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

uneskovogl5 2021-11-16 Answered
Find the area of the region enclosed by one loop of the curve. r=sin4θ
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Expert Answer

Annie Midgett
Answered 2021-11-17 Author has 7 answers

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.
image

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=sin4x is shown below:
We can see that 0 and π4 are two consecutive values for which sin4x 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
image 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(1cos(8θ)4ft)dθ
We used the formula sin2θ=1cos2θ2
[θ4sin(8θ)32]0π4=π16
Area enclosed by the loop is π16
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Philip O'Neill
Answered 2021-11-18 Author has 8 answers

Step 1
The given curve is r=sin4θ
Area of the curve enclosed in the first loop is,
A=ab12r2dθ
Step 2
Two consecutive value of theta for which sin4θ,
sin4θ=0
θ=0, π4
Step 3
Now integrate from 0 to π4 .
Area enclosed by one of the loop is,
A=0π412(sin4θ)2dθ
=120π4(1cos8θ2)dθ[cos2θ=12sin2θ]
=140π4(1cos8θ)dθ
=140π4(1)dθ140π4(cos8θ)dθ
=14([θ]0π4[sin8θ8]0π4)
=14((π40)(sin2πsin08))
=14(π4(008))
=π16

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