I have a double integral:
$$\iint (x+y)\,dx\, dy$$
with circle constraint:
$$x^{2}+y^{2}=x+y$$
I tried to calculate it with transition to polar coordinates:
$$x^{2}+y^{2}=x+y$$
$$\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{2}$$
In polar coordinates:
$$r^{2}(\cos(\varphi))^{2} + r^{2}(\sin(\varphi))^{2} = r\cos(\varphi) + r\sin(\varphi)$$
$$r = \cos(\varphi) + \sin(\varphi)$$
Graph looks like this:
But i don't understand how to find polar radius change interval here. If i separate circle into two, for first half circle for example it will go from $\textbf{some point}$ to $\frac{\pi}{2}$. I don't understand how to find that $\textbf{some point}$, cause it starts from point ($\frac{1}{2}-\frac{1}{\sqrt{2}} = -0.2071$).
Best Answer
There are 3 possible answers I can see, depending on what you meant.
1: What you said—a 2D integral over a 1D set. In this case the answer is 0, no working required.
2: A line integral over the curve $C:x^2+y^2=x+y$. We can parametrise this curve as $(x,y)=r(\theta)=\left(\frac{1}{\sqrt{2}}\cos\theta+\frac{1}{2},\frac{1}{\sqrt{2}}\sin\theta+\frac{1}{2}\right)$ for $\theta\in[0,2\pi]$. A quick check shows that $|r'(\theta)|=\frac{1}{\sqrt{2}}$. So then the integral is $$\int_Cf(x,y)ds = \int_0^{2\pi}f(r(\theta))|r'(\theta)|d\theta = \int_0^{2\pi}\left(\frac{1}{2}(\cos\theta+\sin\theta)+\frac{1}{\sqrt{2}}\right)d\theta,$$ which I'm sure you can work out.
3: A genuine double integral over the disc $D:x^2+y^2\leq x+y$. We can parametrise this in (I believe) a clearer way than has already been given: $(x,y)=\left(r\cos\theta+\frac{1}{2},r\sin\theta+\frac{1}{2}\right)$, for $\theta\in[0,2\pi]$ and $r\in[0,1/\sqrt{2}]$. Recall that the area element in polar coordinates is $dxdy=rdrd\theta$. Thus the integral is: $$\iint_D(x+y)dxdy=\int_0^{2\pi}\int_0^{1/\sqrt{2}}\left(r^2(\cos\theta+\sin\theta)+r\right)drd\theta.$$ This splits into two terms and each term is simply the product of 2 1-variable integrals, so I'm sure you can work this out as well.