Question

Find the area of the region that is bounded by the given curve and lies in the specified sector. The curve is defined by r = e^(-θ/8), where θ ranges from π/2 to π. What is the area of this region?

Answer 1

The area of the region enclosed by the curve r = e^(-θ/8) in the sector π/2 ≤ θ ≤ π is 8(1 - e^(-π/8)).

Step-by-step explanation:

To find the area enclosed by a polar curve r = f(θ) between θ = α and θ = β, we use the following formula:

A = 1/2 ∫(α to β) (f(θ))^2 dθ

In this case, we have:

f(θ) = e^(-θ/8)

α = π/2

β = π

Let's solve for the area A:

A = 1/2 ∫(π/2 to π) (e^(-θ/8))^2 dθ

Expand the exponent:

A = 1/2 ∫(π/2 to π) e^(-θ/4) dθ

Integrate using substitution:

u = -θ/4, du = -dθ/4

∫(π/2 to π) e^(-θ/4) dθ = -4 ∫(π/4 to -π/4) e^(u) du = -4e^u: π/4 to -π/4

= -4(e^(-π/4) - e^(π/4)) = 8(1 - e^(-π/8))

Therefore, the area of the region enclosed by the curve r = e^(-θ/8) in the sector π/2 ≤ θ ≤ π is 8(1 - e^(-π/8)).