Expressing the area of the cross section of a cylinder I was able to solve similar questions where a triangle was involved, but this one has had me looping and pondering on it for the past week.
The Condition:
Let $r$ be a positive constant. Consider the cylinder $x^2+y^2\leq r^2$ and let C be  part of the cylinder that satisfies $0\leq z\leq y$.
The questions:
(1) Consider the cross section of $C$ by the plane $x=t(-r\leq t\leq r)$ and express its area in terms of $r,t$
(2) Calculate the volume of $C$, and express it in terms of $r$.
(3) Let $a$ be the length of the arc along the base of the circle $C$ from the point $(r,0,0)$ to the point $(r cos\theta,r sin\theta,0)(0\leq\theta\leq\pi)$. Let $b$ be the length of the line segment from the point $(r cos\theta, r sin\theta, 0)$ to the point $(r cos\theta, r sin\theta, r sin\theta)$. Express $a$ and $b$ interms of $r,\theta$
(4)Calculate the are of the side of $C$ with $x^2+y^2=r^2$, and express it in terms of $r$.
Here are the expected answers(which I keep failing to get to):
(1) $\frac{1}{2}(r^2-t^2)$
(2) $\frac{2}{3}r^3$
(3) $r\theta$ and $r sin\theta$
(4) $2r^2$




*

*Required area is of triangular form, whose base length is $|y_0 -0|$, where $y_0$ is the coordinate on y-axis at the intersection of plane $y = z$ and $x^2 + y^2 = r^2$, and height is corresponding $z-$ coordinate, say $z_0$ , since $x = t \Rightarrow y = \sqrt{r^2 - t^2} = z$ $$ \Rightarrow A = \cfrac{y_0 z_0}{2} = \cfrac{r^2 - t^2}{2}$$
2.Since, $$V = \iiint dx dy dz$$ And the limits are following: $-\sqrt{r^2 - y^2} \le x \le \sqrt{r^2 - y^2}, 0 \le z \le y$ and $ 0 \le y \le r$, which on evaluating gives $$ V = \cfrac{2}{3} r^3 $$


*Use simple distance formula,
For first part, Simple arc length would suffice, $$a = 2 \pi r \cfrac{\theta}{2 \pi} = r \theta$$ 
and for second part use simple distance formula, $$b = \sqrt{ ( r \cos \theta -  r \cos \theta)^2 + ( r \sin \theta -  r \sin \theta)^2 + (  r \sin \theta - 0)^2 } = r \sin \theta$$
4.Required area is enclosed by two curve and is lying on the cylinder:

$dA$ = Infinitesimal arc length $\times$ height at that point
Here,  Infinitesmial arc length $ = r d \theta$ , hieght $ r \sin \theta$
$$\Rightarrow dA = r \sin \theta r d\theta$$
$$ \Rightarrow A = \int^{\pi}_{0} r \sin \theta r d\theta = 2r^2 $$
