How to find Find the area of the region enclosed by

Frank Guyton

Frank Guyton

Answered question

2021-12-20

How to find
Find the area of the region enclosed by one loop of the curve.
r=sin(4θ)

Answer & Explanation

Neunassauk8

Neunassauk8

Beginner2021-12-21Added 30 answers

Step 1
The given curve is r=sin4θ
Area of the curve enclosed in the first loop is,
A=ab12r2dθ
Step 2
Two conscutive 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(sin4θ)2dθ
=120π4(1cos8θ2)dθ[cos2θ=12sin2θ]
=140π4(1cos8θ)dθ
=140π4(1)dθ140π4cos8θ)dθ
=14([θ]0π4[sin8θ8]0π4)
=14((π40)(sin2πsin08))
=14(π4(008))
=π16
Step 4
Area of the curve enclosed in the first loop is π16

Travis Hicks

Travis Hicks

Beginner2021-12-22Added 29 answers

Step 1
A=120π4(sin4θ)2dθ
=120π4sin24θdθ
=120π412(1cos8θ)dθ(sincesin2u=12(1cos2u))
=140π4(1cos8θ)dθ
=14[θ18sin8θ]0π4
=14[(π418sin2π)(018sin0)]
=14[π4]
=π16
nick1337

nick1337

Expert2021-12-28Added 777 answers

Step 1
r=sin(4θ)
Area enclosed by one of the loops will be simply
abr22dθ
Step 2
To find the limits of integration, we need to find two consecutive values of θ . For which r is zero.
Therefore, the limit of integration is from 0 to π4
Area enclosed is
0π/4r22dθ
0π/4(sin(4θ))22dθ
0π/4sin2(4θ)2dθ
0π/4(1cos(8θ)4)dθ
We used the formula sin2θ=1cos2θ2
[θ4sin(8θ)32]0π/4=π16

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