advertisement

Lecture 17, November 2 • Formula for change of variables. When it comes to double integrals, one has ∫∫ ∫∫ ∂(x, y) du dv. f (x, y) dx dy = f (x(u, v), y(u, v)) · ∂(u, v) Here, the additional factor inside the integral is the absolute value of the Jacobian ∂x ∂x ∂(x, y) ∂u ∂v ∂x ∂y ∂y ∂x = = − . ∂(u, v) ∂y ∂y ∂u ∂v ∂u ∂v ∂u ∂v ..................................................................................... Example 1. Consider the region R in the xy-plane bounded by the lines x + y = 1, x + y = 2, x − y = 0, x − y = 1. If we introduce the variables u = x − y and v = x + y, then we can write ∫∫ ∫ 2∫ 1 x−y u ∂(x, y) dx dy = · du dv. x+y ∂(u, v) 1 0 v R To compute the Jacobian in this case, we ﬁrst need to solve for x and y, namely } } { } { { x = (u + v)/2 u + v = 2x u=x−y . =⇒ =⇒ y = (v − u)/2 v − u = 2y v =x+y This allows us to diﬀerentiate x, y ∂x ∂(x, y) ∂u = ∂(u, v) ∂y ∂u with respect ∂x 1 ∂v 2 = ∂y 1 − 2 ∂v to u, v and we now get 1 1 2 1 1 = + = . 1 4 4 2 2 Keeping this in mind, we can ﬁnally compute the given integral as ∫∫ ∫ 2∫ 1 u x−y dx dy = du dv x+y 0 2v 1 R [ ]2 ∫ 2 [ 2 ]1 ∫ 2 u ln v ln 2 1 = dv = = . dv = 4v u=0 4 1 4 1 1 4v