Final answer:
The integral \(\int\int_R \cos(x^2+y^2) dA\) over the region R bounded by the circle \(x^2+y^2=9\) is evaluated in polar coordinates resulting in the final answer \(\frac{\pi}{4} \sin(9)\).
Step-by-step explanation:
The student has asked to evaluate the integral \(\int\int_R \cos(x^2+y^2) dA\) over the region R in the first quadrant enclosed by the circle \(x^2+y^2=9\) using polar coordinates.
First, we set up the integral in polar coordinates. Since \(x=r\cos\theta\) and \(y=r\sin\theta\), our integrand \(\cos(x^2+y^2)\) becomes \(\cos(r^2)\). The differential area element \(dA\) in polar coordinates is \(r dr d\theta\).
Next, we determine the limits for the integral. The region R is in the first quadrant and enclosed by a circle, so \(\theta\) will range from 0 to \(\frac{\pi}{2}\) and r will range from 0 to 3.
Our integral becomes:
\[ \int_0^{\pi/2} \int_0^3 r \cos(r^2) dr d\theta \]
Performing the integration with respect to r first, we get:
\[ \int_0^{\pi/2} \left(\frac{1}{2} \sin(r^2) \Big|_0^3 \right) d\theta \]
The evaluation of \(\sin(r^2)\) at the bounds gives us \(\sin(9)\), so we integrate with respect to \(\theta\), yielding:
\[ \frac{1}{2} \sin(9) \theta \Big|_0^{\pi/2} = \frac{1}{2} \sin(9) \frac{\pi}{2} \]
This final answer gives us the value of the integral over the region R using polar coordinates.