As a double integral, this is $I=\int_R dA$, where $R$ is the region given by the intersection of the two disks. This integral can be evaluated as an iterated integral in several ways. Picking up where you left off, we can integrate with respect to $y$ first, then with respect to $x$, i.e., $$I=2\int_\frac12^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dy\;dx.$$ You could take further advantage of symmetry by changing the lower bound of $y$ to $0$ and multiplying by $2$, but that doesn’t really make things any simpler.
You could instead integrate with respect to $x$ first:$$I=4\int_0^{\frac{\sqrt{15}}2}\int_\frac12^{\sqrt{4-y^2}} dx\;dy.$$ The key thing in both cases is that the bounds of the inner integral will be variable. I often find it helpful to draw a typical line segment for the inner integral to make sure that I get the bounds right. For example, in this second integral, for each value of $y$, the integral runs over a line segment parallel to the $x$-axis that runs from $x=\frac12$ out to the edge of the left-hand disk.
If you don’t want to deal with iterated integrals at all, this problem is a good candidate for Green’s theorem: $I=\int_R dA=\int_{\partial R}y\;dx$. You can often make the line integral simpler without changing its value by adding the differential of some function to the integrand. If you parametrize the bounding arcs as $\langle 2\cos t,2\sin t\rangle$ and $\langle 1-2\cos t,-2\sin t\rangle$, respectively, and use the differential form $\frac12(x\;dy-y\;dx)$ for the line integral, the integrals end up being very simple to evaluate:$$I=\int_{-\arccos\frac14}^{\arccos\frac14}2\;dt+\int_{-\arccos\frac14}^{\arccos\frac14}2-\cos t\;dt=\int_{-\arccos\frac14}^{\arccos\frac14}4-\cos t\;dt.$$
To evaluate this integral most efficiently, it's better to switch to new variables that better reflect the region of integration (and won't ruin the integrand). Here's an outline of the solution.
Let $u=\frac{y}{x}$. Then two of the bounding equations, $y=\frac{x}{4}$ and $y=2x$ tell us that $u$ ranges in the interval $u\in\left[\frac{1}{4},2\right]$.
Let $v=xy$. Then the other two of the bounding equations, $y=\frac{1}{x}$ and $y=\frac{4}{x}$ tell us that $v$ ranges in the interval $v\in[1,4]$.
Now the given double integral can be rewritten as
$$\iint_R e^{xy/2}\,dA=\int_{1/4}^2\int_1^4 e^{-v/2}\,|J|\,dv\,du,$$
where $J$ is the Jacobian of this change of variables.
Solving $u=\frac{y}{x}$ and $v=xy$ for $x$ and $y$, we find that $x=u^{-1/2}v^{1/2}$ and $y=u^{1/2}v^{1/2}$. Therefore the Jacobian is
$$J=\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}=\begin{vmatrix} -\frac{1}{2}u^{-3/2}v^{1/2} & \frac{1}{2}u^{-1/2}v^{-1/2} \\ \frac{1}{2}u^{-1/2}v^{1/2} & \frac{1}{2}u^{1/2}v^{-1/2} \end{vmatrix}=-\frac{1}{2}u^{-1}.$$
Then the integral can be computed further as
$$\int_{1/4}^2\int_1^4 e^{-v/2}\,|J|\,dv\,du=\int_{1/4}^2\int_1^4 e^{-v/2}\cdot\frac{1}{2}u^{-1}\,dv\,du=\frac{1}{2}\cdot\int_{1/4}^2 u^{-1}\,du\cdot\int_1^4 e^{-v/2}\,dv \\ =\frac{1}{2}\cdot\left[\ln|u|\right]_{1/4}^2\cdot\left[-2e^{-v/2}\right]_1^4=-\left(\ln2-\ln\frac{1}{4}\right)\cdot\left(e^{-2}-e^{-1/2}\right)=\left(e^{-1/2}-e^{-2}\right)\ln8.$$
Best Answer
It will help to draw a picture of your domain. In this case you can choose to integrate over $y$ first, and then (as you will be able to see from a plot), $y$ ranges from 0 to $|x|-1$. You can then split up your integral into 2 domains in x, $(-1,0)$, and $(0,1)$, You then get:
$\int_{-1}^0 \int_0^{x+1} f(x,y) dy dx$ on $(-1,0)$,
$\int_0^1 \int_0^{1-x} f(x,y) dy dx$ on $(0,1)$,
where $f(x,y) = y-2x^2$ and $dy dx$ means you integrate over y first, then x.The idea here is that you integrate up to one variable to get your function in terms of one variable then you have an absolute integral.