Double integral change of variables examples

Example 1

Compute the double integral

_Dg(x,y)dA

where g(x,y) = x² + y² and D is disk of radius 7 centered at origin.

Solution: Since computing this integral in rectangular coordinates is too difficult, we change to polar coordinates. (We chose polar coordinates since the disk is easily described in polar coordinates.) Let

T(r,θ) = (r cos θ,r sin θ).

be the change of variables function from polar coordinates into rectangular coordinates.

Then, by the change of variables formula, the integral becomes

_Dg(x,y)dA =

g(T(r,θ))

drdθ,

where D^* is the region in the rθ-plane that is mapped by T onto the disk D. In polar coordinates, the disk is described by 0 ≤ r ≤ 7 and 0 ≤ θ ≤ 2π, so this is the region D^*.

We calculate

g(T(r,θ))	= g(r cos θ,r sin θ) = r² cos ²θ + r² sin ²θ = r²
(r,θ)	= cos θ
(r,θ)	= sin θ
(r,θ)	= -r sin θ
(r,θ)	= r cos θ
DT(r,θ)	=
det DT(r,θ)	= r cos ²θ - (-r) sin ²θ = r(cos ²θ + sin ²θ) = r

Since r ≥ 0 in D^*, we know that |detDT (r,θ)|

= |r| = r. Therefore the integral is

_Dg(x,y)dA	= ∫ ₀^2π ∫ ₀⁷r²rdrdθ
	= ∫ ₂^2π dθ
	= ∫ ₂^2πdθ =

Note: In case you have trouble believing the formula that dA becomes |det DT (r,θ)| dr,dθ, we could motivate the result that, for polar coordinates, dA becomes rdrdθ as follows.

Chop up the disk into grid corresponding to a grid size in polar coordinates of Δr and Δθ, just as we did for the intro reading. A single “curvy rectangle” with a corner (r,θ) is pictured below.

Let ΔA be area of the “curvy rectangle”. You can see that the area depends on the radius r, since its width is approximately rΔθ and its height is approximately Δr. Hence the area ΔA is approximately rΔrΔθ. I hope that now you can believe that dA becomes rdrdθ when changing variables to polar coordinates.

Example 2

Evaluate

_D(x² - y²)dxdy

where D is the region pictured below.

Solution:

The region can be described simply if we change to coordinates u and v where

u	= y - x
v	= xy.	(1)

With this change of variables, our new region of integration D^* is 0 ≤ u ≤ 1, 1 ≤ v ≤ 2.

Our change of variables as expressed in equation (1) give u and v in terms of x and y. In our change of variables formula, we need to have x and y expressed in terms of u and v using some function (x,y) = T(u,v). So one way to solve this problem is to solve equation (1) for x and y to determine the function T. We could then compute

which is what we need for the change of variables formula.

For this example, we will use another method that skips this step. We will leave u and v expressed in terms of x and y (as in equation (1)) and compute ∂(u,v)-
∂(x,y) instead of ∂(x,y)-
∂(u,v) . Since we know that

we can easily get the correct factor for out change of variables formula.

To compute ∂(u,v-)
∂(x,y ) , we view u and v as functions of x and y: s(x,y) and t(x,y). The partial derivatives of u and v with respect to x and y are

= -1,

= 1,

= y, and

= x.

We compute

	= det
	= -
	= -x - y,

so that

The factor we need for our change of variables formula is then

since x and y are positive in our domain.

Once we change variables, we will need to integrate over the region D^* the expression

(x² - y²)

Since we can factor x² - y² = (x - y)(x + y) the above expression simplifies to x - y.

Since the integral over D^* is an integral in u and v, we need to write x - y in terms of u and v. Glancing at equation (1), we see that x-y = -u, so we simply need to integrate -u over the region D^*.