Show Detailed Solution The original region, D is a rotated ellipse. We're going to do a series of transformations. First we'll use rotation transformation to make D the map of D , a standard ellipse in the pg plane. Second, we'll use a stretching transformation to make D the map of D1, a circle in the uw plane. Finally, we'll use a polar transformation to make D, the map of R, a rectangle in 78 plane. 1) We can rotate a point (p, g) by 8 radians counterclockwise using a rotation . : g transformation: = cos(0)p sin(f)q, y = sin(f)p + cos(6)gq. Using the 7 radian angle, that becomes: The Jacobian of this transformation is 1. Rewriting 4z + 2y in terms of p and g gives (2\\/ + l)p + (\\/ - 2) q. So far we have /[1)4m+2ydA:f/DO((2\\/+1)p+(\\/g_g)q)ldpdq 2) We can stretch a unit circle in uv to the ellipse D using the transformation p = 3u, q = 6. The Jacobian of the that transformation is 18. We rewrite the integral in terms of and v and have f/;h (3(2'/5 + 1)u + 6(~/ - 2)v) S 3) Finally, we integrate over the unit circle D1 using a conversion to polar. /0. /0 (3 (2\\/ o l)rcoS(t')) + 6(\\/ - 2)r sin(B)) 18rdrd T Let D be an ellipse rotated counterclockwise by 5 radians, as pictured below. Before rotation, the horizontal radius was 6 and vertical radius was 2. Convert / / (4x + 4y)d A into an integral over a polar integral over a unit circle. (type theta for 6) D 27 1 T 0 0 Submit