= Solution
Set
$$
u=\frac{x}{y},\qquad v=xy.
$$
Because the region lies in the <positive quadrant>, its four inequalities become
$$
1\leq u\leq\alpha,\qquad 1\leq v\leq\alpha.
$$
Thus this <change of variables formula>[change of variables] sends $D$ to a rectangle in the $uv$-plane. Its <Jacobian determinant> is
$$
\frac{\partial(u,v)}{\partial(x,y)}
=
\begin{vmatrix}
1/y&-x/y^2\\
y&x
\end{vmatrix}
=\frac{2x}{y}=2u,
$$
so
$$
dx\,dy=\frac{du\,dv}{2u}.
$$
Since $x^2=uv$, the <double integral> is
$$
\begin{aligned}
\iint_Dx^2\,dx\,dy
&=\int_1^\alpha\int_1^\alpha
uv\,\frac{du\,dv}{2u}\\
&=\frac12\left(\int_1^\alpha du\right)
\left(\int_1^\alpha v\,dv\right)\\
&=\boxed{\frac{(\alpha-1)(\alpha^2-1)}4}.
\end{aligned}
$$
Back to article page
