Integrate the function $f(x,y) = xy(x^2+y^2)$ over the domain $R : { -3 <= x^2-y^2<3, 1 <=xy <=4 }$
edited Jan 5

+1 vote

Let $x^2-y^2 = u$ and $xy = v$

$x^2+y^2 = \sqrt{(x^2-y^2)^2 + 4 x^2 y^2} = \sqrt{u^2 + 4 v^2}$

$f(u,v) = v\sqrt{u^2 + 4 v^2}$

$I = \int \int f(x,y) dx dy = \int \int f(u,v)|J| du dv$

Now, $|J|^{-1} = \begin{vmatrix} \frac{du}{dx} & \frac{du}{dy} \\ \frac{dv}{dx} & \frac{dv}{dy} \end{vmatrix} \\ = \begin{vmatrix} 2x & -2y\\ y & x \end{vmatrix} = 2x^2 + 2y^2 = 2 \sqrt{u^2 + 4 v^2}$

Thus, $I = \int \int v\sqrt{u^2 + 4 v^2}\ \frac{1}{2\sqrt{u^2 + 4 v^2}} du dv \\ = \frac{1}{2}\int \int v du dv\\ = \frac{1}{2}\int_{v=1}^{4} \int_{u=-3}^{3} v dudv\\ = \frac{1}{2}\int_{v=1}^{4} vu \bigg\rvert_{u= -3}^{3}dv \\ = \frac{1}{2}\int_{v=1}^{4} 6v dv \\ = \frac{6}{4} v^2\bigg\rvert_{1}^{4} \\ = \frac{6}{4}(4^2-1^2)= \frac{45}{2}$

answered Jan 5 by (1,920 points)