How to determine the same moment of inertia in two different ways?

[Tapias5000]

Homework Statement

determine the moment of inertia of the area with respect to the y-axis. with different rectangular elements, solve the problem in two ways: (a) with thickness dx, and (b) with thickness dy.

Relevant Equations

$I_x=\int _{ }^{ }y^2dA$

My solution is

now I am asked for the same result but in this form but I don't know where to start.

 

[anuttarasammyak]

How about
$$2\sigma \int_0^1 x^2 y \ \ dx=32\sigma \int_0^{\pi/2} \sin^2\theta \cos^2\theta d\theta$$
where $\sigma$ is area mass density of board.

 

[Tapias5000]

How about
$$2\sigma \int_0^1 x^2 y \ \ dx=32\sigma \int_0^{\pi/2} \sin^2\theta \cos^2\theta d\theta$$
where $\sigma$ is area mass density of board.

waat, Can you tell me the name of this method?

 

[anuttarasammyak]

I misinterpreted y so
$$2\sigma \int_0^1 x^2 y \ \ dx=2\int_0^1 x^2 (4-4x^2) \ \ dx$$
This a way (b).

Homework Statement:: determine the moment of inertia of the area with respect to the y-axis. with different rectangular elements, solve the problem in two ways: (a) with thickness dx, and (b) with thickness dy.
Relevant Equations:: $I_x=\int _{ }^{ }y^2dA$

This relevant equation seems inappropriate because x should be squared for y-axis rotation inertia.
For (a)

$$2\sigma\int_0^4 dy \int_0^{\frac{\sqrt{4-y}}{2}} x^2 dx$$