How to convert this integral to a polar integral, $\int_{0}^{1}\int_{0}^{x}ydydx$

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.

• 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? May 28, 2020 at 10:55
• Oh yes, I forgot to add that, but we have to convert the integral to a polar form. May 28, 2020 at 10:57

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

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

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:

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

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

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 $$0\le r\le \frac{1}{\cos\theta}.$$