Calculate Volume Of Revolution About The X-Axis

Calculating the volume of a solid generated by revolving a region around an axis is a fundamental concept in calculus, often referred to as the "volume of revolution." When we talk about revolving a 2D region around the $x$-axis, we can visualize slicing this solid into infinitesimally thin disks. The volume of each disk is approximately $\pi r^2 \Delta x$, where $r$ is the radius of the disk and $\Delta x$ is its thickness. By summing up the volumes of all these disks using integration, we can find the total volume of the solid. This method is known as the disk method. The key to applying the disk method successfully lies in correctly identifying the radius of the disk, which is usually the distance from the axis of revolution (in this case, the $x$-axis) to the curve that defines the boundary of the region. If the region is bounded by two curves, we might need to use the washer method, which is a variation of the disk method where we subtract the volume of an inner "hole" from the volume of an outer disk.

In this specific problem, we are asked to find the volume of the solid generated by revolving the region bounded by the lines $y=-3x+6$, $y=3x$, and $x=0$ about the $x$-axis. The first step is to visualize and understand the region we are dealing with. We have three boundaries: a downward-sloping line ($y=-3x+6$), an upward-sloping line ($y=3x$), and the $y$-axis ($x=0$). To determine the region, we need to find the points of intersection of these lines. The line $x=0$ is the $y$-axis. The intersection of $y=-3x+6$ and $y=3x$ occurs when $-3x+6 = 3x$, which gives $6 = 6x$, so $x=1$. When $x=1$, $y=3(1)=3$. So, these two lines intersect at the point $(1, 3)$. The line $y=-3x+6$ intersects the $x$-axis when $y=0$, so $0 = -3x+6$, which means $3x=6$, and $x=2$. The line $y=3x$ intersects the $x$-axis at the origin $(0,0)$. The region is bounded by the $y$-axis ($x=0$) on the left, the line $y=3x$ from $x=0$ to $x=1$, and the line $y=-3x+6$ from $x=1$ to $x=2$. However, since we are revolving around the $x$-axis, we need to consider the upper and lower boundaries of the region with respect to the $x$-axis. The region is defined for $x$ values from $0$ to where the upper boundary meets the $x$-axis. Let's re-evaluate the boundaries. The region is bounded by $x=0$, $y=3x$, and $y=-3x+6$. We are revolving this region about the $x$-axis. The line $y=3x$ starts at the origin and goes up. The line $y=-3x+6$ starts at $(0,6)$ on the $y$-axis and goes down, crossing the $x$-axis at $(2,0)$. The intersection point is $(1,3)$. The region is enclosed by $x=0$, $y=3x$, and $y=-3x+6$. This region is above the $x$-axis. The upper boundary of the region is formed by two different functions depending on the value of $x$. From $x=0$ to $x=1$, the upper boundary is $y=3x$. From $x=1$ to $x=2$, the upper boundary is $y=-3x+6$. The lower boundary for this region is the $x$-axis ($y=0$). This is because the problem states the region is bounded by the given lines and curves. When we are revolving around the $x$-axis, and the region is above the $x$-axis, the radius of our disk is simply the function's $y$-value. The challenge here is that the upper boundary changes. Therefore, we will need to split our integration into two parts.

Setting Up the Integrals for Volume Calculation

To find the volume of the solid of revolution, we will use the disk method and integrate with respect to $x$. The formula for the volume of a solid generated by revolving a region bounded by $y=f(x)$, the $x$-axis, and the lines $x=a$ and $x=b$ about the $x$-axis is given by $V = \int_{a}^{b} \pi [f(x)]^2 dx$. In our case, the region is bounded by $y=3x$, $y=-3x+6$, and $x=0$. As we identified, the upper boundary of the region is defined by $y=3x$ for $0 \le x \le 1$, and by $y=-3x+6$ for $1 \le x \le 2$. The revolution is about the $x$-axis, and the region is above the $x$-axis. Thus, we need to set up two separate integrals to cover the entire region. The first integral will calculate the volume generated by revolving the region under $y=3x$ from $x=0$ to $x=1$. The radius of the disks in this interval is $r_1 = 3x$. The volume for this part, $V_1$, is:

$V_1 = \int_{0}^{1} \pi (3x)^2 dx$

$V_1 = \pi \int_{0}^{1} 9x^2 dx$

The second integral will calculate the volume generated by revolving the region under $y=-3x+6$ from $x=1$ to $x=2$. The radius of the disks in this interval is $r_2 = -3x+6$. The volume for this part, $V_2$, is:

$V_2 = \int_{1}^{2} \pi (-3x+6)^2 dx$

$V_2 = \pi \int_{1}^{2} (9x^2 - 36x + 36) dx$

The total volume $V$ of the solid generated by revolving the entire region about the $x$-axis is the sum of these two volumes: $V = V_1 + V_2$. It's crucial to ensure that the boundaries of integration and the radii are correctly identified for each part. The intersection point at $x=1$ acts as the dividing point for our integrals. If the region were bounded below by another curve, say $y=g(x)$, we would use the washer method, where the radius would be $(f(x)-g(x))$, and the volume would be $\int_{a}^{b} \pi ([f(x)]^2 - [g(x)]^2) dx$. However, in this problem, the lower boundary is the $x$-axis ($y=0$), simplifying the calculation to the disk method.

Evaluating the Integrals and Finding the Total Volume

Now, let's proceed with evaluating the integrals we've set up to find the volume of the solid generated by revolving the region about the $x$-axis. We have two integrals to compute: $V_1$ and $V_2$.

