Let $$\Delta$$ be the area of the region $$\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 21, y^2 \leq 4x, x \geq 1\}$$. Then $$\frac{1}{2}\left(\Delta - 21\sin^{-1}\frac{2}{\sqrt{7}}\right)$$ is equal to

To find the area of the given region, let's first analyze the curves bounding it:

1. A circle: $$x^2 + y^2 \leq 21$$. This is a circle centered at the origin with radius $$r = \sqrt{21}$$.

2. A parabola: $$y^2 \leq 4x$$. This is a right-opening parabola with vertex at $$(0,0)$$.

3. A vertical line: $$x \geq 1$$.

Step 1: Find the points of intersection

We need to find where the parabola $$y^2 = 4x$$ and the circle $$x^2 + y^2 = 21$$ intersect.

Substitute $$y^2 = 4x$$ into the circle equation:

$$x^2 + 4x = 21$$

$$x^2 + 4x - 21 = 0$$

$$(x + 7)(x - 3) = 0$$

Since we are bounded by $$x \geq 1$$, we discard $$x = -7$$ and take $$x = 3$$.

At $$x = 3$$, $$y^2 = 4(3) = 12 \implies y = \pm \sqrt{12} = \pm 2\sqrt{3}$$.

The intersection points in the region $$x \geq 1$$ are $$(3, 2\sqrt{3})$$ and $$(3, -2\sqrt{3})$$.

Step 2: Set up the integral for the area ($$\Delta$$)

The region is symmetric with respect to the x-axis, so we can calculate the area in the first quadrant and multiply by $$2$$.

In the first quadrant, the region is bounded:

From $$x = 1$$ to $$x = 3$$, the upper boundary is the parabola $$y = \sqrt{4x} = 2\sqrt{x}$$.

From $$x = 3$$ to $$x = \sqrt{21}$$, the upper boundary is the circle $$y = \sqrt{21 - x^2}$$.

Therefore, the total area $$\Delta$$ is:

$$\Delta = 2 \left( \int_{1}^{3} 2\sqrt{x} \, dx + \int_{3}^{\sqrt{21}} \sqrt{21 - x^2} \, dx \right)$$

Step 3: Evaluate the integrals

First integral (Parabola):

$$\int_{1}^{3} 2\sqrt{x} \, dx = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{1}^{3} = \frac{4}{3} [x\sqrt{x}]_{1}^{3}$$

$$= \frac{4}{3} (3\sqrt{3} - 1) = 4\sqrt{3} - \frac{4}{3}$$

Second integral (Circle):

Using the standard formula $$\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$$, where $$a = \sqrt{21}$$:

$$\int_{3}^{\sqrt{21}} \sqrt{21 - x^2} \, dx = \left[ \frac{x}{2}\sqrt{21 - x^2} + \frac{21}{2}\sin^{-1}\left(\frac{x}{\sqrt{21}}\right) \right]_{3}^{\sqrt{21}}$$

Upper limit ($$x = \sqrt{21}$$):

$$\frac{\sqrt{21}}{2}(0) + \frac{21}{2}\sin^{-1}(1) = \frac{21}{2}\left(\frac{\pi}{2}\right) = \frac{21\pi}{4}$$

Lower limit ($$x = 3$$):

$$\frac{3}{2}\sqrt{21 - 9} + \frac{21}{2}\sin^{-1}\left(\frac{3}{\sqrt{21}}\right) = \frac{3}{2}\sqrt{12} + \frac{21}{2}\sin^{-1}\left(\sqrt{\frac{9}{21}}\right) = 3\sqrt{3} + \frac{21}{2}\sin^{-1}\left(\sqrt{\frac{3}{7}}\right)$$

So, the second integral evaluates to:

$$\frac{21\pi}{4} - 3\sqrt{3} - \frac{21}{2}\sin^{-1}\left(\sqrt{\frac{3}{7}}\right)$$

We can use a trigonometric identity to relate $$\sin^{-1}\left(\sqrt{\frac{3}{7}}\right)$$to the term in the final expression,$$\sin^{-1}\left(\frac{2}{\sqrt{7}}\right)$$.

Let $$\theta = \sin^{-1}\left(\sqrt{\frac{3}{7}}\right)$$. Then $$\sin\theta = \frac{\sqrt{3}}{\sqrt{7}}$$.

Using a right triangle, the adjacent side is $$\sqrt{7 - 3} = \sqrt{4} = 2$$.

Therefore, $$\cos\theta = \frac{2}{\sqrt{7}}$$, which means $$\theta = \cos^{-1}\left(\frac{2}{\sqrt{7}}\right)$$.

We know that $$\cos^{-1}(z) = \frac{\pi}{2} - \sin^{-1}(z)$$, so:

$$\sin^{-1}\left(\sqrt{\frac{3}{7}}\right) = \frac{\pi}{2} - \sin^{-1}\left(\frac{2}{\sqrt{7}}\right)$$

Substitute this back into the second integral:

$$= \frac{21\pi}{4} - 3\sqrt{3} - \frac{21}{2}\left( \frac{\pi}{2} - \sin^{-1}\left(\frac{2}{\sqrt{7}}\right) \right)$$

$$= \frac{21\pi}{4} - 3\sqrt{3} - \frac{21\pi}{4} + \frac{21}{2}\sin^{-1}\left(\frac{2}{\sqrt{7}}\right)$$

$$= -3\sqrt{3} + \frac{21}{2}\sin^{-1}\left(\frac{2}{\sqrt{7}}\right)$$

Step 4: Calculate $$\Delta$$ and the final expression

Now, sum the results of the two integrals and multiply by 2 to get $$\Delta$$:

$$\Delta = 2 \left( \left(4\sqrt{3} - \frac{4}{3}\right) + \left(-3\sqrt{3} + \frac{21}{2}\sin^{-1}\left(\frac{2}{\sqrt{7}}\right)\right) \right)$$

$$\Delta = 2 \left( \sqrt{3} - \frac{4}{3} + \frac{21}{2}\sin^{-1}\left(\frac{2}{\sqrt{7}}\right) \right)$$

$$\Delta = 2\sqrt{3} - \frac{8}{3} + 21\sin^{-1}\left(\frac{2}{\sqrt{7}}\right)$$

Finally, substitute $$\Delta$$ into the required expression:

$$\frac{1}{2}\left(\Delta - 21\sin^{-1}\frac{2}{\sqrt{7}}\right) = \frac{1}{2}\left( \left(2\sqrt{3} - \frac{8}{3} + 21\sin^{-1}\left(\frac{2}{\sqrt{7}}\right)\right) - 21\sin^{-1}\frac{2}{\sqrt{7}} \right)$$

$$= \frac{1}{2}\left( 2\sqrt{3} - \frac{8}{3} \right)$$

$$= \sqrt{3} - \frac{4}{3}$$