1. Interpret the region

We need the area of

$$\{(x,y)\in \mathbb R^2: y\ge \sqrt{|x+3|},\; 5y\le x+9\le 15\}.$$

The inequalities imply:

• $y\ge \sqrt{|x+3|}$
• $x+9\ge 5y \Rightarrow x\ge 5y-9$
• $x+9\le 15 \Rightarrow x\le 6$

So the region lies:

• above the curve $y=\sqrt{|x+3|}$,
• to the right of the line $x=5y-9$,
• to the left of the line $x=6$.

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2. Rewrite the curve in terms of $x$

Since $y=\sqrt{|x+3|},$ we have the equivalent condition $y^2=|x+3|.$

Thus:

• for $x\ge -3$, $y=\sqrt{x+3}$,
• for $x\le -3$, $y=\sqrt{-x-3}$.

But using horizontal strips is more convenient because of the line $x=5y-9$.

From $y\ge \sqrt{|x+3|}$ we get $y^2\ge |x+3| \Rightarrow -y^2\le x+3\le y^2,$ so $-y^2-3\le x\le y^2-3.$

Hence for a fixed $y$, the curve allows $x\in[-y^2-3,\; y^2-3].$

The line condition gives $x\in[5y-9,\; 6].$

Therefore, for each $y$, the allowed $x$-interval is the intersection $[5y-9,6]\cap[-y^2-3,\; y^2-3].$

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3. Find the range of $y$

Since $x\le 6$, from the curve condition $y\ge \sqrt{|x+3|}$ we must have at least some $x$ satisfying $|x+3|\le y^2$.

Also, from $5y\le x+9\le 15$, we get $5y\le 15 \Rightarrow y\le 3.$

And clearly $y\ge 0$ because $\sqrt{|x+3|}\ge 0$.

So $0\le y\le 3$.

Now determine where the intersection is nonempty.

For fixed $y$, the right endpoint of curve interval is $y^2-3$, and the left endpoint from the line is $5y-9$. We need $5y-9\le y^2-3.$ That is,

$$\Rightarrow (y-2)(y-3)\ge 0.$$

Since $0\le y\le 3$, this gives $0\le y\le 2 \quad \text{or} \quad y=3.$ The single point $y=3$ contributes zero area, so effectively $0\le y\le 2$.

Also, for $0\le y\le 2$, note that $y^2-3\le 1<6,$ so the right boundary is $x=y^2-3$, not $x=6$.

Thus the horizontal width is $\text{width}= (y^2-3)-(5y-9)=y^2-5y+6=(y-2)(y-3).$

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4. Integrate to get the area

Therefore,

$$A=\int_0^2 \big(y^2-5y+6\big)\,dy.$$

Compute:

$$A=\left[\frac{y^3}{3}-\frac{5y^2}{2}+6y\right]_0^2 =\frac{8}{3}-10+12 =\frac{8}{3}+2 =\frac{14}{3}?$$

This seems too large, so let us check carefully.

Actually,

$$$$

so

$$A=\frac{8}{3}+2=\frac{14}{3}.$$

But this contradicts the options, so let us re-check the interval intersection.

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5. Correct interpretation of the region

The condition is $y\ge \sqrt{|x+3|}.$ Since the right-hand side is nonnegative, this means the region is above the curve. For a fixed $y$, points above the curve satisfy

$$$$

so indeed

$$$$

That part is correct.

Now combine with

$$$$

This is

$$$$

So the actual width is

$$$$

For $0\le y\le 2$:

• $y^2-3\le 1<6$, so right end is $y^2-3$.
• Compare $5y-9$ and $-y^2-3$: $(5y-9)-(-y^2-3)=y^2+5y-6=(y+6)(y-1).$ Hence:
  • for $0\le y\le 1$, $5y-9\le -y^2-3$, so left end is $-y^2-3$;
  • for $1\le y\le 2$, $5y-9\ge -y^2-3$, so left end is $5y-9$.

So we must split the integral.

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6. Compute area in two parts

For $0\le y\le 1$

Width:

$$(y^2-3)-(-y^2-3)=2y^2.$$

Area contribution:

$$A_1=\int_0^1 2y^2\,dy=2\cdot \frac{1}{3}=\frac{2}{3}.$$

For $1\le y\le 2$

Width:

$$(y^2-3)-(5y-9)=y^2-5y+6.$$

Area contribution:

$$A_2=\int_1^2 (y^2-5y+6)\,dy.$$

Now,

$$A_2=\left[\frac{y^3}{3}-\frac{5y^2}{2}+6y\right]_1^2.$$

At $y=2$:

$$\frac{8}{3}-10+12=\frac{14}{3}.$$

At $y=1$:

$$\frac{1}{3}-\frac{5}{2}+6=\frac{1}{3}+\frac{7}{2}=\frac{23}{6}.$$

Thus,

$$A_2=\frac{14}{3}-\frac{23}{6}=\frac{28-23}{6}=\frac{5}{6}.$$

Hence total area:

$$A=A_1+A_2=\frac{2}{3}+\frac{5}{6}=\frac{4}{6}+\frac{5}{6}=\frac{9}{6}=\frac{3}{2}.$$

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7. Match with options

$$\boxed{\frac{3}{2}}$$

So the correct option is C.

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8. Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

They agree.