To evaluate the integral, we need to convert the integral from Cartesian coordinates to polar coordinates.
In polar coordinates, we have x = rcos(theta) and y = rsin(theta), where r is the distance from the origin to the point and theta is the angle between the positive x-axis and the line segment from the origin to the point.
The region R is defined by 16 ≤ x^2 + y^2 ≤ 36, which can be rewritten as:
16 ≤ r^2 ≤ 36.
Taking the square root of each inequality, we get:
4 ≤ r ≤ 6.
Therefore, in polar coordinates, the region R is defined by 4 ≤ r ≤ 6.
Next, we need to evaluate the integral ∫∫R sin(x^2+y^2) dA using polar coordinates.
The differential area element in polar coordinates is given by dA = r dr d(theta).
Substituting x = rcos(theta) and y = rsin(theta) into sin(x^2+y^2), we get:
sin(x^2+y^2) = sin(r^2).
Therefore, the integral becomes:
∫∫R sin(x^2+y^2) dA = ∫∫R sin(r^2) r dr d(theta).
The limits of integration for r are from 4 to 6, and the limits of integration for theta are from 0 to 2π (since we need to cover the whole region R).
Therefore, the integral becomes:
∫∫R sin(x^2+y^2) dA = ∫ from 0 to 2π ∫ from 4 to 6 sin(r^2) r dr d(theta).
Evaluating this double integral will give us the final result.
Using polar coordinates, evaluate the integral ∫∫Rsin(x^2+y^2)dA
where R is the region 16≤x^2+y^2≤36
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