Find the area of the region enclosed by one loop of the curve. r=sin⁡4θ

Question

Annie Midgett · Accepted Answer

Step 1 r=sin⁡(4θ) Area enclosed by one of the loops will be simply ∫abr22dθ We will find the limits of integration a, b, and then we will integrate the integral to find the area enclosed by one of the loops.  Step 2 To find the limits of integration, we need to find two consecutive values of θ for which r is zero. Graph of y=sin⁡4x is shown below: We can see that 0 and π4 are two consecutive values for which sin⁡4x is zero Therefore, we can integrate from 0 to π4 You can also integrate from −π4 to 0 The only condition is that you have to integrate between two consecutive values of θ for which r=0  Step 3 Therefore, the limit of integration is from 0 to π4 Area enclosed by one of the loops is ∫0π4r22dθ ∫0π4(sin⁡(4θ))22dθ ∫0π4sin2(4θ)2dθ ∫0π4(1−cos⁡(8θ)4≤ft)dθ We used the formula sin2θ=1−cos⁡2θ2 [θ4−sin⁡(8θ)32]0π4=π16 Area enclosed by the loop is π16

Philip O&#039;Neill · Accepted Answer

Step 1 The given curve is r=sin⁡4θ Area of the curve enclosed in the first loop is, A=∫ab12r2dθ Step 2 Two consecutive value of theta for which sin⁡4θ, sin⁡4θ=0 θ=0, π4 Step 3 Now integrate from 0 to π4 . Area enclosed by one of the loop is, A=∫0π412(sin⁡4θ)2dθ =12∫0π4(1−cos⁡8θ2)dθ[cos⁡2θ=1−2sin2θ] =14∫0π4(1−cos⁡8θ)dθ =14∫0π4(1)dθ−14∫0π4(cos⁡8θ)dθ =14([θ]0π4−[sin⁡8θ8]0π4) =14((π4−0)−(sin⁡2π−sin⁡08)) =14(π4−(0−08)) =π16

Find the area of the region enclosed by one loop of the curve. r=\sin 4 \thet

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