find the area between the x-axis and the graph of the given function over the given interval:
y = sqrt(9-x2^2) over [-3,3]
actually, it's
y = sqrt(9-x^2) over [-3,3]
To find the area between the x-axis and the graph of the function y = sqrt(9-x^2) over the interval [-3,3], you can use the definite integral. The integral represents the area under the curve between the given interval.
First, let's rewrite the function in a more simplified form. The graph of y = sqrt(9-x^2) represents half of a circle with a radius of 3.
To find the area, you can integrate the function from -3 to 3:
A = ∫[-3,3] sqrt(9-x^2) dx
Since the function is only defined for x-values between -3 and 3 due to the square root term, we can use this interval for integration.
To evaluate this integral, you can use trigonometric substitution. Let's substitute x = 3sin(t) and dx = 3cos(t) dt.
Now, the limits of integration change according to the substitution:
When x = -3, t = -π/2
When x = 3, t = π/2
The integral becomes:
A = ∫[-π/2, π/2] sqrt(9-(3sin(t))^2) * 3cos(t) dt
= ∫[-π/2, π/2] sqrt(9-9sin^2(t)) * 3cos(t) dt
= ∫[-π/2, π/2] 3cos^2(t) dt
Now, you can use a trigonometric identity to simplify further. The identity cos^2(t) = 1/2 * (1 + cos(2t)) can be applied.
A = ∫[-π/2, π/2] 3 * (1/2 * (1 + cos(2t))) dt
= 3/2 * ∫[-π/2, π/2] (1 + cos(2t)) dt
Evaluating the integral will give you the area between the x-axis and the graph of the function over the given interval.