# Multivariable Calculate $\int\int_∆(x^4- y^4 )e^{xy} dx dy$

Calculate Double integral $$\iint_∆ (x^4 - y^4 )e^{xy} dxdy$$ where ∆ is a region in the first quadrant bounded by the curves $$xy=1,xy=2, x^2-y^2=1, x^2-y^2=4$$ What i did?

i got the limit = $1 \le u \le 2,1\le v \le4$ and The answer will be 70,05?

Is that correct?

• Assuming you're using the change of variables $u=xy$ and $v=x^2-y^2$, yes, those limits are correct. – user170231 Dec 7 '15 at 22:14
• The answer will be 70,05? – hussi Dec 7 '15 at 22:15
• That answer doesn't seem right to me. What integral are you computing? – user170231 Dec 7 '15 at 22:48

Using the change of variables $\begin{cases}u(x,y)=xy\\[1ex]v(x,y)=x^2-y^2\end{cases}$, you have $$u=xy~\implies~x=\frac{u}{y}\\ v=x^2-y^2~\implies~u^2-y^4=vy^2~\implies~y=\sqrt{\sqrt{u^2+\frac{v^2}{4}}-\frac{v}{2}}~\implies~x=\frac{u}{\sqrt{\sqrt{u^2+\frac{v^2}{4}}-\frac{v}{2}}}$$ Compute the Jacobian (it's quite a mess, so I'm omitting some steps here): $$|J|=\left|\det\begin{pmatrix}\dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\[1ex]\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{pmatrix}\right|=\left|-\frac{1}{2\sqrt{4u^2+v^2}}\right|=\frac{1}{2\sqrt{4u^2+v^2}}$$ So you have $$\iint_\Delta f(x,y)\,\mathrm{d}x\,\mathrm{d}y=\iint_\Delta F(u,v)|J|\,\mathrm{d}u\,\mathrm{d}v$$which reduces the original integral quite nicely. $$x^4-y^4=\left(\frac{u}{\sqrt{\sqrt{u^2+\frac{v^2}{4}}-\frac{v}{2}}}\right)^2-\left(\sqrt{\sqrt{u^2+\frac{v^2}{4}}-\frac{v}{2}}\right)^4=v\sqrt{4u^2+v^2}$$ $$e^{xy}=\exp\left(\frac{u}{\sqrt{\sqrt{u^2+\frac{v^2}{4}}-\frac{v}{2}}}\times\sqrt{\sqrt{u^2+\frac{v^2}{4}}-\frac{v}{2}}\right)=e^u$$ Finally, $$\iint_\Delta (x^4-y^4)e^{xy}\,\mathrm{d}x\,\mathrm{d}y=\frac{1}{2}\iint_{[1,2]\times[1,4]}ve^u\,\mathrm{d}u\,\mathrm{d}v$$where $1\le u\le2$ and $1\le v\le4$ in case the notation of the final integral is confusing.