3.multiple - choice(5 points) medium when computing the area inside the cardioid r = a(1 + cos θ), why is…

3.multiple - choice(5 points) medium when computing the area inside the cardioid r = a(1 + cos θ), why is the integral often evaluated from 0 to π and then doubled? a because the function r = a(1 + cos θ) is only defined for θ ∈ 0, π b because the cardioid is symmetric about the x - axis, and integrating from 0 to π captures half the region c because cos θ is positive only in 0, π d because the derivative dr/dθ changes sign at θ = π

3.multiple - choice(5 points) medium when computing the area inside the cardioid r = a(1 + cos θ), why is the integral often evaluated from 0 to π and then doubled? a because the function r = a(1 + cos θ) is only defined for θ ∈ 0, π b because the cardioid is symmetric about the x - axis, and integrating from 0 to π captures half the region c because cos θ is positive only in 0, π d because the derivative dr/dθ changes sign at θ = π

Answer

Brief Explanations:

The cardioid (r = a(1+\cos\theta)) has symmetry about the x - axis. When we integrate the area formula for polar curves (A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta) from (0) to (\pi), we are calculating the area of the upper - half of the cardioid. Doubling this value gives the total area of the cardioid.

Answer:

B. Because the cardioid is symmetric about the x - axis, and integrating from 0 to (\pi) captures half the region