[IB Math] Applications of Integration: Finding the Volume of Revolution (Washer Method Problem Walkthrough)
Introduction
In the IB Math (Calculus) curriculum, one type of problem that students often find tricky is finding the Volume of Revolution. Particularly when two or more graphs appear, and the region being rotated is separated from the axis of rotation, you must use the Washer Method (where the center is hollowed out like a donut) rather than the simple Disk Method.
Today, through a problem involving rotating a region where a trigonometric function meets a straight line around the $x$-axis, we will review everything together, from setting up the integration interval to applying integration by parts.
1. Problem
As shown in the figure below, there is a closed region $S$ enclosed between the curve $y = x + 2\cos x$ ($0 \le x \le 2\pi$) and the line $y = x$.
Problem: Find the volume $V$ of the solid generated when the region $S$ is rotated by $2\pi$ about the $x$-axis.
(a) Find the coordinates of the points where the two graphs meet. (b) (i) Write down an integral that represents the volume $V$. (ii) Find the value of the volume $V$.
2. Key Concepts
Here are the mathematical tools needed to solve this problem.
1) Washer Method
This is the formula used when the cross-section of the solid of revolution is in the shape of a washer (a disk with a hole in it). You subtract the area of the inner circle from the area of the outer circle.
$ V = \pi \int_{a}^{b} \left( [R(x)]^2 - [r(x)]^2 \right) dx $
- $R(x)$ (Outer Radius): The function that is farther from the axis of rotation (the $x$-axis).
- $r(x)$ (Inner Radius): The function that is closer to the axis of rotation (the $x$-axis).
2) Integration Techniques
- Integration by Parts: Used to solve forms like $\int x \cos x \, dx$. ($\int u dv = uv - \int v du$)
- Half-angle Identity: Used when integrating $\cos^2 x$. ($\cos^2 x = \frac{1+\cos 2x}{2}$)
3. Solution
Step 1: Find Integration Interval (Intersection Points) (Part a)
To find the points where the two graphs meet, set the equations equal to each other.
$x + 2\cos x = x$ $2\cos x = 0 \quad \Rightarrow \quad \cos x = 0$
Within the given range $0 \le x \le 2\pi$, the values where cosine becomes 0 are:
$x$-coordinates of intersection (Integration Interval): $x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2}$
Step 2: Set up the Integral (Part b-i)
In the interval $\frac{\pi}{2} < x < \frac{3\pi}{2}$, the value of $\cos x$ is negative (-). Therefore, $x + 2\cos x$ is smaller than $x$, so the positional relationship of the graphs is as follows:
- Upper (Outer, $R$): $y = x$
- Lower (Inner, $r$): $y = x + 2\cos x$
Substitute these into the volume formula:
$V = \pi \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \left( (x)^2 - (x + 2\cos x)^2 \right) dx$
Step 3: Calculate the Volume (Part b-ii)
1. Simplify the Integrand
Expand the expression inside the brackets to simplify it.
$x^2 - (x + 2\cos x)^2$ $= x^2 - (x^2 + 4x\cos x + 4\cos^2 x)$ $= -4x\cos x - 4\cos^2 x$
If we factor out the constant $-4$, the expression to integrate is: $V = -4\pi \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (x\cos x + \cos^2 x) \, dx$
2. Calculate the Integral
It is best to split this integral into two parts (A and B) for calculation.
(A) $\int x\cos x \, dx$ (Integration by Parts) Set $u = x$ and $dv = \cos x \, dx$. Then $du = dx$ and $v = \sin x$. $\int x\cos x \, dx = x\sin x - \int \sin x \, dx = x\sin x + \cos x$ Applying the bounds $[\frac{\pi}{2}, \frac{3\pi}{2}]$: $\left[ x\sin x + \cos x \right]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} = \left(-\frac{3\pi}{2} + 0\right) - \left(\frac{\pi}{2} + 0\right) = -2\pi$
(B) $\int \cos^2 x \, dx$ (Half-angle Identity) $\int \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2}x + \frac{1}{4}\sin 2x$ Applying the bounds $[\frac{\pi}{2}, \frac{3\pi}{2}]$: $\left[ \frac{1}{2}x + \frac{1}{4}\sin 2x \right]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} = \left(\frac{3\pi}{4} + 0\right) - \left(\frac{\pi}{4} + 0\right) = \frac{\pi}{2}$
3. Combine Final Results
Substitute the values obtained back into the original expression for volume.
$V = -4\pi \times ((A) + (B))$ $V = -4\pi \times \left( -2\pi + \frac{\pi}{2} \right)$ $V = -4\pi \times \left( -\frac{3\pi}{2} \right)$ $\therefore V = 6\pi^2$
4. Conclusion
The final answer to this problem is $6\pi^2$.
The most important point when solving this problem is to grasp the upper/lower relationship of the graphs within the interval so that you do not swap $R$ and $r$. Also, the key to the calculation is remembering to use the half-angle identity without panicking when a squared trigonometric term ($\cos^2 x$) appears.
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