Question

(1 point) Find the area enclosed by the curve r=2 sin(0) + 3 sin(90).

Answer 1

If the 0's are indeed zeroes, then it would appear that you're using degrees, so that the equation

r(θ) = 2 sin(0°) + 3 sin(90°)

describes a circle r = 3 (since sin(0°) = 0 and sin(90°) = 1). In this case, the area would simply be π r ² = 9π.

But I suspect you meant to use θ, so the curve has the far more interesting equation,

r(θ) = 2 sin(θ) + 3 sin(9θ)

Since sin(9θ) has a period of 2π/9 and sin(θ) has a period of 2π, their sum has a period of 2π. (That is, 2π times the LCM of 1 and 1/9, which is 1.) But we observe r(θ) is odd, since

r(-θ) = 2 sin(-θ) + 3 sin(-9θ)

… = -2 sin(θ) - 3 sin(9θ)

… = - r(θ)

which tells us that the curve closes itself over a half-period. So the area bounded by the curve is given by the integral,

$\displaystyle\int_0^\pi\int_0^{r(\theta)}r\,\mathrm dr\,\mathrm d\theta$

If you're not familiar with double integrals yet, all you need to know is that this reduces to the area formula you may/should be familiar with,

$\displaystyle\frac12\int_0^\pi r(\theta)^2\,\mathrm d\theta$

Compute the integral:

$=\displaystyle\frac12\int_0^\pi(2\sin(\theta)+3\sin(9\theta))^2\,\mathrm d\theta$

$=\displaystyle\frac12\int_0^\pi(4\sin^2(\theta)+12\sin(\theta)\sin(9\theta)+9\sin^2(9\theta))\,\mathrm d\theta$

Apply some identities:

sin²(θ) = (1 - cos(2θ)) / 2

sin(θ) sin(9θ) = (cos(9θ - θ) - cos(9θ + θ)) / 2 = (cos(8θ) - cos(10θ)) / 2

sin²(9θ) = (1 - cos(18θ)) / 2

$=\displaystyle\frac14\int_0^\pi(4(1-\cos(2\theta))+12(\cos(8\theta)-\cos(10\theta))+9(1-\cos(18\theta))\,\mathrm d\theta$

$=\displaystyle\frac14\int_0^\pi(13 - 4\cos(2\theta) + 12\cos(8\theta) - 12\cos(10\theta) - 9\cos(18\theta))\,\mathrm d\theta$

$=\dfrac14 \left(13\theta-2\sin(2\theta)+\dfrac32\sin(8\theta)-\dfrac65\sin(10\theta)-\dfrac12\sin(18\theta)\right)\bigg|_0^\pi$

$=\dfrac1{40} \left(130\theta-20\sin(2\theta)+15\sin(8\theta)-12\sin(10\theta)-5\sin(18\theta)\right)\bigg|_0^\pi$

$=\dfrac{130\pi-20\sin(2\pi)+15\sin(8\pi)-12\sin(10\pi)-5\sin(18\pi)}{40}$

= 13π/4