Mathematics, 09.11.2020 17:00, jalenJohnson6t

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

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answered: uniqueesmeraldaangui

$\mathbf{A = \dfrac{8}{3} \bigg (2 \pi + 3\sqrt{3} \bigg ) }$

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:

$D = \bigg \{ (r,\theta )\bigg |-\dfrac{\pi}{3}\leq \theta \leq \dfrac{\pi}{3}, 4 \leq r \leq 8 cos \theta \bigg \}$

Therefore, the are of the required region can now be computed as follows:

$A = \int \int _D \ dA$

$A = \int ^{\pi/3}_{-\pi/3} \int ^{8 \ cos \theta}_{4} \ r dr d \theta$

$A = \int ^{\pi/3}_{-\pi/3} \bigg (\dfrac{r^2}{2} \bigg ) ^{8 \ cos \theta}_{4} \ d \theta$

$A = \int ^{\pi/3}_{-\pi/3} \dfrac{1}{2} \bigg (64 \ cos ^2 \theta - 16 \bigg ) \ d \theta$

$A = \int ^{\pi/3}_{-\pi/3} \dfrac{16}{2} \bigg (4 \ cos ^2 \theta - 1 \bigg ) \ d \theta$

$A = \dfrac{16}{2} * 2 \int ^{\pi/3}_{0} \bigg (4 \bigg ( \dfrac{ 1+ cos 2 \theta}{2} \bigg) - 1 \bigg ) \ d \theta$

$A = 16 \int^{\pi/3}_{0} (1 + 2 cos 2 \theta ) d \theta$

$A =16 ( \theta + sin2 \theta )^{\pi/3}_{0}$

$A = 16 \begin {bmatrix} \bigg (\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2} \bigg ) - (0-0) \end {bmatrix}$

$A = 16 \begin {bmatrix} \bigg (\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2} \bigg ) \end {bmatrix}$

$\mathbf{A = \dfrac{8}{3} \bigg (2 \pi + 3\sqrt{3} \bigg ) }$

Use a double integral in polar coordinates to find the area of the region. The region inside the cir

answered: Guest

(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

step-by-step explanation:

simplify the following:

x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)

combine powers. x/x^2 = x^(1 - 2):

x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)

1 - 2 = -1:

x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)

put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):

(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

3 x^2 (x + 2) = 3 x^3 + 6 x^2:

(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

2 x (x + 2) = 2 x^2 + 4 x:

(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))

-x^2 (x + 1) = -x^2 - x^3:

(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))

grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:

((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

3 x^3 - x^3 = 2 x^3:

(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

6 x^2 + 2 x^2 - x^2 = 7 x^2:

(2 x^3 + 7 x^2 + (x + 4 x) + 2)/(x (x + 2))

x + 4 x = 5 x:

answer: (2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

answered: Guest

17 x 256=4352

sorry if this is wrong, i'm new.

step-by-step explanation:

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Use a double integral in polar coordinates to find the area of the region. The region inside the cir...