Is the value of these two same double Integrals of different upper limit of y equal when i shift to polar coordinates and use Jacobian?

Question

In a example question its given that $$\int\int(x^2+y^2)dydx$$ where x=0 to $\sqrt{1-y^2}$ and y=0 to 1. In an exercise question same question is given but the only difference is that y=0 to 1/2. But after shifting to polar coordinates, with r=0 to 1 and $\theta$=0 to pi/2 and Taking Jacobian it seems that answer of both are same and matching with the bookish answer given. Is it really so ? Image has been added for convenience

Answer 1

At the intersection of $y = r \sin\theta = \frac{1}{2}$ and circle with $r = 1$,

$\sin\theta = \frac{1}{2} \implies \theta = \frac{\pi}{6}$

a) So for $0 \leq \theta \leq \frac{\pi}{6}$, $0 \leq r \leq 1$

b) For $\frac{\pi}{6} \leq \theta \leq \frac{\pi}{2}$, the upper bound of $r$ is defined by the line $y = \frac{1}{2}$ and not the circle.

$y \leq \frac{1}{2} \implies r \sin\theta \leq \frac{1}{2}$

So you have, $0 \leq r \leq \frac{\csc \theta}{2}$

So you need to split the integral into two. Draw a diagram of the region to understand it better.