Find the domain of the function. Write your answer in interval notation. (If you need to enter ∞ or -∞, type INFINITY or -INFINITY.)
h(u,v) = sqrt(4-u^2-v^2)
u^2+v^2 = ?
To find the domain of the given function, we need to determine the possible values for the variables u and v.
To find the expression u^2 + v^2, we add the squares of u and v together:
u^2 + v^2
Now let's focus on the domain of the function.
In this case, the expression u^2 + v^2 represents the sum of the squares of u and v, which is always a non-negative value. The square root function, denoted by sqrt, is defined for non-negative real numbers. Therefore, the expression sqrt(4 - u^2 - v^2) is valid whenever the expression inside the square root is non-negative.
To determine the domain, we set the expression inside the square root greater than or equal to zero:
4 - u^2 - v^2 ≥ 0
Rearranging the inequality, we get:
-u^2 - v^2 + 4 ≥ 0
Now, the domain is the set of all values (u,v) that satisfy this inequality.
The inequality -u^2 - v^2 + 4 ≥ 0 represents an equation of a circle centered at the origin (0,0) with a radius of √4 = 2. The circle includes all points inside the circle as well as the boundary.
Using the equation of a circle, we can rewrite the inequality as:
u^2 + v^2 ≤ 4
This inequality means that the values of u and v should be within a circle with a radius of 2 centered at the origin.
The domain of the function h(u,v) = sqrt(4-u^2-v^2) is all the points (u,v) that lie within or on the boundary of the circle with radius 2 centered at the origin.
In interval notation, the domain can be written as:
D = { (u,v) | u^2 + v^2 ≤ 4 }, which can also be represented as: D = { (u,v) | -2 ≤ u ≤ 2, -2 ≤ v ≤ 2 }
So, the domain of the function h(u,v) in interval notation is: D = [-2, 2] × [-2, 2], where × represents the Cartesian product.