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}

We consider the function $f$ defined on $\mathbb { R }$ by:
$$f ( x ) = \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { x } + 1 }$$
Let A be a point on $\mathscr { C }$ with positive abscissa $a$. The rotation around the x-axis applied to the part of $\mathscr { C }$ bounded by points I and A generates a surface modeling the flute container, taking 1 cm as the unit.
The real number $a$ being strictly positive, we admit that the volume $V ( a )$ of this solid in $\mathrm { cm } ^ { 3 }$ is given by the formula:
$$V ( a ) = \pi \int _ { 0 } ^ { a } ( f ( x ) ) ^ { 2 } \mathrm {~d} x$$

1. Verify, for all real numbers $x \geqslant 0$, the equality: $$( f ( x ) ) ^ { 2 } = 4 \left( \frac { \mathrm { e } ^ { x } } { \mathrm { e } ^ { x } + 1 } + \frac { - \mathrm { e } ^ { x } } { \left( \mathrm { e } ^ { x } + 1 \right) ^ { 2 } } \right) .$$
2. Determine a primitive on $\mathbb { R }$ of each of the functions: $$g : x \longmapsto \frac { \mathrm { e } ^ { x } } { \mathrm { e } ^ { x } + 1 } \quad \text { and } \quad h : x \longmapsto \frac { - \mathrm { e } ^ { x } } { \left( \mathrm { e } ^ { x } + 1 \right) ^ { 2 } }$$
3. Deduce that for all real $a > 0$: $$V ( a ) = 4 \pi \left[ \ln \left( \frac { \mathrm { e } ^ { a } + 1 } { 2 } \right) + \frac { 1 } { \mathrm { e } ^ { a } + 1 } - \frac { 1 } { 2 } \right] .$$
4. Using a calculator, determine an approximate value of $a$ to 0.1, knowing that a flute must contain $12.5 \mathrm { cL }$, that is $125 \mathrm {~cm} ^ { 3 }$. No justification is required.

Find the volume of the solid obtained when the region bounded by $y = \sqrt{x}$, $y = -x$ and the line $x = 9$ is revolved around the $x$-axis. (It may be useful to draw the specified region.)

When a solid of revolution is created by rotating the figure enclosed by the curve $y = a \left( 1 - x ^ { 2 } \right)$ and the $x$-axis around the $y$-axis, and the volume of the solid of revolution is $16 \pi$, find the positive value of $a$. [3 points]

The volume of the solid of revolution created by rotating the region enclosed by the two curves $y = \sqrt { x } , y = \sqrt { - x + 10 }$ and the $x$-axis around the $x$-axis is $a \pi$. Find the value of $a$. [3 points]

As shown in the figure, there is a line $l : x - y - 1 = 0$ and a hyperbola $C : x ^ { 2 } - 2 y ^ { 2 } = 1$ with one focus at point $\mathrm { F } ( c , 0 )$ (where $c < 0$).
When the region enclosed by the line $l$ and the hyperbola $C$ is rotated about the $y$-axis, what is the volume of the solid of revolution? [3 points]
(1) $\frac { 5 } { 3 } \pi$
(2) $\frac { 3 } { 2 } \pi$
(3) $\frac { 4 } { 3 } \pi$
(4) $\frac { 7 } { 6 } \pi$
(5) $\pi$

Consider the function $$f ( x ) = \begin{cases} | 5 x ( x + 2 ) | & ( x < 0 ) \\ | 5 x ( x - 2 ) | & ( x \geq 0 ) \end{cases}$$ On the closed interval $[ 0,1 ]$, what is the volume of the solid of revolution generated by rotating the region enclosed by the graph of $y = f ( x )$, the $x$-axis, and the line $x = 1$ about the $x$-axis? [3 points]
(1) $\frac { 65 } { 6 } \pi$
(2) $\frac { 35 } { 3 } \pi$
(3) $\frac { 25 } { 2 } \pi$
(4) $\frac { 40 } { 3 } \pi$
(5) $\frac { 85 } { 6 } \pi$

As shown in the figure, there is a solid figure with base formed by the curve $y = \sqrt { x } + 1$, the $x$-axis, the $y$-axis, and the line $x = 1$. When the cross-section of this solid figure cut by a plane perpendicular to the $x$-axis is always a square, what is the volume of this solid figure? [3 points]
(1) $\frac { 7 } { 3 }$
(2) $\frac { 5 } { 2 }$
(3) $\frac { 8 } { 3 }$
(4) $\frac { 17 } { 6 }$
(5) 3

As shown in the figure, for a positive number $k$, the region enclosed by the curve $y = \sqrt { \frac { e ^ { x } } { e ^ { x } + 1 } }$, the $x$-axis, the $y$-axis, and the line $x = k$ is the base of a solid figure. When the cross-section perpendicular to the $x$-axis is always a square and the volume is $\ln 7$, what is the value of $k$? [3 points]
(1) $\ln 11$
(2) $\ln 13$
(3) $\ln 15$
(4) $\ln 17$
(5) $\ln 19$

