Find the area bounded by the arc of the parabola y² =

Find the area bounded by the arc of the parabola $y^2 = 4a(a+x)$, the arc of the circle $x^2 + y^2 = 4a^2$, and the tangent to the circle at the point $(2a, 0)$.

We will work in the upper half-plane using symmetry, integrating with respect to $y$.

Step 1: Analyze the curves

Circle: $x^2 + y^2 = 4a^2$ is centered at the origin with radius $2a$.

Parabola: $y^2 = 4a(a+x)$ can be rewritten as:

\[y^2 = 4a^2 + 4ax\]

This is a parabola opening to the right with vertex at $(-a, 0)$.

Tangent at $(2a, 0)$: The circle has equation $x^2 + y^2 = 4a^2$. At the point $(2a, 0)$, the tangent to the circle is vertical (perpendicular to the radius). The tangent line is: $x = 2a$.

Step 2: Find intersection of parabola and circle

Setting $y^2 = 4a(a+x)$ into $x^2 + y^2 = 4a^2$:

\[x^2 + 4a(a+x) = 4a^2\]

\[x^2 + 4a^2 + 4ax = 4a^2\]

\[x^2 + 4ax = 0\]

\[x(x + 4a) = 0\]

So $x = 0$ or $x = -4a$. Since the circle has radius $2a$, points must satisfy $|x| \leq 2a$. Therefore, the relevant intersection is at $x = 0$.

When $x = 0$: $y^2 = 4a \cdot a = 4a^2$, so $y = \pm 2a$. The parabola and circle intersect at $(0, 2a)$ and $(0, -2a)$.

Step 3: Set up the region

By symmetry, we can find the area in the upper half-plane and double it. For the upper half ($y \geq 0$):

From parabola: $y^2 = 4a(a+x) \implies x = \frac{y^2}{4a} - a$

From circle: $x^2 + y^2 = 4a^2 \implies x = \sqrt{4a^2 - y^2}$ (right half)

From tangent: $x = 2a$

Step 4: Determine the region

The region is bounded by:

• Parabola arc from $(0, 2a)$ to some point
• Circle arc from some point to $(2a, 0)$
• Tangent line $x = 2a$ from $(2a, 0)$ back to the parabola

Let's find where the parabola intersects $x = 2a$:

\[y^2 = 4a(a + 2a) = 4a \cdot 3a = 12a^2\]

\[y = 2\sqrt{3}a \quad \text{(taking positive root)}\]

So the parabola intersects the tangent at $(2a, 2\sqrt{3}a)$.

Step 5: Set up integral with respect to $y$

For $y \in [0, 2a]$: the region extends from the parabola to the circle.

Parabola: $x_p = \frac{y^2}{4a} - a$. Circle: $x_c = \sqrt{4a^2 - y^2}$

Area of lower region (between parabola and circle from $y = 0$ to $y = 2a$):

\[A_1 = \int_0^{2a} \left(\sqrt{4a^2 - y^2} - \left(\frac{y^2}{4a} - a\right)\right) dy\]

For $y \in [2a, 2\sqrt{3}a]$: the region extends from the parabola to $x = 2a$.

\[A_2 = \int_{2a}^{2\sqrt{3}a} \left(2a - \left(\frac{y^2}{4a} - a\right)\right) dy\]

\[= \int_{2a}^{2\sqrt{3}a} \left(3a - \frac{y^2}{4a}\right) dy\]

Step 6: Evaluate $A_1$

\[A_1 = \int_0^{2a} \sqrt{4a^2 - y^2} \, dy - \int_0^{2a} \frac{y^2}{4a} \, dy + \int_0^{2a} a \, dy\]

First integral: This is a quarter circle of radius $2a$:

\[\int_0^{2a} \sqrt{4a^2 - y^2} \, dy = \frac{\pi (2a)^2}{4} = \pi a^2\]

Second integral:

\[\int_0^{2a} \frac{y^2}{4a} \, dy = \frac{1}{4a}\left[\frac{y^3}{3}\right]_0^{2a} = \frac{1}{4a} \cdot \frac{8a^3}{3} = \frac{2a^2}{3}\]

Third integral:

\[\int_0^{2a} a \, dy = a \cdot 2a = 2a^2\]

Therefore:

\[A_1 = \pi a^2 - \frac{2a^2}{3} + 2a^2 = \pi a^2 + \frac{4a^2}{3}\]

Step 7: Evaluate $A_2$

\[A_2 = \int_{2a}^{2\sqrt{3}a} \left(3a - \frac{y^2}{4a}\right) dy = \left[3ay - \frac{y^3}{12a}\right]_{2a}^{2\sqrt{3}a}\]

At $y = 2\sqrt{3}a$:

\[3a \cdot 2\sqrt{3}a - \frac{(2\sqrt{3}a)^3}{12a} = 6\sqrt{3}a^2 - \frac{8 \cdot 3\sqrt{3}a^3}{12a} = 6\sqrt{3}a^2 - 2\sqrt{3}a^2 = 4\sqrt{3}a^2\]

At $y = 2a$:

\[3a \cdot 2a - \frac{(2a)^3}{12a} = 6a^2 - \frac{8a^3}{12a} = 6a^2 - \frac{2a^2}{3} = \frac{16a^2}{3}\]

Therefore:

\[A_2 = 4\sqrt{3}a^2 - \frac{16a^2}{3}\]

Step 8: Total area

\[A = A_1 + A_2 = \pi a^2 + \frac{4a^2}{3} + 4\sqrt{3}a^2 - \frac{16a^2}{3}\]

\[= \pi a^2 + 4\sqrt{3}a^2 + \frac{4a^2 - 16a^2}{3} = \pi a^2 + 4\sqrt{3}a^2 - 4a^2\]

\[\boxed{A = a^2(\pi + 4\sqrt{3} - 4) \text{ square units}}\]

Or:

\[\boxed{A = a^2(4\sqrt{3} + \pi - 4) \approx a^2 \cdot 3.07 \text{ square units}}\]

Summary

The area bounded by the parabola $y^2 = 4a(a+x)$, the circle $x^2+y^2=4a^2$, and the tangent line $x = 2a$ is:

\[\boxed{A = a^2(4\sqrt{3} + \pi - 4) \approx 3.07\,a^2 \text{ square units}}\]

Key steps:

1. Find where the parabola meets the circle: $(0, \pm 2a)$
2. Find where the parabola meets the tangent line: $(2a, \pm 2\sqrt{3}a)$
3. Split the upper-half region at $y = 2a$ into two vertical strips in $y$
4. Evaluate each strip: $A_1 = \pi a^2 + \frac{4a^2}{3}$, $A_2 = 4\sqrt{3}a^2 - \frac{16a^2}{3}$
5. Add: $A = a^2(\pi + 4\sqrt{3} - 4)$

Key insight:

Because the tangent to a circle at a point on the $x$-axis is vertical, the region naturally splits into a circle-bounded piece and a line-bounded piece, both most easily integrated with respect to $y$.