Cross Sections Perpendicular To The X Axis

Understanding how to calculate the volume of a solid using cross sections perpendicular to the x axis is a fundamental application of integral calculus. This method, often called the slicing method or the method of cross sections, allows us to find the volume of complex three-dimensional shapes that are not standard solids of revolution. By visualizing the solid as a stack of infinitely thin slices, each with a known geometric area, we can sum these areas using a definite integral to determine the total volume Nothing fancy..

The Core Concept: Slicing a Solid

Imagine a loaf of bread sitting on a cutting board. Now, if you slice the loaf perpendicular to its length, each slice has a specific shape and thickness. The volume of the entire loaf is essentially the sum of the volumes of all individual slices. In calculus, we take this concept to the limit: we make the slices infinitely thin ($dx$), calculate the area of the face of each slice ($A(x)$), and integrate these areas along the x-axis over a specific interval $[a, b]$.

Easier said than done, but still worth knowing Easy to understand, harder to ignore..

The general formula for volume using this technique is:

$V = \int_{a}^{b} A(x) , dx$

Where:

• $V$ is the total volume. Practically speaking, * $a$ and $b$ are the bounds of the solid along the x-axis. * $A(x)$ is the area of a single cross section expressed as a function of $x$.
• $dx$ represents the infinitesimal thickness of the slice.

The critical challenge—and the key to mastering this topic—lies in correctly determining the function $A(x)$ based on the geometry of the cross section.

Defining the Base Region and Cross Section Shape

Problems involving cross sections perpendicular to the x axis typically provide two vital pieces of information:

1. In real terms, The Base Region: A two-dimensional region in the xy-plane (usually bounded by curves $y = f(x)$, $y = g(x)$, and vertical lines $x=a$, $x=b$). The Cross Section Shape: The geometric shape of the slice taken perpendicular to the x-axis (e.This region acts as the "footprint" of the solid. So g. 2. , squares, equilateral triangles, semicircles, isosceles right triangles).

The dimension of the cross section (side length, base, height, or diameter) is almost always determined by a vertical distance within the base region at a specific $x$-value. Because the slices are perpendicular to the x-axis, we measure vertically (parallel to the y-axis) to find the length of the segment that defines the slice But it adds up..

Determining the "Length" of the Slice

At any specific $x = x_i$ within the interval $[a, b]$, the cross section intersects the base region along a vertical line segment. The length of this segment, let's call it $L(x)$, is the difference between the upper bounding curve and the lower bounding curve:

$L(x) = f_{\text{top}}(x) - f_{\text{bottom}}(x)$

This length $L(x)$ serves as the primary variable (side, base, diameter, or diagonal) for the area formula $A(x)$.

Common Cross Section Geometries and Area Formulas

Once $L(x)$ is established, the area $A(x)$ depends entirely on the specified shape. Here are the most frequent scenarios encountered in calculus courses:

1. Squares

If the cross sections are squares with bases lying in the xy-plane (perpendicular to the x-axis), the side length of the square is $s = L(x)$. $A(x) = [L(x)]^2 = [f_{\text{top}}(x) - f_{\text{bottom}}(x)]^2$

2. Equilateral Triangles

If the cross sections are equilateral triangles with bases in the xy-plane, the side length is $s = L(x)$. The area of an equilateral triangle with side $s$ is $\frac{\sqrt{3}}{4}s^2$. $A(x) = \frac{\sqrt{3}}{4} [L(x)]^2$

3. Isosceles Right Triangles

Two variations exist here, and reading the problem carefully is essential:

• Leg (Base) in the plane: The leg length is $L(x)$. Area = $\frac{1}{2}[L(x)]^2$.
• Hypotenuse in the plane: The hypotenuse length is $L(x)$. For an isosceles right triangle, legs are $\frac{L(x)}{\sqrt{2}}$. Area = $\frac{1}{2}(\frac{L(x)}{\sqrt{2}})^2 = \frac{1}{4}[L(x)]^2$.

4. Semicircles

If the cross sections are semicircles with diameters lying in the xy-plane, the diameter is $d = L(x)$. The radius is $r = \frac{L(x)}{2}$. $A(x) = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi \left(\frac{L(x)}{2}\right)^2 = \frac{\pi}{8} [L(x)]^2$

5. Rectangles (Constant Height)

Sometimes the cross section is a rectangle where the base is $L(x)$ but the height $h$ is a given constant. $A(x) = h \cdot L(x)$

Step-by-Step Problem Solving Strategy

To consistently solve these problems, follow this structured workflow:

1. Sketch the Base Region: Draw the xy-plane. Plot the bounding curves $y = f(x)$ and $y = g(x)$. Shade the enclosed region. Identify the intersection points to find limits of integration $a$ and $b$.
2. Identify the Slice Orientation: Confirm the slices are perpendicular to the x axis. This means you integrate with respect to $x$ ($dx$). The representative rectangle in your sketch should be vertical.
3. Determine the Slice Length $L(x)$: Draw a vertical segment inside the base region at a generic $x$. Label the top curve $y_{\text{top}}$ and bottom curve $y_{\text{bottom}}$. Write $L(x) = y_{\text{top}} - y_{\text{bottom}}$.
4. Select the Area Formula: Based on the problem's description of the cross section (square, semicircle, etc.), write the formula for $A(x)$ in terms of $L(x)$.
5. Set Up the Integral: Substitute $L(x)$ into $A(x)$. Write the definite integral $V = \int_{a}^{b} A(x) , dx$.
6. Evaluate: Compute the integral using the Fundamental Theorem of Calculus. Simplify the result and include cubic units.

Worked Example: Squares and Semicircles

Let’s apply this strategy to a concrete example.

Problem: Find the volume of a solid whose base is the region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 4$. Cross sections perpendicular to the x-axis are (a) squares and (b) semicircles.

Step 1: Sketch and Bounds

The region is bounded above by $y = \sqrt{x}$, below by the x-axis ($y=0$), and on the right by the vertical line $x=4$. The curves intersect at $x=0$. Thus, the limits of integration are $a=0$ and $b=4$.

Step 2: Slice Length $L(x)$

A vertical slice at position $x$ runs from $y=0$ to $y=\sqrt{x}$. $L(x) = \sqrt{x}