1. Interpret the region

We need the area of

$$\{(x,y): x^2+y^2\le 21,\; y^2\le 4x,\; x\ge 1\}.$$

This is the common region of:

• the circle $x^2+y^2=21$,
• the parabola $y^2=4x$ (region inside/right of the parabola since $y^2\le 4x \iff x\ge y^2/4$),
• and the half-plane $x\ge 1$.

Because the figure is symmetric about the $x$-axis, we can compute using vertical strips.

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2. Find the intersection of circle and parabola

At intersection,

$$x=\frac{y^2}{4}.$$

Substitute into the circle:

$$\left(\frac{y^2}{4}\right)^2+y^2=21.$$

Let $t=y^2$. Then

$$\frac{t^2}{16}+t=21 \quad\Rightarrow\quad t^2+16t-336=0.$$

So,

$$t=12 \quad (\text{since } t\ge 0).$$

Hence

$$y=\pm 2\sqrt{3}, \qquad x=\frac{12}{4}=3.$$

So the circle and parabola meet at $(3,\pm 2\sqrt3)$.

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3. Describe the horizontal extent of the region

For a fixed $y$, the conditions give:

• from the circle: $x\le \sqrt{21-y^2}$,
• from the parabola: $x\ge y^2/4$,
• from $x\ge 1$: $x\ge 1$.

Thus the left boundary is

$$x=\max\left(1,\frac{y^2}{4}\right),$$

and the right boundary is

$$x=\sqrt{21-y^2}.$$

Now:

• if $|y|\le 2$, then $y^2/4\le 1$, so left boundary is $x=1$,
• if $2\le |y|\le 2\sqrt3$, then $y^2/4\ge 1$, so left boundary is $x=y^2/4$.

Also, the region exists only until the circle meets the parabola, i.e. $|y|\le 2\sqrt3$.

Therefore,

$$\Delta = 2\left[\int_0^2 \left(\sqrt{21-y^2}-1\right)dy + \int_2^{2\sqrt3}\left(\sqrt{21-y^2}-\frac{y^2}{4}\right)dy\right].$$

Combine:

$$\Delta=2\left[\int_0^{2\sqrt3}\sqrt{21-y^2}\,dy - \int_0^2 1\,dy - \int_2^{2\sqrt3}\frac{y^2}{4}\,dy\right].$$

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4. Evaluate $\int_0^{2\sqrt3}\sqrt{21-y^2}\,dy$

Use the standard formula:

$$\int \sqrt{a^2-y^2}\,dy = \frac{y}{2}\sqrt{a^2-y^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{y}{a}\right).$$

Here $a=\sqrt{21}$. So

$$I=\int_0^{2\sqrt3}\sqrt{21-y^2}\,dy.$$

At $y=2\sqrt3$,

$$\sqrt{21-12}=3.$$

Hence

$$I=\left[\frac{y}{2}\sqrt{21-y^2}+\frac{21}{2}\sin^{-1}\left(\frac{y}{\sqrt{21}}\right)\right]_0^{2\sqrt3}$$ $$= \frac{2\sqrt3}{2}\cdot 3 + \frac{21}{2}\sin^{-1}\left(\frac{2\sqrt3}{\sqrt{21}}\right).$$

Now

$$\frac{2\sqrt3}{\sqrt{21}}=\frac{2}{\sqrt7}.$$

Therefore,

$$I=3\sqrt3 + \frac{21}{2}\sin^{-1}\left(\frac{2}{\sqrt7}\right).$$

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5. Evaluate the other integrals

First,

$$\int_0^2 1\,dy=2.$$

Next,

$$\int_2^{2\sqrt3}\frac{y^2}{4}\,dy=\frac14\left[\frac{y^3}{3}\right]_2^{2\sqrt3} =\frac{1}{12}\left((2\sqrt3)^3-8\right).$$

Now,

$$(2\sqrt3)^3=24\sqrt3.$$

So,

$$\int_2^{2\sqrt3}\frac{y^2}{4}\,dy=\frac{24\sqrt3-8}{12}=2\sqrt3-\frac23.$$

Hence,

$$\Delta=2\left[3\sqrt3 + \frac{21}{2}\sin^{-1}\left(\frac{2}{\sqrt7}\right)-2-\left(2\sqrt3-\frac23\right)\right].$$

Simplify inside the bracket:

$$3\sqrt3-2\sqrt3=\sqrt3,$$

so

$$\Delta=2\left[\sqrt3 + \frac{21}{2}\sin^{-1}\left(\frac{2}{\sqrt7}\right)-2+\frac23\right].$$

Since $-2+\frac23=-\frac43$,

$$\Delta=2\left[\sqrt3-\frac43+\frac{21}{2}\sin^{-1}\left(\frac{2}{\sqrt7}\right)\right].$$

Thus,

$$\Delta=2\sqrt3-\frac83+21\sin^{-1}\left(\frac{2}{\sqrt7}\right).$$

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6. Compute the required expression

We need

$$\frac12\left(\Delta-21\sin^{-1}\frac{2}{\sqrt7}\right).$$

Substitute $\Delta$:

$$\frac12\left(2\sqrt3-\frac83+21\sin^{-1}\frac{2}{\sqrt7}-21\sin^{-1}\frac{2}{\sqrt7}\right)$$ $$=\frac12\left(2\sqrt3-\frac83\right)=\sqrt3-\frac43.$$

So the correct option is

$$\boxed{\sqrt3-\frac43}.$$

This is Option C.

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7. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.