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I am a bit lost on how to evaluate double integrals over a region. I am asked to evaluate the following integral

$$\iint\frac{y}{(x^2+y^2} dA$$

over the region R: triangle bounded by $y=x, y=2x, x=2$

I got $0<y<4, y<x<2$ horizontally and $ 0<x<2, x<y<2x$ vertically.

I wish to know if I am correct or not.

Regards,

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  • $\begingroup$ Can you show briefly your work? I don't think the answer should be $0$. $\endgroup$
    – KittyL
    Apr 2 '15 at 12:38
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Note that $$ R = \{(x,y)\in \def\R{\mathbf R}\R^2 \mid x\le y \le 2x, 0 \le x \le 2\} $$ Hence \begin{align*} \int_R \frac{y}{x^2 + y^2} \, d(x,y) &= \int_0^2 \int_x^{2x} \frac{y}{x^2+y^2}\, dy\, dx\\ &= \int_0^2 \left[\frac12 \log(x^2 + y^2)\right]_{x}^{2x}\, dx\\ &= \int_0^2 \frac 12\log (5x^2) - \frac 12 \log(2x^2)\, dx\\ &= \int_0^2 \frac 12 \bigl(\log 5 - \log 2\bigr)\, dx\\ &= \log 5 - \log 2 & (\ne 0). \end{align*}

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Your integration limits should be just $0<x<2, x<y<2x$ as you said vertically. You don't need the other set of limits.

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