Animated Solution for Mathematics - Definite Integration: Area of the region $\{(x,y):x^2+(y-2{)}^2\le 4,x^2\ge 2y\}$ is

First, let's ground ourselves. The equation $x^2+(y-2{)}^2=4$ describes a circle with center $(0,2)$ and radius $2$. It sits perfectly on the origin $(0,0)$ and reaches up to $(0,4)$.

Now, consider the parabola $y=\frac{x^2}{2}$. It starts at the origin and opens upwards.

The problem asks for the region where $x^2+(y-2{)}^2\le 4$ (inside the circle) and $x^2\ge 2y$ (which is equivalent to $y\le \frac{x^2}{2}$, or the region below the parabola). When you sketch these, you will see a lens-like shape trapped between the parabola and the bottom arc of the circle.

To calculate the area, we must find the intersection points. We substitute $x^2=2y$ into the circle's equation:

Expanding this, we get $2y+y^2-4y+4=4$. The constants cancel out, leaving us with $y^2-2y=0$.

Factoring this gives $y(y-2)=0$. Our intersection points occur at $y=0$ and $y=2$. When $y=0$, $x=0$. When $y=2$, $x^2=4$, so $x=\pm 2$.

The area $A$ between two curves is the integral of the upper curve minus the lower curve. Our upper boundary is the parabola $y=\frac{x^2}{2}$.

From $(y-2{)}^2=4-x^2$, we get $y-2=\pm \sqrt{4-x^2}$. Since we need the lower arc of the circle, we take $y=2-\sqrt{4-x^2}$.

Thus, our integral becomes:

We can split this into two manageable pieces:

Let's tackle the first integral, $I_1={\int }_{-2}^2\sqrt{4-x^2}dx$. This represents the area of a semicircle with radius $2$.

Using the geometric formula for the area of a semicircle, $\frac{1}{2}\pi r^2$, with $r=2$, we get:

Now for the second integral, $I_2={\int }_{-2}^2(2-\frac{x^2}{2})dx$. Integrating term by term, we get:

Evaluating at the limits:

Finally, we subtract the two results:

The final area is $2\pi -\frac{16}{3}$.