# find the area of the part of the circle r=9sinθ+cosθ in the fourth quadrant?

area of the part = ...........

Update:

how would be 0 for example for 2sinθ+cosθ its 0.07956 and for4sinθ+cosθ its 0.04116

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This is pretty easy to convert to Cartesian coordinates. Multiply both sides by r to get

r² = 9r sin θ + r cos θ

x² + y² = 9y + x

x² - x + y² - 9y = 0

x² - x + (1/2)² + y² - 9y + (9/2)² = 1/4 + 81/4

(x - 1/2)² + (y - 9/2)² = 82/4

That's a circle centered at (1/2, 9/2) with a radius of (√82)/2, slightly larger than 9/2. Set y=0 to solve for the x-intercepts of the circle:

(x - 1/2)² + 81/4 = 82/4

(x - 1/2)² = 1/4

|x - 1/2| = 1/2

x = 0 or 1

So, none of the area in the circle and below the x-axis is in the 3rd quadrant. Find y as a function of x from the circle equation above:

(y - 9/2) = - √[82/4 - (x - 1/2)²] .... negative square root on right since lower arc of circle is wanted

y = 9/2 - √[82/4 - (x - 1/2)²]

Integrate y dx from x=0 to 1 and you have your answer.

• its correct thanks
its
0.01847303406486075

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• Has a diameter from (0,9) to (1,0), and the circle passes through (0,0).

Area of the sector including this segment: (41/4)arccos((2(41/2) - 1)/(2*(41/2))

Area of the isosceles triangle in the sector including this segment: (9/4)

Area of the segment (in Qiv) is (41/4)arccos(40/41) - 9/4 ≈ 0.018 units²

Note: This is a very small protrusion into the 4th quadrant. Please visit the graph below, and zoom in around (0.5,0) for more detail: https://www.desmos.com/calculator/nxcpwsic9b

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• A = ∫1/2(9sinθ+cosθ)^2 dθ [3pi/2, 2pi] = (41pi )/4 - 9/2 ≈ 27.7

I hope this helps.

• I am glad you got the right answer. All the best.

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• Was this a "trick" question or am I missing the meaning?

Here is the graph of that circle and as you can see it is entirely in the first two quadrants.

https://www.wolframalpha.com/input/?i=r%3D9sin%CE%...

So, the answer would be zero.

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