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I was browsing through some past final exams, and I ran into this integral:

$\int_{0}^{1}\int_{0}^{x}ydydx,$

The question wants us to convert this integral to a polar integral.

I'm wondering how we convert this integral? I started drawing the region, which I got as a triangle in quadrant 1 with vertices (0,0), (1,0), (1,1). Then I tried using $y=rcos(\theta)$ and $dA = rdrd\theta$. But now I'm stuck with the limits of integration.

I tried converting the vertices of the triangle into polar coordinates and working them but that got me nowhere. So I'm not sure what to do next.

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  • $\begingroup$ Has the question asked you to convert to polar coordinates? Otherwise I don't see the advantage. Wouldn't it be easier just to first integrate wrt y, then x? $\endgroup$ May 28, 2020 at 10:55
  • $\begingroup$ Oh yes, I forgot to add that, but we have to convert the integral to a polar form. $\endgroup$ May 28, 2020 at 10:57

3 Answers 3

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To convert the given triangle to polar coordinates, draw a line at angle $\theta$, and the range of $r$ that falls within the region. For a line at $\theta$ with x-axis, the length of hypotenuse would be $\sec \theta$ . Also, the maximum angle you can have is $\theta = \frac{\pi}{4}$.

Hence

$$0 \leq \theta \leq \frac{\pi}{4} \\ 0 \leq r \leq \sec \theta$$

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One can rearrange the order of integration as well to get an easier integral to do. The bounds can be given by

$$\begin{cases}y = x \\ y = 0 \\ x = 1 \\\end{cases} \implies \begin{cases}\theta = \frac{\pi}{4} \\ \theta = 0 \\ r\cos\theta = 1 \\\end{cases}$$

Then we can arrange the integral to do $\theta$ first:

$$I = \int_0^1 \int_0^{\frac{\pi}{4}} r^2\sin\theta \:d\theta \:dr + \int_1^{\sqrt{2}} \int_{\sec^{-1}(r)}^{\frac{\pi}{4}} r^2 \sin\theta \:d\theta \:dr$$

$$ = \frac{\sqrt{2}-1}{6} + \int_1^{\sqrt{2}} r - \frac{r^2}{\sqrt{2}}\:dr = \frac{\sqrt{2}-1}{3}$$

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Hint. You need to write the line $x=1$ in polar coordinates to get $r\cos\theta=1.$ Then the region is determined by the inequalities $$0\le \theta \le π/4$$ and $$0\le r\le \frac{1}{\cos\theta}.$$

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