G is quarter of circle ring in frist quadrant: $G=\left\{ \begin{pmatrix} x \\ y \end{pmatrix} x\in \mathbb R ^2 : x\geq 0, y\geq 0, 9\leq x^2+y^2\leq 49\right\} $
Calculate its area: $\int_{G}^{} \! 1 \, d(x,y) $.
Can someone help me to determine the boundaries of the integral?
On
Kaj Hansen addresses the simpler option. If you insist on using Cartesian coordinates, then
$$\int_G1\ \mathrm{d}x\ \mathrm{d}y=\int_0^7\int_{\sqrt{9-x^2}}^\sqrt{49-x^2}1\ \mathrm{d}y\ \mathrm{d}x.$$
Here you do not have to know about the Jacobian determinant. However, you should be pretty good at manual integration involving roots.
Limits on $x$
Since we are in the first quadrant and the larger circle's $x$-intercept at $x=7$, then
$$0\le x\le7.$$
Limits on $y$
Solve the equations $$x^2+y^2=9, \ \ \ \ x^2+y^2=49$$
for $y$, taking into account that $y\ge0$.
Also note that $G$ is called a quarter-annulus (more familiar than quarter-circle ring).
The easiest way to approach this would be to use polar coordinates. We have $3 \leq r \leq 7$ and $0 \leq \theta \leq \pi/2$. The integrand changes$^\dagger$ from $1 \ dx \ dy$ into $r \ dr \ d \theta$, so:
$$\int_G 1 \ dx \ dy = \int_0^{\pi/2} \int_3^7 r \ dr \ d\theta$$
$^\dagger$To elaborate, if changing to polar coordinates is unfamiliar, the substitution is $(x, y) = f(r, \theta) = \langle r \cos(\theta), r\sin(\theta) \rangle$. If we proceed according to this, the differential $dx \ dy$ transforms as: $$dx \ dy = \det \left( \left[ \begin{array}{ c c } \frac{\partial f_1}{\partial r} & \frac{\partial f_1}{\partial \theta} \\ \frac{\partial f_2}{\partial r} & \frac{\partial f_2}{\partial \theta} \end{array} \right] \right) dr \ d \theta = \det \left( \left[ \begin{array}{ c c } \cos(\theta) & -r \sin(\theta) \\ \sin(\theta) & r\cos(\theta) \end{array} \right] \right) dr \ d \theta \\ = \ r(\cos^2(\theta) + \sin^2(\theta)) \ dr \ d \theta \ = \fbox{$ \ r \ dr \ d \theta \ $} $$