The collection of lines parallel to the y-axis have the form $x =a$. Let $I=I_y$ be the usual moment of inertia around the y-axis $$I=\iint_R x^2 \delta \, \mathrm dA$$
Let $\bar{I}$ be the moment of inertia around the axis $x=\bar x$, where $(\bar x, \bar y)$ is the center of mass. Show that $$I = \bar I + M \bar x^2$$
Here was my attempt:
$I = \bar I + M \bar x^2$ means $I - \bar I = M \bar x^2$. We compute the LHS below:
$$\iint_R x^2 \delta \, \mathrm dA - \iint_R (x-\bar x)^2 \delta \, \mathrm dA$$
$$\iint_R x^2 \delta \, \mathrm dA - \bigg(\iint_R x^2 \delta \, \mathrm dA - \iint_R 2x\bar x \, \delta \, \mathrm dA + \iint_R \bar x^2 \delta \, \mathrm dA \bigg )$$
$$\iint_R 2x\bar x \, \delta \, \mathrm dA - \iint_R \bar x^2 \delta \, \mathrm dA $$
$$\iint_R (2x\bar x - \bar x^2)\, \delta \, \mathrm dA $$
But I don't see why $\iint_R (2x\bar x - \bar x^2)\, \delta \, \mathrm dA = M \bar x^2$. Integrating with respect to $\mathrm dx \mathrm dy$ doesn't give me anything useful either. Did I make a mistake, or is there an intuitive property of $2x\bar x + \bar x^2$ that I fail to see?
Is there a better way to solve this problem?
You are almost there. You just need to use the formula for the center of mass. And also remember that $\bar x$ can be taken outside of the integral. $$\iint_R (2x\bar x - \bar x^2)\, \delta \, \mathrm dA=2\bar x \iint_R x \delta \, \mathrm dA-\bar x^2\iint_R \delta \, \mathrm dA$$ Now $$\iint_R \delta \, \mathrm dA=M$$ and $$\bar x=\frac{1}{M}\iint_R x \delta \, \mathrm dA$$ Therefore $$\iint_R (2x\bar x - \bar x^2)\, \delta \, \mathrm dA=2\bar x(\bar xM)-\bar x^2M=M\bar x^2$$