Draw a graph of the curve y =sqrt 9-x^2 ,x greater than equal to 0 and...
Draw a graph of the curve y =sqrt 9-x^2 ,x greater than equal to 0 and...
Graph of the curve y = sqrt(9 - x^2)
To draw the graph of the curve y = sqrt(9 - x^2), we can start by considering the domain of the function, which is x greater than or equal to 0.
Let's plot some points to understand the shape of the curve:
When x = 0, y = sqrt(9 - 0^2) = 3. So, the point (0, 3) is on the curve.
When x = 1, y = sqrt(9 - 1^2) = sqrt(8) ≈ 2.83. So, the point (1, 2.83) is on the curve.
When x = 2, y = sqrt(9 - 2^2) = sqrt(5) ≈ 2.24. So, the point (2, 2.24) is on the curve.
By plotting more points and connecting them, we can obtain the graph of the curve y = sqrt(9 - x^2).
Area bounded by the curve and coordinate axes
The area bounded by the curve y = sqrt(9 - x^2) and the coordinate axes can be found by integrating the function over the given domain.
The curve is a semicircle with a radius of 3 centered at the origin (0,0). To find the area, we can split the curve into two parts - the upper half and the lower half.
Upper half:
In the upper half of the curve, y is positive. We need to find the area bounded by the curve, x-axis, and the y-axis. This can be calculated by integrating the function y = sqrt(9 - x^2) with respect to x from 0 to 3.
∫[0,3] sqrt(9-x^2) dx
We can evaluate this integral using a trigonometric substitution. Let x = 3sinθ, dx = 3cosθ dθ.
Substituting the values, we get:
∫[0,π/2] sqrt(9-9sin^2θ) (3cosθ) dθ
Simplifying further:
∫[0,π/2] 3cos^2θ dθ
Using the trigonometric identity cos^2θ = (1 + cos2θ)/2, we can rewrite the integral as:
(3/2) ∫[0,π/2] (1 + cos2θ) dθ
Integrating each term separately:
(3/2) [θ + (1/2)sin2θ] evaluated from 0 to π/2
Substituting the values, we get:
(3/2) [(π/2) + (1/2)sin(π)] - [(0) + (1/2)sin(0)]
(3/2) [(π/2) + (1/2)(0)] - [(0) + (1/2)(0)]
(3/2) (π/2)
3π/4
Therefore, the area bounded by the upper half of the curve and the coordinate axes is 3π/4 square units.
Lower half:
In the lower