Can someone check if my solution to the following problem is correct.

(a) Evaluate

$$\int_{0}^{\alpha\sin\beta} \int_{y\cot\beta}^{\sqrt{{a}^{2}-{y}^{2}}} \log({x}^{2}+{y}^{2}) \,dx\,dy$$

(b) Change the order of integration in the integral in (a)

For part (a), I let $x=\alpha cos(\beta)$ and $y=\alpha sin(\beta)$, then applying the change of variable formula

$\int_{0}^{y} \int_{x}^{-x} \alpha\log({\alpha}^{2}) \,d\alpha\,d\beta$

$\int_{x}^{-x} \alpha\log({\alpha}^{2}) \,d\alpha\,d\beta=0$

So $\int_{0}^{y} \int_{x}^{-x} \alpha\log({\alpha}^{2}) \,d\alpha\,d\beta=0$

For part (b), the reverse order of integration becomes

$\int_{0}^{\alpha\cos\beta} \int_{\sqrt{{a}^{2}-{x}^{2}}}^{x\tan\beta} \alpha\log({\alpha}^{2}) \,d\beta\,d\alpha$

thank you in advance

  • $\begingroup$ The problem statement is confusing. You have x and y as the variables of integration, while $\alpha\ and \ \beta$ are used to describe integration limits. Then x and y get defined in terms of the other two. I suggest you use $x=rcos(\theta ) \ and\ y=rsin(\theta)$ when making the switch to polar coordinates. $\endgroup$ – herb steinberg Apr 23 '18 at 2:24
  • $\begingroup$ @herbsteinberg, what about $\alpha$ and $\beta$, do i change them to $r$ and $\theta$ respectively? Like this $$\int_{0}^{r\sin\theta} \int_{r\cot\theta}^{\sqrt{{r}^{2}-{(r\sin(\theta))}^{2}}} \log({x}^{2}+{y}^{2}) \,dx\,dy$$ $\endgroup$ – Seth Mai Apr 23 '18 at 2:26
  • $\begingroup$ @herb steinberg, I meant: $$\int_{0}^{r\sin\theta} \int_{r\sin\theta\cot\theta}^{\sqrt{{r}^{2}-{(r\sin(\theta))}^{2}}} \log({x}^{2}+{y}^{2}) \,dx\,dy$$ $\endgroup$ – Seth Mai Apr 23 '18 at 2:33
  • $\begingroup$ @mr_e_man, is the way I did it incorrect? $\endgroup$ – Seth Mai Apr 23 '18 at 3:02
  • $\begingroup$ @mr_e_man, so how would I go about both parts. I am now stumped how to do it. Because I thought i can treat $\alpha$ and $\beta$ as $r$ and $\theta$ $\endgroup$ – Seth Mai Apr 23 '18 at 3:10

So the region of integration is a sector of a disk $x^2+y^2\le a^2$ with angle $\beta$ at the center. So in polar coordinates the integration becomes $$\int_0^{\beta}\int_0^ar\log r^2 dr d\theta=\beta(\int_0^ar\log r^2 dr)=\beta(\frac{a^2}2(2\log a-1))$$

Change of variables is as follows $$\int_0^{a\cos \beta}\int_0^{x\tan\beta}\log(x^2+y^2)dydx+\int_{a\cos\beta}^a\int_0^{\sqrt{a^2-y^2}}\log(x^2+y^2)dydx$$

  • $\begingroup$ Thank you for helping me to clear up the liimits of integration Much appreciate it. $\endgroup$ – Seth Mai Apr 23 '18 at 4:20

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