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) $\frac { 100 } { 9 } \left( 2 - \sqrt { 3 } + \frac { \pi } { 3 } \right)$

As shown in the figure, there is a solid figure with base formed by the curve $y = \sqrt{(1-2x)\cos x}$ ($\frac{3}{4}\pi \leq x \leq \frac{5}{4}\pi$) and the $x$-axis and the two lines $x = \frac{3}{4}\pi$ and $x = \frac{5}{4}\pi$. When this solid figure is cut by a plane perpendicular to the $x$-axis, all cross-sections are squares. Find the volume of this solid figure. [3 points]
(1) $\sqrt{2}\pi - \sqrt{2}$
(2) $\sqrt{2}\pi - 1$
(3) $2\sqrt{2}\pi - \sqrt{2}$
(4) $2\sqrt{2}\pi - 1$
(5) $2\sqrt{2}\pi$

The three views of a certain solid are shown in the figure. The volume of this solid is
(A) $\frac { 1 } { 3 } + 2 \pi$
(B) $\frac { 13 \pi } { 6 }$
(C) $\frac { 7 \pi } { 3 }$
(D) $\frac { 5 \pi } { 2 }$

5. The three-view drawing of a certain solid is shown in the figure. The volume of this solid is
[Figure]
Front view
[Figure]
Left view
[Figure]
Top view
A. $\frac { 1 } { 3 } + \pi$
B. $\frac { 2 } { 3 } + \pi$
C. $\frac { 1 } { 3 } + 2 \pi$
D. $\frac { 2 } { 3 } + 2 \pi$

10. The three-view drawing of a workpiece is shown in Figure 3. The workpiece is to be machined by cutting into a rectangular parallelepiped with the largest possible volume, with one face of the new workpiece lying on a face of the original workpiece. Then the material utilization rate of the original workpiece is (material utilization rate = volume of new workpiece / volume of original workpiece) [Figure]
A. $\frac { 8 } { 9 \pi }$
B. $\frac { 8 } { 27 \pi }$
C. $\frac { 24 ( \sqrt { 2 } - 1 ) ^ { 3 } } { \pi }$
D. $\frac { 8 ( \sqrt { 2 } - 1 ) ^ { 3 } } { \pi }$
II. Fill-in-the-Blank Questions: This section has 5 questions, each worth 5 points, for a total of 25 points

10. The three-view drawing of a workpiece is shown in Figure 3. The workpiece is to be machined by cutting into a rectangular solid with the largest possible volume, with one face of the new workpiece lying on a face of the original workpiece. The material utilization rate of the original workpiece is (Material utilization rate = Volume of new workpiece / Volume of original workpiece.) [Figure]
A. $\frac { 8 } { 9 }$
B. $\frac { 16 } { 9 }$
C. $\frac { 4 ( \sqrt { 2 } - 1 ) ^ { 2 } } { 9 }$
D. $\frac { 12 ( \sqrt { 2 } - 1 ) ^ { 2 } } { 9 }$
II. Fill-in-the-Blank Questions: This section has 5 questions, each worth 5 points, for a total of 25 points

As shown in the figure, a circular piece of paper has center $O$ and radius $5$ cm. An equilateral triangle $ABC$ on this paper has center at $O$. Points $D, E, F$ are on circle $O$. Triangles $DBC, ECA, FAB$ are isosceles triangles with $BC, CA, AB$ as their bases respectively. After cutting along the dashed lines and folding triangles $DBC, ECA, FAB$ along $BC, CA, AB$ respectively so that $D, E, F$ coincide, a triangular pyramid is formed. As the side length of $\triangle ABC$ varies, the maximum volume (in $\mathrm { cm } ^ { 3 }$) of the resulting triangular pyramid is \_\_\_\_

Points $A, B, C, D$ are on the surface of a sphere with radius 4. $\triangle ABC$ is an equilateral triangle with area $9 \sqrt { 3 }$. The maximum volume of the tetrahedron $D$-$ABC$ is
A. $12 \sqrt { 3 }$
B. $18 \sqrt { 3 }$
C. $24 \sqrt { 3 }$
D. $54 \sqrt { 3 }$

8. Two unit vectors $e _ { 1 } , e _ { 2 }$ have an angle of $60 ^ { \circ }$ between them. Vector $m = t e _ { 1 } + 2 e _ { 2 } ( t < 0 )$. Then
A. The maximum value of $\frac { | m | } { t }$ is $\frac { \sqrt { 3 } } { 2 }$
B. The minimum value of $\frac { | m | } { t }$ is $- 2$
C. The minimum value of $\frac { | m | } { t }$ is $\frac { \sqrt { 3 } } { 2 }$
D. The maximum value of $\frac { | m | } { t }$ is $- 2$

