# Question #c27db

Sep 29, 2017

The answer is $4 + \sqrt{2} + \frac{9 \pi}{4} + \frac{9}{2} \arcsin \left(\frac{1}{3}\right) \approx 14.0121$

#### Explanation:

First, write the integral as a sum:

${\int}_{- 3}^{1} \left(1 + \sqrt{9 - {x}^{2}}\right) \mathrm{dx} = {\int}_{- 3}^{1} 1 \mathrm{dx} + {\int}_{- 3}^{1} \sqrt{9 - {x}^{2}} \mathrm{dx}$.

The area under the graph of $y = 1$ for $- 3 \le x \le 1$ is the area of a rectangle: $b a s e \cdot h e i g h t = 4 \cdot 1 = 4$. Hence, ${\int}_{- 3}^{1} 1 \mathrm{dx} = 4$.

Next, write ${\int}_{- 3}^{1} \sqrt{9 - {x}^{2}} \mathrm{dx} = {\int}_{- 3}^{0} \sqrt{9 - {x}^{2}} \mathrm{dx} + {\int}_{0}^{1} \sqrt{9 - {x}^{2}} \mathrm{dx}$.

The graph of $y = \sqrt{9 - {x}^{2}}$ is a semicircle of radius 3 centered at the origin. For $- 3 \le x \le 0$, we get a quarter circle of area $\frac{1}{4} \pi \cdot {3}^{2} = \frac{9 \pi}{4}$.

Hence, ${\int}_{- 3}^{0} \sqrt{9 - {x}^{2}} \mathrm{dx} = \frac{9 \pi}{4}$.

The remaining region to calculate the area is the area under the graph of $y = \sqrt{9 - {x}^{2}}$ for $0 \le x \le 1$. This region can be split into a triangle and a sector of the circle. See the figure below. The triangle has $b a s e = 1$ and $h e i g h t = \sqrt{9 - {1}^{2}} = \sqrt{8} = 2 \sqrt{2}$. Therefore, its area is $\frac{1}{2} \cdot 1 \cdot 2 \sqrt{2} = \sqrt{2}$.

The angle for the sector of the circle is congruent to the upper angle for the triangle, which can be written as $\theta = \arcsin \left(\frac{1}{3}\right)$ since the hypotenuse of the triangle has length 3. The area of the sector is therefore $\frac{1}{2} \cdot {r}^{2} \cdot \theta = \frac{1}{2} \cdot {3}^{2} \cdot \arcsin \left(\frac{1}{3}\right) = \frac{9}{2} \arcsin \left(\frac{1}{3}\right)$.

Therefore, ${\int}_{0}^{1} \sqrt{9 - {x}^{2}} \mathrm{dx} = \sqrt{2} + \frac{9}{2} \arcsin \left(\frac{1}{3}\right)$.

Putting these things all together gives the final answer of $4 + \sqrt{2} + \frac{9 \pi}{4} + \frac{9}{2} \arcsin \left(\frac{1}{3}\right) \approx 14.0121$