Step-by-step explanation:
The problem asks to calculate the integral by using area of geometric shapes, but it is possible to integrate using trig substitution.
∫₀³ √(9 − x²) dx
If x = 3 sin u, then dx = 3 cos u du.
When x = 0, u = 0. When x = 3, u = π/2.
∫₀ᵖⁱ`² √(9 − 9 sin²u) (3 cos u du)
∫₀ᵖⁱ`² 3 √(1 − sin²u) (3 cos u du)
∫₀ᵖⁱ`² 3 cos u (3 cos u du)
∫₀ᵖⁱ`² 9 cos²u du
Use power reduction formula.
9/2 ∫₀ᵖⁱ`² (cos(2u) + 1) du
9/2 ∫₀ᵖⁱ`² cos(2u) du + 9/2 ∫₀ᵖⁱ`² du
9/4 ∫₀ᵖⁱ`² 2 cos(2u) du + 9/2 ∫₀ᵖⁱ`² du
[ 9/4 sin(2u) + 9/2 u ] |₀ᵖⁱ`²
[ 9/4 sin(π) + 9/2 (π/2) ] − [ 9/4 sin(0) + 9/2 (0) ]
9π/4
Using geometry instead, the function is the top half of a circle centered at the origin with a radius of 3. The limits are 0 ≤ x ≤ 3, so we're looking at the first quadrant only. So the area is:
A = πr² / 4
A = π (3)² / 4
A = 9π/4
