Find the area of the region enclosed by

Answered question

2022-05-01

Find the area of the region enclosed by one loop of the curve v=3sin2θ

Answer & Explanation

star233

star233

Skilled2022-05-13Added 403 answers

0π212(3sin2(2θ))dθ

Write the problem as a mathematical expression.

0π212(3sin2(2θ))dθ

Simplify.

0π23sin2(2θ)2dθ

Since 32 is constant with respect to θ, move 32 out of the integral.

320π2sin2(2θ)dθ

Let u1=2θ. Then du1=2dθ, so 12du1=dθ. Rewrite using u1 and du1.

320πsin2(u1)12du1

Combine sin2(u1) and 12.

320πsin2(u1)2du1

Since 12 is constant with respect to u1, move 12 out of the integral.

32(120πsin2(u1)du1)

Simplify.

Multiply 12 by 32.

3220πsin2(u1)du1

Multiply 2 by 2.

340πsin2(u1)du1

Use the half-angle formula to rewrite sin2(u1) as 1-cos(2u1)2.

340π1-cos(2u1)2du1

Since 12 is constant with respect to u1, move 12 out of the integral.

34(120π1-cos(2u1)du1)

Simplify.

Multiply 12 by 34.

3240π1-cos(2u1)du1

Multiply 2 by 4.

380π1-cos(2u1)du1

Split the single integral into multiple integrals.

38(0πdu1+0π-cos(2u1)du1)

Apply the constant rule.

38(u1]0π+0π-cos(2u1)du1)

Since -1 is constant with respect to u1, move -1 out of the integral.

38(u1]0π-0πcos(2u1)du1)

Let u2=2u1. Then du2=2du1, so 12du2=du1. Rewrite using u2 and du2.

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38(u1]0π-02πcos(u2)12du2)

Combine cos(u2) and 12.

38(u1]0π-02πcos(u2)2du2)

Since 12 is constant with respect to u2, move 12 out of the integral.

38(u1]0π-(1202πcos(u2)du2))

The integral of cos(u2) with respect to u2 is sin(u2).

38(u1]0π-12sin(u2)]02π)

Substitute and simplify.

38(π-12(sin(2π)-sin(0)))

Simplify.

38(π-sin(2π)2)

Simplify.

3π8

The result can be shown in multiple forms.

3π8

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