By rotating the cross-section figure around the x-axis, we obtain a model of the light bulb. We decompose it into three parts. We recall that:

• the volume of a cylinder is given by the formula $\pi r^2 h$ where $r$ is the radius of the base disk and $h$ is the height;
• the volume of a sphere of radius $r$ is given by the formula $\frac{4}{3}\pi r^3$.

We also admit that, for every real number $x$ in the interval $[0;4]$, $f(x) = 2 - \cos\left(\frac{\pi}{4}x\right)$.
The points are $\mathrm{A}(-1;1)$, $\mathrm{B}(0;1)$, $\mathrm{C}(4;3)$, $\mathrm{D}(7;0)$, $\mathrm{E}(4;-3)$, $\mathrm{F}(0;-1)$, $\mathrm{G}(-1;-1)$.

1. Calculate the volume of the cylinder with cross-section the rectangle $ABFG$.
2. Calculate the volume of the hemisphere with cross-section the half-disk with diameter $[CE]$.
3. To approximate the volume of the solid with cross-section the shaded region BCEF, we divide the segment $[OO']$ into $n$ segments of equal length $\frac{4}{n}$ then we construct $n$ cylinders of equal height $\frac{4}{n}$. a. Special case: in this question only we choose $n = 5$. Calculate the volume of the third cylinder, shaded in the figures, then give its value rounded to $10^{-2}$. b. General case: in this question, $n$ denotes any non-zero natural number. We approximate the volume of the solid with cross-section BCEF by the sum of the volumes of the $n$ cylinders thus created by choosing a sufficiently large value of $n$. Copy and complete the following algorithm so that at the end of its execution, the variable $V$ contains the sum of the volumes of the $n$ cylinders created when $n$ is entered. \begin{verbatim} $V \leftarrow 0$ For $k$ going from...to ... : $\mid V \leftarrow \ldots$ Fin For \end{verbatim}

As shown in the figure, a sector ABD with center A and central angle $90 ^ { \circ }$ is drawn in a square ABCD with side length 5. Let $\mathrm { A } _ { 1 }$ be the point that divides segment AD in the ratio $3 : 2$, and let $\mathrm { B } _ { 1 }$ be the point where the line passing through $\mathrm { A } _ { 1 }$ and parallel to segment AB meets arc BD. A square $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 } \mathrm { C } _ { 1 } \mathrm { D } _ { 1 }$ is drawn with segment $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 }$ as one side and meeting segment DC, and then a sector $\mathrm { D } _ { 1 } \mathrm {~A} _ { 1 } \mathrm { C } _ { 1 }$ with center $\mathrm { D } _ { 1 }$ and central angle $90 ^ { \circ }$ is drawn. Let $\mathrm { E } _ { 1 } , \mathrm {~F} _ { 1 }$ be the points where segment DC meets arc $\mathrm { A } _ { 1 } \mathrm { C } _ { 1 }$ and segment $\mathrm { B } _ { 1 } \mathrm { C } _ { 1 }$ respectively. The region enclosed by two segments $\mathrm { DA } _ { 1 } , \mathrm { DE } _ { 1 }$ and arc $\mathrm { A } _ { 1 } \mathrm { E } _ { 1 }$, and the region enclosed by two segments $\mathrm { E } _ { 1 } \mathrm {~F} _ { 1 } , \mathrm {~F} _ { 1 } \mathrm { C } _ { 1 }$ and arc $\mathrm { E } _ { 1 } \mathrm { C } _ { 1 }$ are shaded to obtain the figure $R _ { 1 }$.
In figure $R _ { 1 }$, a sector $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 } \mathrm { D } _ { 1 }$ with center $\mathrm { A } _ { 1 }$ and central angle $90 ^ { \circ }$ is drawn in square $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 } \mathrm { C } _ { 1 } \mathrm { D } _ { 1 }$. Let $\mathrm { A } _ { 2 }$ be the point that divides segment $\mathrm { A } _ { 1 } \mathrm { D } _ { 1 }$ in the ratio $3 : 2$, and let $\mathrm { B } _ { 2 }$ be the point where the line passing through $\mathrm { A } _ { 2 }$ and parallel to segment $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 }$ meets arc $\mathrm { B } _ { 1 } \mathrm { D } _ { 1 }$. A square $\mathrm { A } _ { 2 } \mathrm {~B} _ { 2 } \mathrm { C } _ { 2 } \mathrm { D } _ { 2 }$ is drawn with segment $\mathrm { A } _ { 2 } \mathrm {~B} _ { 2 }$ as one side and meeting segment $\mathrm { D } _ { 1 } \mathrm { C } _ { 1 }$, and then a shaded figure is drawn and colored in square $\mathrm { A } _ { 2 } \mathrm {~B} _ { 2 } \mathrm { C } _ { 2 } \mathrm { D } _ { 2 }$ in the same way as obtaining figure $R _ { 1 }$ to obtain the figure $R _ { 2 }$. Continuing this process, let $S _ { n }$ denote the area of the shaded part in the $n$-th obtained figure $R _ { n }$. What is the value of $\lim _ { n \rightarrow \infty } S _ { n }$? [4 points]
(1) $\frac { 50 } { 3 } \left( 3 - \sqrt { 3 } + \frac { \pi } { 6 } \right)$
(2) $\frac { 100 } { 9 } \left( 3 - \sqrt { 3 } + \frac { \pi } { 3 } \right)$
(3) $\frac { 50 } { 3 } \left( 2 - \sqrt { 3 } + \frac { \pi } { 3 } \right)$
(4) $\frac { 100 } { 9 } \left( 3 - \sqrt { 3 } + \frac { \pi } { 6 } \right)$
(5) $\frac { 100 } { 9 } \left( 2 - \sqrt { 3 } + \frac { \pi } { 3 } \right)$

