Answer:
[tex]\mathbf{A = \dfrac{8}{3} \bigg (2 \pi + 3\sqrt{3} \bigg ) }[/tex]
Step-by-step explanation:
From the information given:
Lets first change the given relations into polar coordinates
So, the region of the inner circle into polar coordinates is as follows:
(x - 4)² + y² = 16
x² + y² - 8x = 0
r² - 8r cos θ = 0
r = 8 cos θ
For the outside circle:
x² + y² = 16
r² = 16
Thus, the intersection point are:
64cos²θ = 16
cos²θ = 1/4
θ = -(π/3) , (π/3)
Now: if we are to represent the sketch on a graph, the region of the integration D will be:
[tex]D = \bigg \{ (r,\theta )\bigg |-\dfrac{\pi}{3}\leq \theta \leq \dfrac{\pi}{3}, 4 \leq r \leq 8 cos \theta \bigg \}[/tex]
Therefore, the are of the required region can now be computed as follows:
[tex]A = \int \int _D \ dA[/tex]
[tex]A = \int ^{\pi/3}_{-\pi/3} \int ^{8 \ cos \theta}_{4} \ r dr d \theta[/tex]
[tex]A = \int ^{\pi/3}_{-\pi/3} \bigg (\dfrac{r^2}{2} \bigg ) ^{8 \ cos \theta}_{4} \ d \theta[/tex]
[tex]A = \int ^{\pi/3}_{-\pi/3} \dfrac{1}{2} \bigg (64 \ cos ^2 \theta - 16 \bigg ) \ d \theta[/tex]
[tex]A = \int ^{\pi/3}_{-\pi/3} \dfrac{16}{2} \bigg (4 \ cos ^2 \theta - 1 \bigg ) \ d \theta[/tex]
[tex]A = \dfrac{16}{2} * 2 \int ^{\pi/3}_{0} \bigg (4 \bigg ( \dfrac{ 1+ cos 2 \theta}{2} \bigg) - 1 \bigg ) \ d \theta[/tex]
[tex]A = 16 \int^{\pi/3}_{0} (1 + 2 cos 2 \theta ) d \theta[/tex]
[tex]A =16 ( \theta + sin2 \theta )^{\pi/3}_{0}[/tex]
[tex]A = 16 \begin {bmatrix} \bigg (\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2} \bigg ) - (0-0) \end {bmatrix}[/tex]
[tex]A = 16 \begin {bmatrix} \bigg (\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2} \bigg ) \end {bmatrix}[/tex]
[tex]\mathbf{A = \dfrac{8}{3} \bigg (2 \pi + 3\sqrt{3} \bigg ) }[/tex]
The area of the region is equal to [tex]8\sqrt{3} + \frac{16\pi}{3}[/tex] square units.
The double integral formula for the area of a given curve is described below:
[tex]A = \iint\,r(\theta) \,dr\,d\theta[/tex] (1)
Where:
We notice that [tex]r \in [4, 8\cdot \cos \theta][/tex] and [tex]\theta \in \left[-\frac{\pi}{3}, \frac{\pi}{3} \right][/tex], then we have the following integral:
[tex]A = \int\limits_{-\frac{\pi}{3} }^{+\frac{\pi}{3} } \int \limits_{r= 4}^{r = 8\cdot \cos \theta}\,r\,dr\,d\theta[/tex]
Now we proceed to integrate the expression:
[tex]A = \frac{1}{2}\int\limits_{-\frac{\pi}{3} }^{+\frac{\pi}{3} } \,(64\cdot \cos^{2}\theta -16)\,d\theta = 32 \int\limits^{+\frac{\pi}{3} }_{-\frac{\pi}{3} } \,\cos^{2}\theta \, d\theta - 8\,\int\limits^{+\frac{\pi}{3} }_{-\frac{\pi}{3} } \, d\theta[/tex]
[tex]A = 16\int\limits^{+\frac{\pi}{3} }_{-\frac{\pi}{3} } {\cos 2\theta} \, d\theta + 8\int\limits^{+\frac{\pi}{3} }_{-\frac{\pi}{3} } \, d\theta[/tex]
[tex]A = 8\sqrt{3}+\frac{16\pi}{3}[/tex]
The area of the region is equal to [tex]8\sqrt{3} + \frac{16\pi}{3}[/tex] square units.
We kindly invite to check this question on double integrals: https://brainly.com/question/2289273
