4. calculate the volume of the ramp in figure 17 in three ways by integrating the area of the cross sections: (a) perpendicular to the x-axis (rectangles) (b) perpendicular to the y-axis (triangles) (c) perpendicular to the z-axis (rectangles) figure 17 ramp of length 6, width 4, and height 2

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To solve the volume of the ramp using different cross - section orientations, we can use the formula for the volume of a prism: $V=\text{Area of cross - section}\times\text{length of the prism}$ (where the "length" here is the distance along the axis perpendicular to the cross - section). We can also use integration, but since the ramp is a prism (a polyhedron with two congruent polygonal bases connected by parallelogram faces), we can use the formula for the volume of a prism which is $V = A\times l$, where $A$ is the area of the base and $l$ is the length of the lateral edge.

### Part (a): Perpendicular to the $x$ - axis (rectangles)
#### Step 1: Determine the dimensions of the cross - section
The ramp has a height of $h = 2$ and a width of $w=4$ (the dimension along the $y$ - axis). When we take cross - sections perpendicular to the $x$ - axis, the cross - section is a rectangle. The length of the ramp along the $x$ - axis is $L = 6$. The height of the rectangle (in the $z$ - direction) varies linearly from $z = 0$ to $z = 2$ as $x$ goes from $0$ to $6$. The equation of the line for the height $z$ as a function of $x$ is $z=\frac{2}{6}x=\frac{1}{3}x$? Wait, no. Actually, the ramp is a prism with a triangular base in the $y - z$ plane? Wait, looking at the figure, the ramp has length $6$ (along $x$), width $4$ (along $y$) and height $2$ (along $z$).

Wait, maybe a better approach: The ramp is a triangular prism. The volume of a triangular prism is $V=\frac{1}{2}\times\text{base of triangle}\times\text{height of triangle}\times\text{length of the prism}$. But when we take cross - sections perpendicular to the $x$ - axis:

The cross - section perpendicular to the $x$ - axis is a rectangle. The height of the rectangle (in the $z$ - direction) at a position $x$ (where $0\leq x\leq6$): The slope of the ramp in the $x - z$ plane is $\frac{2}{6}=\frac{1}{3}$. So $z(x)=\frac{1}{3}x$? No, wait, if at $x = 0$, $z = 0$ and at $x=6$, $z = 2$, then $z(x)=\frac{2}{6}x=\frac{1}{3}x$. The width of the rectangle (in the $y$ - direction) is $4$ (constant). So the area of the cross - section $A(x)=4\times z(x)=4\times\frac{1}{3}x=\frac{4}{3}x$? No, that can't be right. Wait, maybe I misinterpret the figure. The ramp has length $6$ (along $x$), width $4$ (along $y$) and height $2$ (along $z$). So it's a prism with a triangular base in the $x - z$ plane with base $6$ and height $2$, and the length along the $y$ - axis is $4$.

So when we take cross - sections perpendicular to the $x$ - axis, the cross - section is a rectangle with height $z$ (where $z$ goes from $0$ to $2$ as $x$ goes from $0$ to $6$) and width $4$ (along $y$). The relationship between $z$ and $x$ is linear: $z=\frac{2}{6}x=\frac{1}{3}x$ (when $x = 0$, $z = 0$; when $x = 6$, $z = 2$).

The volume is given by the integral $\int_{0}^{6}A(x)dx$, where $A(x)$ is the area of the cross - section at position $x$. Since $A(x)=4\times z(x)=4\times\frac{1}{3}x$? No, wait, if the cross - section is a rectangle, and the height in the $z$ - direction is $z$, and the width in the $y$ - direction is $4$, and the length in the $x$ - direction is being integrated. Wait, maybe a simpler way: The volume of a prism is also equal to the area of the base (in the $y - z$ plane) times the length in the $x$ - direction. The base in the $y - z$ plane is a rectangle? No, the ramp is a triangular prism. The base triangle has base $2$ (height in $z$) and length $6$ (in $x$)? No, I think I made a mistake. Let's re - examine the figure: "Ramp of length 6, width 4, and height 2". So length $l = 6$ (along $x$), width $w = 4$ (along $y$), height $h=2$ (along $z$). The ramp is a right triangular prism? No, it's a prism with a triangular base. The triangular base has base $b = 4$ (width) and height $h = 2$ (height), and the length of the prism (along $x$) is $6$. Wait, no.

