Animated Solution for Mathematics - Definite Integration: Find the area bounded by the curves, $x^2+y^2=25,4y=\left|4-x^2\right|$ and $x=0$ above the $x$-axis.

Imagine you are standing on the Cartesian plane, looking at two distinct mathematical entities. On one hand, we have the circle $x^2+y^2=25$, a perfect, unchanging loop of radius $5$ centered at the origin.

On the other, we have the parabola $4y=|4-x^2|$, a shape that seems to change its behavior exactly where it crosses the $x$-axis. Our goal is to find the area trapped between these two curves, specifically above the $x$-axis.

The modulus function is often the source of student anxiety, but let us treat it as a simple switch. The equation $4y=|4-x^2|$ tells us that the parabola behaves differently depending on whether $x^2$ is less than or greater than $4$.

When $|x|\le 2$, the term $4-x^2$ is non-negative, so the parabola is defined as: This represents a downward-opening curve.

Once we cross the threshold of $|x|=2$, the expression $4-x^2$ becomes negative, and the modulus flips the sign, giving us: This represents an upward-opening curve. It is like a path that dips into a valley and then climbs a mountain.

To find the area, we must know where these curves meet. We focus on the right side, where $x>2$. We substitute the parabola's equation $y=\frac{x^2}{4}-1$ into the circle's equation $x^2+y^2=25$:

Expanding this, we get:

Multiplying by $16$ to clear the fractions, we arrive at the biquadratic equation:

Factoring it, we find:

Since $x^2$ cannot be negative, we must have $x^2=16$, which means $x=4$. Plugging this back into the parabola equation, we find $y=3$. Thus, our intersection point in the first quadrant is $(4,3)$.

Now, we set up our integral. The area $A$ is the integral of the upper curve minus the lower curve. For the right side, the upper curve is the circle $y=\sqrt{25-x^2}$, and the lower curve is the parabola.

Because the parabola changes definition at $x=2$, we split the integral:

The circle integral is a classic form:

Evaluating this from $0$ to $4$ gives $6+\frac{25}{2}{\sin }^{-1}(\frac{4}{5})$. The parabola integrals are simpler: the first part yields $\frac{4}{3}$, and the second yields $\frac{8}{3}$. Their sum is exactly $4$.

Subtracting the parabola's area from the circle's area, we get $2+\frac{25}{2}{\sin }^{-1}(\frac{4}{5})$ for the right side. Finally, we invoke the symmetry of the $y$-axis to account for the left side.

The total area is simply twice this value:

The final result is $4+25{\sin }^{-1}(\frac{4}{5})$. This is clean, precise, and deeply satisfying. You have navigated the modulus, solved the biquadratic, and mastered the integration.