Problems with cylinder about finding area and volume. 
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.
 A: 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 $\theta = \pi$:  $$A = \int_{\theta = 0}^\pi r \sin \theta \, d\theta = 2r.$$
