]$.

Step3: Calculate enclosed area

The curve forms a shape where the area is equivalent to the area under the path of the circle's center plus the area of the circle itself? No—correctly, this is a cycloid-like swept area, but more simply: for each $x$, the vertical extent of the circle is $y = x^{2/3} \pm \sqrt{a^2 - x^2}$. The total enclosed area is the integral over $x=-a$ to $x=a$ of the vertical length of the circle at each $x$, which is $2\sqrt{a^2 - x^2}$.

$$ \text{Area} = \int_{-a}^{a} 2\sqrt{a^2 - x^2} dx $$

We know $\int_{-a}^{a} \sqrt{a^2 - x^2} dx$ is the area of a semicircle of radius $a$, so $\int_{-a}^{a} \sqrt{a^2 - x^2} dx = \frac{1}{2}\pi a^2$.

Step4: Compute final value

$$ \text{Area} = 2 \times \frac{1}{2}\pi a^2 = \pi a^2 $$

Answer:

$\pi a^2$ (matches the fourth option)