Question

Find the area in the first quadrant bounded by the arc of the circle described by the polar equation r = 2 sin θ + 4 cos θ. The circle is graphed in the accompanying figure.

A. 5π/2 B. 5 π C. 5π/2 + 4 D. 5 π + 8

Answer 1

The first quadrant restricts us to the domain
`0\le\theta\le\frac\pi2`. So the area of the region, call it
`R`, is

`\displaystyle\iint_R\mathrm dA=\int_(\theta=0)^(\theta=\pi/2)\int_(r=0)^(r=2\sin\theta+4\cos\theta)r\,\mathrm dr\,\mathrm d\theta=\frac12\int_(\theta=0)^(\theta=\pi/2)(2\sin\theta+4\cos\theta)^2\,\mathrm d\theta`

Expanding the integrand yields

`(2\sin\theta+4\cos\theta)^2=4\sin^2\theta+16\sin\theta\cos\theta+16\cos^2\theta=4+8\sin\theta\cos\theta+12\cos^2\theta`

Apply the double angle identities:

`\sin2\theta=2\sin\theta\cos\theta\implies16\sin\theta\cos\theta=8\sin2\theta`

`\cos^2\theta=\frac{1+\cos2\theta}2`

So the integral is

`\displaystyle\int_0^(\pi/2)(5+4\sin2\theta+3\cos2\theta)\,\mathrm d\theta`

`=5\theta-2\cos2\theta+\frac32\sin2\theta\bigg|_0^(\pi/2)`

`=\left(\frac{5\pi}2-2\cos\pi+\frac32\sin\pi\right)-\left(0-2\cos0+0\right)=\frac{5\pi}2+4`

so the answer is C.