The picture is I think clearer now.
You are looking for the volume of the solid of rotation generated by sweeping a quarter circle around the center of the beaker.
Part of the problem is easy and part of it is subtler. If we take the distance (outer radius - inner radius) to be unit, the area of the quarter circle is just $\pi/4$.
The volume is simply that area multiplied by the distance it travels, which is $2 \pi R$, and the only question is, what to use for R? Pappus' Theorem says that we calculate that distance using the geometric centroid, whose coordinates are found as in the edit below (the value is the same in both dimensions, by symmetry).
So the coordinates of the centroid of the quarter circle are $\{ \frac{4}{3\pi}, \frac{4}{3\pi} \}$.
The radius of revolution should be: R = r_inner + $4/3\pi$ (r_outer - r_inner).
So finally the formula is, using R' as r_inner and r = r_outer - r_inner:
$V = A*2 \pi R = r^2 \frac{\pi} {4}2\pi (r \frac{4}{3 \pi}+ R') $
Edit: some detail about the centroid.
We have the area A from above. For $M_x,M_y$ we have
$M_y = \int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} x dydx$ = 1/3.
$M_x = \int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} y dydx$ = 1/3.
$x_{centroid} = M_y/A = (1/3) /( Pi/4) = 4/3\pi$
$y_{centroid} = M_x/A = (1/3) /( Pi/4) = 4/3\pi$
The expected distance from a unit circle to a randomly chosen point inside the circle is given by
$$
\frac{1}{\pi}\int_{0}^{1}(1-r)(2\pi r)dr=\left(r^2-\frac{2}{3}r^3\right)\bigg\vert_{0}^{1}=\frac{1}{3};
$$
as a function of the area of the circle, then, the expected distance is
$$
{d}_{\text{circle}}(A)=\frac{1}{3\sqrt{\pi}}\sqrt{A} \approx 0.1881 \sqrt{A}
$$
(which holds for circles of any size). For a square of side length $2$, the expected distance from the square to a random interior point (which can be calculated by considering a single quadrant) is
$$
\int_{0}^{1}y(2-2y)dy = \left(y^2-\frac{2}{3}y^3\right)\bigg\vert_{0}^{1}=\frac{1}{3}
$$
as well, so
$$
d_{\text{square}}(A)=\frac{1}{6}\sqrt{A} \approx 0.1667 \sqrt{A}
$$
for an arbitrary square. Finally, for a $2\times 4$ rectangle, you need to consider the short and long sides differently. Essentially you have two $2 \times 1$ end caps that behave like the square (so the average distance is $1/3$), and a $2\times 2$ central block for which the average distance is just $1/2$. The two components have equal areas, so the overall average distance is the average of $1/2$ and $1/3$, or $5/12$. Since the area of the entire rectangle is $8$, we have
$$
d_{\text{rect}}(A)=\frac{5}{24\sqrt{2}}\sqrt{A} \approx 0.1473 \sqrt{A}
$$
for any rectangle with aspect ratio $2$.
If it helps, you can think of "unrolling" each shape, while preserving its area, so that the set of points at distance $d$ from the perimeter lies along $y=d$. For the circle and the square (and for any regular polygon), this gives a triangle with base equal to the original shape's perimeter. For the rectangle, though, it gives a trapezoid, because the points maximally distant from the perimeter are a line segment, not a single point.
Best Answer
Required volume$=\pi{l}[(.307)^2-(.3)^2]$ therefore the nearest estimate will be $\frac{\pi{l}[(.307)^2-(.3)^2)]}{\pi{l}[(.307)^2}=1-\frac{3^2}{.307^2}$
sir if someone says what is the length without giving you any clue you got to say it is half of full length of the bridge.