As be seen here, I have problems with it because my answer didn't compatible to the solutions (only the question $(3)$ that is exactly the same as in the solution)
$(1)$ I found out that the cross section is rectangle with the side $2\sqrt{r^2-t^2}$ and $\sqrt{r^2-t^2}$ the former follows from Pythagoras at the segment of cycle and the latter is from substituting $x=t$ in the circle inequality and get $y=\pm\sqrt{r^2-t^2}$ and as $z=y$ we have the height of rectangle be $\sqrt{r^2-t^2}$ and then my answer is $2(r^2-t^2)$, but the solution is $\dfrac{1}{2}\left(r^2-t^2\right)$
$(2)$ I calculate the volume by integrating all area of $(1)$ from $0$ to $r$, and thus, my answer to this is $\displaystyle \int_{0}^r 2(r^2-t^2)dt=\dfrac{4r^3}{3}$ while the solution is $\dfrac{2r^3}{3}$
$(3)$ this is fine as I corrected.
$(4)$ I calculate the side area by accumulating all the area of rectangle from $(3)$ which have area $=r^2\theta\sin\theta$ with the help of integration from $0$ to $\pi$ of the variable $\theta$, which is $\displaystyle \int_{0}^\pi r^2\theta\sin\theta d\theta=\pi r^2$ and the solution is $2r^2$, actually I don't understand this question genuinely, in particular with the term "with $x^2+y^2=r^2$" in teh question.
So please give me the method to calculate all of these. PS. I'm not sure what to tag this question.
For (1), this is not at all correct. The cross section is a right triangle. To understand why, first observe that the volume is bounded below by $z \ge 0$, which is the $xy$-plane, and above by $z \le y$. Therefore, the solid is a semicircular wedge shape in which, for a given $x = t$ with $|t| \le r$ satisfies the simultaneous relationships $$t^2 + y^2 \le r^2 \\ 0 \le z \le y$$ which upon solving for $y$ in terms of $r$ and $t$, and solving $z$ in terms of $y$, yields $$0 \le z \le y \le \sqrt{r^2 - t^2}.$$ This means, for a given "slice" of the solid at the plane $x = t$, the region is enclosed by the lines $$y = \sqrt{r^2 - t^2}, \\ z = y, \\ 0 = z.$$ This is a right triangle with vertices at $$(t,0,0), \left(t,\sqrt{r^2-t^2},0\right), \left(t,\sqrt{r^2-t^2},\sqrt{r^2-t^2}\right),$$ and the right angle is at the second vertex in this list. Consequently the area of this triangle is simply $$A(t) = \frac{1}{2} \left(\sqrt{r^2 - t^2}\right)^2 = \frac{r^2 - t^2}{2}.$$ It is extremely useful to sketch the solid of interest here.
Next, the differential volume for such a triangle is $$dV(t) = A(t) \, dt = \frac{r^2 - t^2}{2} \, dt$$ and the total volume is obtained by integrating from $t = -r$ to $t = r$: $$V = \int_{t=-r}^r \frac{r^2-t^2}{2} \, dt = \left[\frac{r^2 t}{2} - \frac{t^3}{6}\right]_{t=-r}^r = \frac{2r^3}{3}.$$
The last part is simply asking you to find the lateral surface area of the portion of the cylinder that lies between the planes $z = 0$ and $z = y$. To do this, it is immediately clear that the question suggests computing this area as a function of the angle $\theta$. The height of the cylinder given this angle is $r \sin \theta$, so the differential surface area of a strip of width $d\theta$ along this circumference is $$dA(\theta) = r \sin \theta \, d\theta.$$ Consequently the total area is found by integrating from $\theta = 0$ to $\theta = \pi$: $$A = \int_{\theta = 0}^\pi r \sin \theta \, d\theta = 2r.$$