Finding this integral!

Calculus Level 4

$\iint4(x^{4}-y^{4})\sin(2xy) \, dx \; dy$

Calculate the integral above over the region bounded by the following 4 curves

$\begin{cases} x^{2}-y^{2}=1 \\ x^{2}-y^{2}=2 \\ 2xy = \dfrac \pi 2 \\ 2xy = \pi \end{cases}$

1 solution

Kushal Bose
Apr 17, 2017

First substitute $x^2-y^2=u$ and $2xy=v$ .So, new limits will be $u: 1 \to 2$ and $v:\pi/2 \to \pi$

Now integral becomes $\displaystyle \int_{1}^{2} \int_{\pi/2}^{\pi} 4 (u)(\sqrt{u^2+v^2}) \sin (v) |J| du dv$

The Jacobian can be calculated which is $\dfrac{1}{4 \sqrt{u^2+v^2}}$

So, finally the integration becomes $\displaystyle \int_{1}^{2} \int_{\pi/2}^{\pi} (u) \sin (v) du dv=1.5$