Final answer:
To address the question, the parallelogram-shaped region R is sketched by plotting the given lines, and the integral is computed by changing variables to polar coordinates after plotting the parallelogram, which simplifies the exponential term.
Step-by-step explanation:
Sketching Region R and Evaluating the Integral with a Change of Variables
To sketch region R, we need to plot the lines y = -3x, y = -3x + 3, y = 3x, and y = 3x - 3 on the coordinate axis. These lines form a symmetric parallelogram around the origin. The lines y = -3x and y = 3x intersect the y-axis at the origin and have slopes of -3 and 3, respectively. The lines y = -3x + 3 and y = 3x - 3 intersect the y-axis at y = 3 and y = -3, respectively.
For the integral ∫∫(3x+y)e-9x^2-y^2dA, a change of variables to polar coordinates (x = rcos(\theta) and y = rsin(\theta)) simplifies the exponential term, exploiting the identity r^2 = x^2 + y^2. This substitution maps the parallelogram into a region in the r-\theta plane. The Jacobian of the transformation, r, accounts for the change in area element from dA to r dr d\theta. After the substitution, the integral takes the form of a polar integral which can be evaluated over the new bounds corresponding to the transformed region R.