Assume Line $l$ is the axis of the ball, and the radius of the ball is $R$.
I want to calculate the Moment of Inertia for the ball with cylinder:
I take cylinder in the ball as is shown in the img:
Assume the angle noted on the img is $\theta$, ranging from $0$ to $\dfrac{\pi}{2}$.
For each $\theta$, the area of the surface for the cylinder is $2\pi R\sin{\theta}\cdot 2R\cos{\theta}$.
And then the moment of inertia for the cylinder is $\rho(2\pi R\sin{\theta}\cdot 2R\cos{\theta})\cdot (R\sin{\theta})^2$
Finally I have $I = \int_0^{\frac{\pi}{2}} \rho(2\pi R\sin{\theta}\cdot 2R\cos{\theta})\cdot (R\sin{\theta})^2 d\theta$.
This is not correct, for that I know the $I$ should be $\frac{8\pi}{15}\rho R^5$. It contains a $R^5$, not the $R^4$ in my integration.
Where have I gone wrong in my method? And how to correct this method with 'cylinder'?
Your moment of inertia for a thin-walled cylindrical tube is incorrect: you are forgetting the differential thickness of the tube wall. If the tube has radius $r$, height (or length) $L$, and wall thickness $dr$, then its moment of inertia about its symmetry axis is $dI_{zz} = 2 \pi \rho r^3 L dr$. Here, $r = R \sin \theta$ as you've correctly identified, which also implies $dr = R \cos \theta d\theta$. The height (or length) of the tube is $L = 2 R \cos \theta$. Therefore,
$$dI_{zz} = 2 \pi \rho (R \sin \theta)^3 (2 R \cos \theta) (R \cos \theta d\theta)$$
$$= 4 \pi \rho R^5 \sin^3 \theta \cos^2 \theta d\theta$$
Integrating,
$$I_{zz} = \int_{\theta = 0}^{\theta = \frac{\pi}{2}} dI_{zz} = \int_0^\frac{\pi}{2} 4 \pi \rho R^5 \sin^3 \theta \cos^2 \theta d\theta$$
$$= \rho R^5 \left[ \int_0^\frac{\pi}{2} 4 \pi \sin^3 \theta \cos^2 \theta d\theta \right]$$
The integral evaluates to $\frac{8 \pi}{15}$, as you can check.