Computing the polar moment of inertia (calculus)

[Leo Liu]

Homework Statement

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Relevant Equations

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Question:

Diagram:

So the common approach to this problem is using polar coordinates.
The definition of infinitesimal rotational inertia at O is $dI_O=r^2\sigma\, dA$. Therefore the r. inertia of the triangle is
$$I_O=\int_{0}^{\pi/3}\int_{0}^{\sec\theta}r^2r\,drd\theta$$
whose value is $=\sqrt 3 /2$.

But when I used rectangular coordinates to solve this problem, I got a different answer. The steps are shown below.
$$\xcancel{\begin{aligned}
I_O&=\int_{0}^{1}\int_{0}^{2x}x^2+y^2\,dydx\\
&=\int_0^1x^2y+\frac{y^3} 3\Bigg|_0^{2x}dx\\
&=\int_0^12x^3+\frac 8 3x^3\,dx\\
&=\frac{x^4} 2+\frac 2 3x^4\Bigg|_0^1=\frac 7 6
\end{aligned}}$$
Can someone please tell me where my mistakes are? Thanks!

Edit:
$$\begin{aligned}
I_O&=\int_{0}^{1}\int_{0}^{\sqrt 3 x}x^2+y^2\,dydx\\
&=\int_0^1x^2y+\frac{y^3} 3\Bigg|_0^{\sqrt 3 x}dx\\
&=\int_0^1 \sqrt 3 x^3+\sqrt 3 x^3\,dx\\
&=\frac{2\sqrt 3}{4}=\frac{\sqrt 3}2
\end{aligned}$$

 

[BvU]

Perhaps the slope of the line is not 2 but a little less ...