Use a double integral in polar coordinates to find the area of the region. The region inside the circle (x − 4)2 + y2 = 16 and outside the circle x2 + y2 = 16

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

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