Evaluate ∫_0^a sqrt(a^2 - x^2) dx. Which value is correct in terms of a?

Evaluate ∫_0^a sqrt(a^2 - x^2) dx. Which value is correct in terms of a?

• A. (π a^2)/2
• B. (π a^2)/4
• C. 0
• D. π a^2

Answer: (B)

This integral computes the area under the upper half of the circle x^2 + y^2 = a^2 from x = 0 to x = a, which is a quarter of the circle. Since the full circle has area πa^2, a quarter has area (πa^2)/4. If you do it by substitution, let x = a sin θ, so dx = a cos θ dθ and sqrt(a^2 − x^2) = a cos θ. The bounds go from θ = 0 to θ = π/2. The integral becomes a^2 ∫_0^{π/2} cos^2 θ dθ = a^2 · (π/4). So the value is (π a^2)/4.

This integral computes the area under the upper half of the circle x^2 + y^2 = a^2 from x = 0 to x = a, which is a quarter of the circle. Since the full circle has area πa^2, a quarter has area (πa^2)/4.

If you do it by substitution, let x = a sin θ, so dx = a cos θ dθ and sqrt(a^2 − x^2) = a cos θ. The bounds go from θ = 0 to θ = π/2. The integral becomes a^2 ∫_0^{π/2} cos^2 θ dθ = a^2 · (π/4).

So the value is (π a^2)/4.