# Evaluating double integral by change in variable.

Evaluate the following integral by changing to polar coordinates: $$\iint x dxdy$$, where $$0\leq x \leq y$$ and $$0 \leq y \leq 1$$. Above integral can be evaluated directly without changing the variables. I am getting the answer $$\frac{1}{6}$$. But I have to evaluate it by changing in polar coordinates. So let $$x=r\cos\theta$$ and $$y=r\sin\theta$$. Then $$dxdy=rdrd\theta$$. But what will be the limits on $$r$$ and $$\theta$$ ?

• You want the region between the lines y=x and y=0 and your radius is going between 0 and 1. Try drawing it out and see what u get Jan 12, 2021 at 3:22
• @HenryLee I think the limit of $\theta$ will be $\frac{\pi}{4}$ to $\frac{\pi}{2}$. I am not sure whether I am correct or not and I have no idea for $r$. Jan 12, 2021 at 3:26
• @HenryLee I drew the region and the limits of r will be $0\leq r \leq \frac {1}{\sin\theta}$ Jan 12, 2021 at 3:46

Hint: Draw the region is key to understand the integral. If you do the graph you will find $$\frac{\pi}{4}\leq \theta\leq \frac{\pi}{2}$$ and $$0\leq y\leq 1$$ $$0\leq r \text{sin}(x)\leq 1$$ $$0\leq r\leq\frac{1}{\text{sin}(x)}$$
• How $0\leq r \ leq 1$? Jan 12, 2021 at 3:31
• @Mathfun, remember $r=\sqrt{(x^2+y^2)}$ and $0\leq x\leq y\leq 1$
• I drew the region and the limits of r will be $0\leq r \leq \frac {1}{\sin\theta}$ Jan 12, 2021 at 3:46