5. A right square frustum has upper and lower base edges of lengths 2 and 4 respectively, and lateral edge length 2. Its volume is ( )
A. $20 + 12 \sqrt { 3 }$
B. $28 \sqrt { 2 }$
C. $\frac { 56 } { 3 }$
D. $\frac { 28 \sqrt { 2 } } { 3 }$
【Answer】D 【Solution】 【Analysis】Using the geometric properties of the frustum, calculate the height of the solid and the areas of the upper and lower bases, then use the volume formula for a frustum.
【Detailed Solution】Draw a figure and connect the centers of the upper and lower bases of the right square frustum, as shown in the diagram. [Figure]
Since the upper and lower base edges of the frustum are 2 and 4 respectively, and the lateral edge length is 2, the height of the frustum is $h = \sqrt { 2 ^ { 2 } - ( 2 \sqrt { 2 } - \sqrt { 2 } ) ^ { 2 } } = \sqrt { 2 }$, the area of the lower base is $S _ { 1 } = 16$, and the area of the upper base is $S _ { 2 } = 4$, so the volume of the frustum is $V = \frac { 1 } { 3 } h \left( S _ { 1 } + S _ { 2 } + \sqrt { S _ { 1 } S _ { 2 } } \right) = \frac { 1 } { 3 } \times \sqrt { 2 } \times ( 16 + 4 + \sqrt { 64 } ) = \frac { 28 } { 3 } \sqrt { 2 }$ . Therefore, the answer is: D.

4. The South-to-North Water Diversion Project has alleviated water shortages in some northern regions, with part of the water stored in a certain reservoir. When the water level of the reservoir is at an elevation of 148.5 m, the corresponding water surface area is $140.0 \mathrm {~km} ^ { 2 }$; when the water level is at an elevation of 157.5 m, the corresponding water surface area is $180.0 \mathrm {~km} ^ { 2 }$. Treating the shape of the reservoir between these two water levels as a frustum, the volume of water added when the water level rises from an elevation of 148.5 m to 157.5 m is approximately ( $\sqrt { 7 } \approx 2.65$ )
A. $1.0 \times 10 ^ { 9 } \mathrm {~m} ^ { 3 }$
B. $1.2 \times 10 ^ { 9 } \mathrm {~m} ^ { 3 }$
C. $1.4 \times 10 ^ { 9 } \mathrm {~m} ^ { 3 }$
D. $1.6 \times 10 ^ { 9 } \mathrm {~m} ^ { 3 }$

8. A regular square pyramid has lateral edge length $l$, and all its vertices lie on the same sphere. If the volume of the sphere is $36 \pi$ and $3 \leqslant l \leqslant 3 \sqrt { 3 }$, then the range of the volume of the regular square pyramid is
A. $\left[ 18 , \frac { 81 } { 4 } \right]$
B. $\left[ \frac { 27 } { 4 } , \frac { 81 } { 4 } \right]$
C. $\left[ \frac { 27 } { 4 } , \frac { 64 } { 3 } \right]$
D. $[ 18,27 ]$
II. Multiple Choice Questions: This section contains 4 questions, each worth 5 points, for a total of 20 points. For each question, there may be multiple correct options. Full marks are awarded for selecting all correct options, 2 points for partially correct selections, and 0 points if any incorrect option is selected.

A cylinder and a cone have equal base radii and equal lateral surface areas, and both have height $\sqrt { 3 }$ . Then the volume of the cone is
A. $2 \sqrt { 3 } \pi$
B. $3 \sqrt { 3 } \pi$
C. $6 \sqrt { 3 } \pi$
D. $9 \sqrt { 3 } \pi$

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 volume of the region $S = \{ ( x , y , z ) : | x | + 2 | y | + 3 | z | \leq 6 \}$ is
(A) 36 .
(B) 48 .
(C) 72 .
(D) 6 .

Q2 For $\mathbf{K} \sim \mathbf{ZZ}$ in the following statements, choose the appropriate answer from among (0) $\sim$ (9) at the bottom of this page.
Let $a$ and $t$ be positive real numbers. Let $D$ denote the region of a plane bounded by the graph of the quadratic function in $x$
$$y = \frac{1}{t^2}\left(x - at^2\right)^2$$
the $x$-axis, and the $y$-axis. Let $V_1$ denote the volume of the solid obtained by rotating $D$ once about the $x$-axis, and $V_2$ denote the volume of the solid obtained by rotating $D$ once about the $y$-axis. Now, let us show that for a certain value of $a$, $V_1 = V_2$, independent of the value of $t$.
First, the value of $V_1$ is
$$\begin{aligned} V_1 &= \pi \int_{\mathbf{K}}^{\mathbf{L}} \frac{1}{t^{\mathbf{M}}}\left(x - at^2\right)^{\mathbf{N}} dx \\ &= \frac{\pi}{\mathbf{O}} a^{\mathbf{P}} t^{\mathbf{Q}} \end{aligned}$$
Next, the value of $V_2$ is
$$\begin{aligned} V_2 &= \pi \int_{\mathbf{R}}^{\mathbf{S}} (\cdots) \\ &= \frac{\pi}{\mathbf{T}} a^{\mathbf{W}} t^{\mathbf{W}} \end{aligned}$$
Hence, when $a = \frac{\mathbf{Y}}{\mathbf{Y}}$, then $V_1 = V_2$, independent of the value of $t$.
Options: (0) 0
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5 (6) 6 (7) $t$ (8) $at^2$ (9) $a^2t^2$