First, a comment. Presumably you have drawn a picture. Note the symmetry about the $y$-axis of the region we are rotating. It is definitely more pleasant to go from $x=0$ to $x=2$, and then double the result.
We still have to split the integral, but into two parts only. And we get a lot fewer minus signs.
Secondly, the formula you are using is not quite the appropriate one. It would be correct if we were rotating a region entirely above the $x$-axis about the $x$-axis. Take a tiny $dx\times dy$ rectangle in your region, with say lower left corner at $(x,y)$. The distance of this rectangle from the line $y=-2$ is roughly $2+y$. (Note this also works for our negative $y$, since $y=-2$ is below the region we are rotating.) so the volume swept out by the little rectangle is approximately $2\pi(2+y)\,dx\,dy$. That is what you need to integrate.
Remark: Or else you can use the formula you were given. Then we must make the the $x$-axis as the axis of rotation.
Lift everything up by $2$. We are rotating the region between $y=|x^2-1|+2$ and $y=1$ about the $x$-axis. For simplicity use symmetry. Let $D^+$ be the part of the modified region that is in the first quadrant. Then our volume is $2\iint_{D^+} y\,dx\,dy$.
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
Try a change of variables by letting $u=x^2/y$ and $v=xy$. Then the region $R$ is transformed into the rectangle $[1/4,1]\times [1,5]$. Since the Jacobian determinant is $$\det\left(\begin{bmatrix} \frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y} \end{bmatrix}\right) =\det\left(\begin{bmatrix} 2x/y& -x^2/y^2\\ y& x \end{bmatrix}\right)=\frac{3x^2}{y},$$ it follows that $$\iint_R\frac{y^2}{x}dydx=\frac{1}{3}\int_{u=1/4}^1\int_{v=1}^5\frac{y^3}{x^3}dvdu= \frac{1}{3}\int_{1/4}^1\frac{du}{u^2}\cdot\int_{1}^5vdv.$$ Can you take it from here?