Animated Solution for Mathematics - Definite Integration: Let $\Delta$ be the area of the region $\{(x,y)\in {\mathbb{R}}^2:x^2+y^2\le 21,y^2\le 4x,x\ge 1\}$. Then $\frac{1}{2}\left(\Delta -21{\sin }^{-1}\frac{2}{\sqrt{7}}\right)$ is equal to

Welcome, future engineer. Today, we are not just solving an area problem; we are exploring the boundaries of a geometric landscape. Imagine standing on a coordinate plane.

You have a circle, $x^2+y^2=21$, a massive, smooth curve centered at the origin. Then, you have a parabola, $y^2=4x$, cutting through the plane like a sharp blade. Finally, the vertical line $x=1$ acts as a gatekeeper, restricting our movement.

The region we are interested in is the intersection of these three constraints. It is a beautiful, symmetric shape. Before we touch a single integral, we must find the 'hinge'—the point where the parabola hands off the boundary duty to the circle.

By substituting $y^2=4x$ into the circle equation, we get $x^2+4x-21=0$. Solving this quadratic gives us $x=3$ (we ignore the negative root because our region starts at $x=1$). This point, $x=3$, is the heartbeat of our problem.

Symmetry and Splitting

Now, look at the symmetry. The region is perfectly mirrored across the $x$-axis. This is a gift!

Instead of calculating the entire area $\Delta$, we calculate the area in the first quadrant and multiply by two. But wait—the boundary changes.

From $x=1$ to $x=3$, the upper boundary is the parabola, $y=2\sqrt{x}$. From $x=3$ to $x=\sqrt{21}$, the boundary is the circle, $y=\sqrt{21-x^2}$. We must respect this transition.

Our total area $\Delta$ is defined as:

This is our master equation. It looks intimidating, but we will dismantle it piece by piece.

Integration

Let us tackle the first integral, $I_1={\int }_1^{3}2\sqrt{x}dx$. This is a straightforward power rule application.

The integral of $x^{1/2}$ is $\frac{2}{3}{x}^{3/2}$. Multiplying by the constant 2, we get $\frac{4}{3}{x}^{3/2}$.

Evaluating from 1 to 3, we get $\frac{4}{3}(3\sqrt{3}-1)=4\sqrt{3}-\frac{4}{3}$. Simple, clean, and effective.

Now, the second integral, $I_2={\int }_3^{\sqrt{21}}\sqrt{21-x^2}dx$. This is the classic standard form:

Here, $a^2=21$. When we plug in the upper limit $\sqrt{21}$, the first term vanishes because $\sqrt{21-21}=0$. We are left with $\frac{21}{2}{\sin }^{-1}(1)=\frac{21\pi }{4}$.

When we plug in the lower limit $x=3$, we get $\frac{3}{2}\sqrt{12}+\frac{21}{2}{\sin }^{-1}\left(\frac{3}{\sqrt{21}}\right)$. This simplifies to $3\sqrt{3}+\frac{21}{2}{\sin }^{-1}\left(\sqrt{\frac{3}{7}}\right)$.

The Trig Identity

Here is where the magic happens. The problem asks for an expression involving ${\sin }^{-1}\left(\frac{2}{\sqrt{7}}\right)$, but we have ${\sin }^{-1}\left(\sqrt{\frac{3}{7}}\right)$. Do not panic.

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

Using the Pythagorean identity, $\cos \theta =\sqrt{1-\frac{3}{7}}=\sqrt{\frac{4}{7}}=\frac{2}{\sqrt{7}}$. Thus, $\theta ={\cos }^{-1}\left(\frac{2}{\sqrt{7}}\right)=\frac{\pi }{2}-{\sin }^{-1}\left(\frac{2}{\sqrt{7}}\right)$.

When we substitute this back into our expression for $I_2$, the $\frac{21\pi }{4}$ terms cancel out beautifully, leaving us with $I_2=-3\sqrt{3}+\frac{21}{2}{\sin }^{-1}\left(\frac{2}{\sqrt{7}}\right)$.

Finally, we combine everything. $\Delta =2(I_1+I_2)$. Substituting our values:

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

The question asks for $\frac{1}{2}\left(\Delta -21{\sin }^{-1}\frac{2}{\sqrt{7}}\right)$. When we plug in our $\Delta$, the inverse sine terms vanish entirely, leaving us with $\frac{1}{2}\left(2\sqrt{3}-\frac{8}{3}\right)=\sqrt{3}-\frac{4}{3}$.

And there it is—the elegance of mathematics. We navigated the geometry, conquered the integrals, and utilized the trig identities to reach a clean, satisfying conclusion. Keep this persistence, and no problem will ever be too complex for you.