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I have been trying to integrate the two following integrands; $$\int \int_{D}(x^{2}+y^{2})dxdy$$ where $D=\{{x^{2}+xy+y^{2}\leq 1}\}$ and $$\int \int_{D}\sqrt{x^{2}+y^{2}}dxdy$$ where $D=\{{x^{2}+y^{2}\leq x}\}$. Now, I have been tempted to use change of variables (polar coordinates) in both cases. For example, in the first case, I completed the square so: $$D=\{{(x+1/2y)^{2}+3/4y^{2}\leq 1}\}$$ I then set $u=r\cos{t}-(1/2)r\sin{t}$ and $v=r\sin{t}$, performed partial differentiation, formed the jacobin matrix, and calculated the determinant. I then set the new boundaries $0\leq r\leq$1 and $0\leq t \leq 2\pi$ before calculating the whole expression. Now, the answer I obtain is $(53\pi)/96$, which is blatantly wrong as the textbook gives $(4\pi)/(3\sqrt{3})$. Since my approach to the second problem is similar, I fear, that the answer I obtain there is wrong as well. I would be exceedingly grateful if you could help me.

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You've been trying to integrate the integrands, not the integrals. – joriki Feb 13 '13 at 0:06
Also note that your new bounds describe the closed unit disk. – 1015 Feb 13 '13 at 0:08

First, do coordinates transform $$ x = \frac u{\sqrt 3} - v; \qquad y = \frac u{\sqrt 3} + v $$ It will reduce your domain $D$ to the inner region of the circle: $$ x^2+xy+y^2= \frac {u^2}3 - 2\frac{uv}{\sqrt 3} + v^2 + \frac {u^2}3 - v^2 + \frac{u^2}3 + 2 \frac{uv}{\sqrt 3} + v^2 = u^2 + v^2 \leq 1 $$ Now change the system to "polar": $$ u = r \cos \phi; \qquad v = r \sin \phi $$ so $D = \left \{ \left ( r, \phi\right )\ \left |\ 0 \le r \le 1, 0 \le \phi \le 2\pi \right .\right \}$

As for the integrand $$ x^2+y^2 = \frac {u^2}3 - 2\frac{uv}{\sqrt 3} + v^2 + \frac{u^2}3 + 2 \frac{uv}{\sqrt 3} + v^2 = \frac {2u^2}3 + 2v^2 $$ Also $dx\ dy = J_1\ du\ dv$, where $J$ is a Jacobian of first $(x,y) \rightarrow (u, v)$ transformation. On the other hand $du\ dv = J_2\ dr\ d\phi$, where $J_2$ is a Jacobian of second, $(u, v) \rightarrow (r, \phi)$, transformation, and we know that $J_2 = r$. Let's find $J_1$: $$ J_1 = \text{det} \left [ \begin{array}{cc} \frac 1{\sqrt 3} & -1 \\ \frac 1{\sqrt 3} & 1 \end{array} \right ] = \frac 2{\sqrt 3} $$ so $dx\ dy = \frac {2r}{\sqrt 3} dr\ d\phi$. And finally $$ I_1 = \frac 2{\sqrt 3}\int_0^{2\pi} \int_0^1 \left ( \frac {2r^2 \cos^2 \phi}3 + 2 r^2 \sin^2 \phi\right) r dr d\phi = \frac 4{3\sqrt 3}\int_0^{2\pi} \left ( \left . \frac {r^4}4 \right |_0^1\right )\left( cos^2 \phi + 3 \sin^2 \phi \right)d\phi = \frac 1{3 \sqrt 3} \int_0^{2\pi} \left ( 2 - \cos 2\phi\right ) = \frac {4\pi}{3\sqrt 3} $$

Same can be done for 2 part of the problem.

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Well, maybe not exactly the same, but similar. For example, you can easily see that if you take $u = \frac {u+1}2$ and $v = \frac y2$, then your domain is a unit circle again. Integrating itself might be challenging though. – Kaster Feb 13 '13 at 6:43

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