Wait, the volume of a prism is $V=\text{Area of the base}\times\text{length of the lateral edge}$. If we take the base as the triangle in the $y - z$ plane with base $4$ and height $2$, then the area of the base $A=\frac{1}{2}\times4\times2 = 4$. The length of the lateral edge (along $x$) is $6$. So $V=4\times6 = 24$.

But let's do it using integration for part (a):

#### Step 1: Define the cross - section area function
When cross - sections are perpendicular to the $x$ - axis, the cross - section is a rectangle. The height of the rectangle (in the $z$ - direction) at a given $x$ (where $0\leq x\leq6$): The slope of the ramp in the $x - z$ plane is $\frac{2}{6}=\frac{1}{3}$. So $z(x)=\frac{1}{3}x$ (when $x = 0$, $z = 0$; when $x = 6$, $z = 2$). The width of the rectangle (in the $y$ - direction) is $4$ (constant). So the area of the cross - section $A(x)=4\times z(x)=4\times\frac{1}{3}x$? No, that's wrong. Wait, if the cross - section is perpendicular to the $x$ - axis, the dimensions of the cross - section are in the $y$ and $z$ directions. The length along the $x$ - axis is the variable of integration.

Actually, the correct way: The ramp is a prism with a triangular base. The base triangle has vertices at $(0,0,0)$, $(6,0,0)$, $(6,0,2)$, $(0,0,2)$ in the $x - z$ plane? No, the width is along $y$. So the ramp extends from $y = 0$ to $y = 4$, $x = 0$ to $x = 6$, and $z$ goes from $0$ to $2$ as $x$ goes from $0$ to $6$ (and $y$ is constant). So for a fixed $x$, the cross - section perpendicular to $x$ is a rectangle with $y$ from $0$ to $4$ and $z$ from $0$ to $z(x)$, where $z(x)=\frac{2}{6}x=\frac{1}{3}x$. So the area of the cross - section $A(x)=4\times z(x)=4\times\frac{1}{3}x$? No, $z(x)$ is the height in the $z$ - direction, and the width in the $y$ - direction is $4$, so $A(x)=4\times z(x)$. But when $x = 6$, $z(6) = 2$, so $A(6)=4\times2 = 8$. When $x = 0$, $A(0)=0$.

#### Step 2: Integrate to find the volume
The volume $V=\int_{0}^{6}A(x)dx=\int_{0}^{6}4\times\frac{1}{3}xdx$? No, wait, the correct linear relationship: If at $x = 0$, $z = 0$ and at $x = 6$, $z = 2$, then $z(x)=\frac{2}{6}x=\frac{1}{3}x$ is incorrect. Wait, it should be $z(x)=\frac{2}{6}x$ is wrong. The correct slope is $\frac{\text{rise}}{\text{run}}=\frac{2}{6}=\frac{1}{3}$, but actually, the height $z$ as a function of $x$ is $z(x)=\frac{2}{6}x$ when $x$ is the run. But the cross - section area when perpendicular to $x$ is a rectangle with length $4$ (along $y$) and height $z(x)$ (along $z$). So $A(x)=4\times z(x)$. But we know that the volume of the ramp should be the same as the volume of a triangular prism. The volume of a triangular prism is $V=\frac{1}{2}\times\text{base}\times\text{height}\times\text{length}$. If we consider the base as a triangle with base $6$ (along $x$) and height $2$ (along $z$), and the length along $y$ is $4$, then $V=\frac{1}{2}\times6\times2\times4=24$.