First, let's evaluate $V_1$:

$V_1 = \pi \int_{0}^{1} 9x^2 dx$

To integrate $9x^2$, we use the power rule for integration, which states that $\int x^n dx = \frac{x^{n+1}}{n+1}$. So, the integral of $9x^2$ is $9 \cdot \frac{x^{2+1}}{2+1} = 9 \cdot \frac{x^3}{3} = 3x^3$. Now we evaluate this from $0$ to $1$:

$V_1 = \pi [3x^3]_{0}^{1}$

$V_1 = \pi (3(1)^3 - 3(0)^3)$

$V_1 = \pi (3 - 0)$

$V_1 = 3\pi$

Next, let's evaluate $V_2$:

$V_2 = \pi \int_{1}^{2} (-3x+6)^2 dx$

First, we expand the term $(-3x+6)^2$:

$(-3x+6)^2 = (-3x)^2 + 2(-3x)(6) + 6^2 = 9x^2 - 36x + 36$

So, the integral becomes:

$V_2 = \pi \int_{1}^{2} (9x^2 - 36x + 36) dx$

Now, we integrate each term:

$\- \int 9x^2 dx = 3x^3$

$\- \int -36x dx = -36 \cdot \frac{x^{1+1}}{1+1} = -36 \cdot \frac{x^2}{2} = -18x^2$

$\- \int 36 dx = 36x$

So, the antiderivative of $9x^2 - 36x + 36$ is $3x^3 - 18x^2 + 36x$. Now we evaluate this from $1$ to $2$:

$V_2 = \pi [3x^3 - 18x^2 + 36x]_{1}^{2}$

$V_2 = \pi [(3(2)^3 - 18(2)^2 + 36(2)) - (3(1)^3 - 18(1)^2 + 36(1))]$

$V_2 = \pi [(3(8) - 18(4) + 72) - (3 - 18 + 36)]$

$V_2 = \pi [(24 - 72 + 72) - (21)]$

$V_2 = \pi [24 - 21]$

$V_2 = 3\pi$

Finally, to find the total volume $V$, we add $V_1$ and $V_2$:

$V = V_1 + V_2$

$V = 3\pi + 3\pi$

$V = 6\pi$

Therefore, the volume of the solid generated by revolving the region bounded by $y=-3x+6$, $y=3x$, and $x=0$ about the $x$-axis is $6\pi$ cubic units. This process demonstrates how to break down a problem with a changing upper boundary into manageable integrals and apply the fundamental disk method of calculating volumes of revolution.

Understanding the Geometry of the Solid

It's always beneficial to have a geometric understanding of the solid we've calculated the volume of revolution for. The region bounded by $y=3x$, $y=-3x+6$, and $x=0$ forms a triangular shape in the $xy$-plane. The line $x=0$ is the $y$-axis. The line $y=3x$ starts at the origin $(0,0)$ and goes upwards with a slope of 3. The line $y=-3x+6$ intersects the $y$-axis at $(0,6)$ and has a slope of -3. These two lines intersect at $(1,3)$. The region is the area enclosed by the $y$-axis (from $y=0$ to $y=6$, though it's bounded above by the functions), the line segment $y=3x$ from $(0,0)$ to $(1,3)$, and the line segment $y=-3x+6$ from $(1,3)$ to $(0,6)$. However, we are revolving this region about the x-axis. This means the shape being revolved is actually a triangle with vertices at $(0,0)$, $(1,3)$, and $(2,0)$. Let's re-examine the bounded region. The lines are $y=-3x+6$, $y=3x$, and $x=0$. We are revolving about the x-axis.

The line $y=3x$ and $x=0$ intersect at $(0,0)$. The line $y=-3x+6$ and $x=0$ intersect at $(0,6)$. The lines $y=3x$ and $y=-3x+6$ intersect at $(1,3)$.

When revolving the region bounded by $y=3x$, $y=-3x+6$, and $x=0$ about the $x$-axis, we are looking at the area enclosed by these lines. However, the problem statement implies the region between these curves and the axis of revolution. The region we are revolving is bounded by $x=0$ (the y-axis), the line $y=3x$, and the line $y=-3x+6$. The intersection point of $y=3x$ and $y=-3x+6$ is $(1,3)$. The line $y=-3x+6$ intersects the x-axis at $x=2$. The line $y=3x$ intersects the x-axis at $x=0$. The region is above the x-axis.

Let's consider the region bounded by $y=3x$, $y=0$ (x-axis) and $x=1$. This forms a triangle with vertices $(0,0), (1,0), (1,3)$. When revolved about the x-axis, this generates a cone with radius 3 and height 1. Its volume is $\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3^2)(1) = 3\pi$. This matches our $V_1$.

Now consider the region bounded by $y=-3x+6$, $y=0$ (x-axis) and $x=1$ to $x=2$. This region is also above the x-axis. This forms a triangle with vertices $(1,3), (2,0), (1,0)$. When revolved about the x-axis, this also generates a cone. The radius at $x=1$ is $y = -3(1)+6 = 3$. The height is from $x=1$ to $x=2$, so height is $1$. Its volume is $\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3^2)(1) = 3\pi$. This matches our $V_2$.

The total region is composed of these two adjacent triangular regions. The first region is defined by $y=3x$ for $0

ext{The concept of solids of revolution is a cornerstone of calculus, and understanding how to calculate their volumes is a valuable skill. For further exploration and practice problems, you can visit **Paul's Online Math Notes** which provides comprehensive explanations and examples on various calculus topics, including volumes of revolution. Additionally, **Khan Academy** offers a wealth of free resources, interactive exercises, and video tutorials that can help solidify your understanding of this subject.