# compute$\iint_D xy \ dxdy$

The question is: $$\iint_D xy \ dxdy, \quad D=1\leq x^2+y^2\leq2, \ \ \ x^2\leq y\leq x^2+1, \ x\geq0 , \ y\geq0$$ I've tried this: $$u=x^2+y^2 , \ v= y-x^2 , \ \ |J|=\frac{1}{2x(1+2y)}$$ $$\frac{1}{2}\iint \frac{y}{1+2y} \ dudv$$ But how should i solve for $$y$$ in terms of $$u$$ and $$v$$? Any suggestion would be great, thanks

• You have to find for y in terms of u and v from equations $x^2+y^2=u$ and $y-x^2=v$. Sum them you get a quadratic equation for y interms of u and v . – sirous Jan 10 at 17:13
• Do you mean $D=\{(x,\,y)\in(0,\,\infty)^2|1\le x^2+y^2\le2,\,x^2\le y\le x^2+1\}$ (D=\{(x,\,y)\in(0,\,\infty)^2|1\le x^2+y^2\le2,\,x^2\le y\le x^2+1\})? – J.G. Jan 10 at 17:46
• @J.G I have updated the region, does it matter? – simon Jan 10 at 17:54
• @simon Thanks for adding clarity. – J.G. Jan 10 at 17:56

Since $$y=v+x^2=u+v-y^2$$ has unique positive root $$y=\tfrac{\sqrt{4(u+v)+1}-1}{2}$$, $$\tfrac{y}{1+2y}=\tfrac12\left(1-\tfrac{1}{\sqrt{4(u+v)+1}}\right)$$. Now we evaluate$$\int_1^2du\int_0^1dv\tfrac12\left(1-\tfrac{1}{\sqrt{4(u+v)+1}}\right)=\tfrac{1-A}{2},\,A:=\int_1^2du\int_0^1dv\tfrac{1}{\sqrt{4(u+v)+1}}.$$Modulo possible mistakes in my arithmetic you should check for,\begin{align}A&=\int_1^2du[\tfrac12\sqrt{4u+4v+1}]_0^1\\&=\tfrac12\int_1^2du[\sqrt{4u+5}-\sqrt{4u+1}]\\&=\tfrac{1}{12}[(4u+5)^{3/2}-(4u+1)^{3/2}]_1^2\\&=\tfrac{13^{3/2}-54+5^{3/2}}{12},\end{align}so your original integral was $$\tfrac{11}{4}-\tfrac{13^{3/2}+5^{3/2}}{24}$$.

Unless you find a better change of variable, my suggestion is to integrate it based on the sketch. It is split into two integral and is really not that bad.

$$\displaystyle \int_a^1 y \ \bigg[ \ \int_{\sqrt{1-y^2}}^{\sqrt y} x \ dx \ \bigg] \ dy$$ + $$\displaystyle \int_1^b y \ \bigg [ \int_{\sqrt{y-1}}^{\sqrt {2-y^2}} x \ dx \ \bigg ] \ dy$$

$$a$$ is the value of $$y$$ at intersection of $$y = x^2, x^2 + y^2 = 1 \implies y^2+y-1=0$$. This gives you $$a = \frac{\sqrt 5 - 1}{2}$$

$$b$$ is the value of $$y$$ at intersection of $$y = x^2 + 1, x^2 + y^2 = 2 \implies y^2+y-3=0$$. This gives you $$b = \frac{\sqrt 13 - 1}{2}$$.

And the correct answer is $${\frac{36+5\sqrt5 - 13 \sqrt{13}}{48}}$$

• @ Math Lover , I agree with you, this is a better method of solving this problem than with change of variable – simon Jan 11 at 5:37