Question

For these types of formatted integrals, why can’t we use the basic integral formula to calculate it?

Answer 1

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

Answer 2

9π/4

Explanation:

Consider this approach.

We have a circle, rather a semicircle, covering quadrants I and II. This is as it is simply the 'upper part of the circle.' Now then we have our limits from 0 to 3, and therefore our radius will be 3, as the difference between 0 and 3 on the coordinate is 3. Well then the shaded region for consideration here would be the area under the curve of the circle in the first quadrant, as remember our limits are from 0 to 3 (positive).

What would be the area of this shaded region? Consider the area formula πr^2. If radius = 3, area = π(3)^2 = 9π. The shaded region is only 1/4ths of the whole circle, so the area would be 9π/4. Done!

Check:

y = √9 - x^2,

y^2 = 9 - x^2,

x^2 + y^2 = 0,

x^2 + y^2 = 3^2

As you can see this is a circle equation, where radius = 3, centered at (0,0). Respectively √9 - x^2 is our upper semicircle.