A water pitcher has a hemispherical bottom and a neck in the shape of two truncated cones of the same size. The vertical cross-section of the pitcher with relevant dimensions is shown in the figure. Suppose that the pitcher is filled with water to the brim. If a solid cylinder with diameter 24 cm and height greater than 60 cm is inserted vertically into the pitcher as far down to the bottom as possible, how much water would remain in the pitcher?
(A) $6316\pi \text{ cm}^3$
(B) $6116\pi \text{ cm}^3$
(C) $6336\pi \text{ cm}^3$
(D) $6136\pi \text{ cm}^3$

The position of two iron balls in the shape of spheres with radius 3 units placed inside a right circular cylinder with radius 6 units is shown in Figure 1.
The cylinder is filled with water until both balls are completely submerged in water and the view in Figure 2 is obtained.
Accordingly, what is the volume of water in the cylinder in Figure 2 in cubic units?
A) $96 \pi$
B) $108 \pi$
C) $120 \pi$
D) $132 \pi$
E) $144 \pi$

A square right prism with edge lengths 10, 10, 25 units is divided into unit cubes. Then, using all of these cubes, a square right prism with height 1 unit is formed with no gaps between them.
Accordingly, what is the surface area of this square right prism in square units?
A) 5200 B) 5400 C) 5600 D) 5800 E) 6000

A right circular cone with height 10 units is placed inside a hollow right circular cylinder with height 10 units as shown in Figure 1. Water with volume $\mathrm { V } _ { 1 }$ cubic units is poured between this cylinder and cone, and the water height becomes 5 units. Then this object is inverted as shown in Figure 2, and after adding more water, the water volume becomes $\mathrm { V } _ { 2 }$ cubic units and the height becomes 5 units.
Accordingly, what is the ratio $\frac { \mathrm { V } _ { 1 } } { \mathrm {~V} _ { 2 } }$?
(During this process, water does not enter the cone.)
A) $\frac { 3 } { 7 }$ B) $\frac { 5 } { 11 }$ C) $\frac { 8 } { 15 }$ D) $\frac { 10 } { 21 }$ E) $\frac { 15 } { 31 }$