Solution

For a one-to-one differentiable change of variables

(x, y) \mapsto (u, v)

\iint_{D} f (x, y) d x d y = \iint_{D^{'}} f (x (u, v), y (u, v)) \frac{\partial ( x , y )}{\partial ( u , v )} d u d v,

where

\frac{\partial ( x , y )}{\partial ( u , v )} = \frac{1}{∣ u _{x} v _{y} - u _{y} v _{x} ∣} .

Here

u = x + y

v = x^{2} + y^{2}

, and

\frac{\partial ( u , v )}{\partial ( x , y )} = 2 (y - x) .

At fixed

v

, write

x = v cos θ

y = v sin θ

with

0 \leq θ \leq π

. Then

u = v (cos θ + sin θ)

, so

D^{'} = {(u, v) : 0 \leq v \leq 1, - v \leq u \leq 2 v} .

The constant-

v

curves are semicircles and the constant-

u

curves are lines

x + y = u

The map is two-to-one where

v < u < 2 v

, because both

(x, y)

and

(y, x)

lie in the upper half-disc, and one-to-one elsewhere. Since

∣ y - x ∣ = 2 v - u^{2}

, accounting for this multiplicity gives

I = \int_{0}^{1} [\int_{- v}^{v} \frac{u d u}{2 2 v - u ^{2}} + 2 \int_{v}^{2 v} \frac{u d u}{2 2 v - u ^{2}}] d v = \int_{0}^{1} v d v = \frac{2}{3} .

Directly, symmetry makes

\iint_{D} x d A = 0

\iint_{D} y d A = \int_{0}^{1} r^{2} d r \int_{0}^{π} sin θ d θ = 2/3

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