Fdtxjr

$$intint_D(1+xy)mathrm dx;mathrm dy$$ where $D$ is the region given by $ygeq 0$ and $1leq x^2+y^2leq 2$.

I have been told to evaluate this integral but I am struggling to do so. I have changed the inequality for $1<(x^2)+(y^2)<2$ so it shows the limits for $x$ but the limits for $y$ tend to $0$ and an unknown value which confuses me. What would be the right way to answer this question?

The easiest way is to use polar coordinates... The integration region is given by
$$
D = (x,y)in mathbbR^2: y ge 0 wedge 1 leq x^2+y^2 leq 2
$$

or, in polar coordinates, by

$$
D^* = (rho, theta): 1 leq rho leq 2 wedge 0 leq theta leq pi
$$

now you have that
$$
iint_D (1+xy) dx dy = iint_D^* rho (1+(rho cos theta)(rho sin theta))d rho dtheta = int_0^pi int_1^2rho(1+frac 12 rho^2 sin(2 theta))drho dtheta = cdots = frac3 pi2
$$

Hint: this integral is equal to $$
int_0^pi , int_1^2 ,(1+r^2sinthetacostheta)r mathrm dr mathrm dtheta
$$

The domain $D$ is the area between disks of radius $1$ and $2$, above the $y=0$ (the horizontal axis). In addition to the solutions in the polar coordinates, you can observe that the integral of the $xy$ term is zero (for every positive $y$ you integrate symmetrically around $0$). Then $$iint_D(1+xy)dxdy=iint_D dxdy$$
This is just the area of $D$: $$frac 12 (pi 2^2-pi 1^2)=frac 32pi$$

$$I=iint_D(1+xy)dxdy$$
First lets think about what the region $D$ represents. since $yge0$ and $1le x^2+y^2le2$ we know that we are only looking at the top two quadrants. we also know that the formula $x^2+y^2=r^2$ represents a circle of radius, $r$. Since we have a lower limit of $1$ rather than $0$, we have a half-donut shape as our region. To evaulate this integral it is easiest to express in polar coordinates, which have the following conditions:
$$dxdy=rdrdtheta$$
$$x=rcostheta$$
$$y=rsintheta$$
$$x^2+y^2=r^2$$
so we know that our integral will be of the form:
$$I=iint_Omegaleft(1+r^2sinthetacosthetaright)rdrdtheta$$
To determine the upper and lower bounds we must think about the polar angle, $theta$ that we will cover. Since it is defined as going from the $+x$ $+y$ quadrant round anticlockwise, and we only want the positive region, we can see that this must be $0lethetalepi$. The limits for $r$ are easy since we already have them in the right form so we can say that $1le rle2$ . If we now substitute these in we obtain:
$$I=int_0^piint_1^2left[r+r^3sinthetacostheta right]drdtheta$$
$$I=int_0^pileft[fracr^22+fracr^44sinthetacosthetaright]_r=1^2dtheta=int_0^pileft[frac32+frac154sinthetacosthetaright]dtheta=int_0^pileft[frac32+frac158sin2thetaright]dtheta=left[frac32theta-frac1516cos2thetaright]_theta=0^pi=frac32pi$$
And this is your final answer. Normally for any circular region defined in this way polar coordinates are the easiest way to proceed.