How do you find the area of the region shared by the circles $r = 2 cos (θ)$ $r=2cos(theta)$ and $r = 2 sin (θ)$ $r=2sin(theta)$ ?

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How do you find the area of the region shared by the circles $r = 2 cos (θ)$ $r=2cos(theta)$ and $r = 2 sin (θ)$ $r=2sin(theta)$ ?

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Answer 1 · 2015-10-24T01:40:54

$\frac{π}{2} - 1$ $pi/2-1$

Explanation:

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Intersection:
$2 sin θ = 2 cos θ \Rightarrow \frac{sin θ}{cos θ} = 1 \Rightarrow tan θ = 1 \Rightarrow θ = \frac{π}{4}$ $2sintheta=2costheta => sintheta/costheta=1 => tantheta=1 => theta=pi/4$

$A = \int \int_{A} r d r d θ$ $A=intint_A rdrd theta$

$A = \int_{0}^{\frac{π}{4}} d θ \int_{0}^{2 sin θ} r d r + \int_{\frac{π}{4}}^{\frac{π}{2}} d θ \int_{0}^{2 cos θ} r d r$ $A=int_0^(pi/4) d theta int_0^(2sintheta) rdr + int_(pi/4)^(pi/2) d theta int_0^(2costheta) rdr$

$A_{1} = \frac{1}{2} \int_{0}^{\frac{π}{4}} d θ \cdot 4 {sin}^{2} θ = \int_{0}^{\frac{π}{4}} (1 - cos 2 θ) d θ$ $A_1=1/2 int_0^(pi/4) d theta * 4sin^2 theta = int_0^(pi/4) (1-cos2theta) d theta$

$A_{1} = (θ - \frac{1}{2} sin 2 θ) ∣_{0}^{\frac{π}{4}} = \frac{π}{4} - \frac{1}{2}$ $A_1 = (theta-1/2sin2theta) |_0^(pi/4) = pi/4-1/2$

$A_{2} = \frac{1}{2} \int_{\frac{π}{4}}^{\frac{π}{2}} d θ \cdot 4 {cos}^{2} θ = \int_{\frac{π}{4}}^{\frac{π}{2}} (1 + cos 2 θ) d θ$ $A_2=1/2 int_(pi/4)^(pi/2) d theta * 4cos^2 theta = int_(pi/4)^(pi/2) (1+cos2theta) d theta$

$A_{2} = (θ + \frac{1}{2} sin 2 θ) ∣_{\frac{π}{4}}^{\frac{π}{2}} = \frac{π}{4} - \frac{1}{2}$ $A_2 = (theta+1/2sin2theta) |_(pi/4)^(pi/2) = pi/4-1/2$

$A = \frac{π}{4} - \frac{1}{2} + \frac{π}{4} - \frac{1}{2} = \frac{π}{2} - 1$ $A=pi/4-1/2+pi/4-1/2 = pi/2-1$

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How do you find the area of the region shared by the circles $r = 2 cos (θ)$ $r=2cos(theta)$ and $r = 2 sin (θ)$ $r=2sin(theta)$ ?

1 Answer

Explanation:

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How do you find the area of the region shared by the circles r=2cos(theta)r=2cos(θ)r=2cos(theta) and r=2sin(theta)r=2sin(θ)r=2sin(theta)?

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Explanation:

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How do you find the area of the region shared by the circles $r = 2 cos (θ)$ $r=2cos(theta)$ and $r = 2 sin (θ)$ $r=2sin(theta)$ ?