Let's do the integration correctly. The function $z(x)$ should be $z(x)=\frac{2}{6}x=\frac{1}{3}x$ is wrong. Wait, when $x = 0$, $z = 0$; when $x = 6$, $z = 2$. So the equation of the line is $z=\frac{2}{6}x=\frac{1}{3}x$. Then $A(x)=4\times z(x)=4\times\frac{1}{3}x$. Then $\int_{0}^{6}\frac{4}{3}xdx=\frac{4}{3}\times\frac{x^{2}}{2}\big|_{0}^{6}=\frac{4}{3}\times\frac{36}{2}=\frac{4}{3}\times18 = 24$.

### Part (b): Perpendicular to the $y$ - axis (triangles)
#### Step 1: Determine the dimensions of the cross - section
When cross - sections are perpendicular to the $y$ - axis, the cross - section is a triangle. The base of the triangle (in the $x$ - direction) is $6$ (constant, since the ramp has length $6$ along $x$) and the height of the triangle (in the $z$ - direction) is $2$ (constant, since the ramp has height $2$ along $z$)? No, wait, for a fixed $y$ (where $0\leq y\leq4$), the cross - section perpendicular to $y$ is a triangle with base $6$ (along $x$) and height $2$ (along $z$)? No, that can't be. Wait, the ramp extends from $y = 0$ to $y = 4$, $x = 0$ to $x = 6$, and $z$ from $0$ to $2$ as $x$ goes from $0$ to $6$. When we take a cross - section perpendicular to $y$, we are looking at a slice at a fixed $y$. The shape of this slice is a triangle with base $6$ (along $x$) and height $2$ (along $z$)? No, the area of a triangle is $\frac{1}{2}\times\text{base}\times\text{height}$. The base of the triangle (in the $x$ - direction) is $6$ and the height (in the $z$ - direction) is $2$. So the area of the cross - section $A(y)=\frac{1}{2}\times6\times2 = 6$ (constant for all $y$ in $[0,4]$).

#### Step 2: Integrate to find the volume
The volume $V=\int_{0}^{4}A(y)dy=\int_{0}^{4}6dy=6y\big|_{0}^{4}=24$.

### Part (c): Perpendicular to the $z$ - axis (rectangles)
#### Step 1: Determine the dimensions of the cross - section
When cross - sections are perpendicular to the $z$ - axis, the cross - section is a rectangle. The length of the rectangle (in the $x$ - direction) at a given $z$ (where $0\leq z\leq2$): The relationship between $x$ and $z$ is $x = \frac{6}{2}z=3z$ (since when $z = 0$, $x = 0$; when $z = 2$, $x = 6$). The width of the rectangle (in the $y$ - direction) is $4$ (constant). So the area of the cross - section $A(z)=4\times x(z)=4\times3z = 12z$? No, wait, when $z = 0$, $x = 0$; when $z = 2$, $x = 6$, so $x(z)=3z$. The area of the cross - section (rectangle) is $A(z)=x(z)\times4=3z\times4 = 12z$? No, that's wrong. Wait, the length along $x$ for a given $z$ is $x(z)$, and the width along $y$ is $4$. So $A(z)=x(z)\times4$. Since $x(z)$ goes from $0$ to $6$ as $z$ goes from $0$ to $2$, $x(z)=3z$.

#### Step 2: Integrate to find the volume
The volume $V=\int_{0}^{2}A(z)dz=\int_{0}^{2}12zdz=6z^{2}\big|_{0}^{2}=6\times4 - 0=24$.

# Answer:
The volume of the ramp is $\boldsymbol{24}$ (for all three methods of integration with different cross - section orientations).

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Explanation:

Part (a): Perpendicular to the \(x\) - axis (rectangles)

Step 1: Determine the dimensions of the cross - section

Answer: