Volume of revolution, disk method (HL)

This involves a similar set up as finding an area.

The area of an annulus (ring with a width) is π(R2−r2)\pi(R^2 - r^2), ie the difference between the outer circle and the inner circle.

In the disk/washer method, we integrate with respect to the axis of rotation, by adding up each annulus slice.

Contents

About xx-axis

V=π∫x1x2g(x)2−f(x)2 dxV = \pi\int_{x_1}^{x_2} g(x)^2 - f(x)^2\d x

with gg further from xx-axis than ff, and x1x_1 and x2x_2 are the leftmost and rightmost xx values for both functions of xx. That is, both functions must have the same leftmost and rightmost endpoints.

About yy-axis

V=π∫y1y2g(y)2−f(y)2 dyV = \pi\int_{y_1}^{y_2} g(y)^2 - f(y)^2\d y

with gg further from yy-axis than ff, and y1y_1 and y2y_2 are the bottommost and topmost yy values for both functions of yy. That is, both functions must have the same lowest and highest heights.

Setting up the integral

Rotating across xx-axis means we need functions of xx and if rotating across yy-axis means we need functions of yy. This may necessitate finding the inverse functions, and/or splitting up the domain so that each slice is between two functions.

Very often, the second function is 00 because the region is bounded on the bottom by y=0y = 0 if revolving around xx-axis, or bounded on the left by x=0x = 0 if revolving around the yy-axis. This should not be taken for granted though.

On a graphing calculator, you should definitely learn how to evaluate composite functions and reuse function definitions.

Example: (May 2015 HL TZ1 Paper 2 #1) The region RR is enclosed by the graph of y=e−x2y = \e^{-x^2}, the xx-axis, and the lines x=−1x = -1, and x=1x = 1. Find the volume of the solid of revolution that is formed when RR is rotated through 2π2π about the xx-axis

Vx=π∫−11(e−x2)2−02 dx≈3.76■V_x = \pi\int_{-1}^{1}\left(\e^{-x^2}\right)^2 - 0^2\d x \approx 3.76 \qed

using graphing calculator

Some candidates got too clever and tried to simplify the integrand to e−x4\e^{-x^4} but exponents rule says e−2x2\e^{-2x^2}. Learn your calculator!

Example: (May 2013 HL TZ1 Paper 2 #4) Find the volume of the solid formed when the region bounded by the graph of y=sin⁡(x−1)y=\sin(x−1), and the lines x=0x = 0, y=0y = 0 and y=1y = 1 is rotated by 2π2π about the yy-axis.

Transformations say this is y=sin⁡xy = \sin x shifted one unit to the right (also this is paper 2). So it’s the region left of the curve, right of the yy axis.

arcsin⁡\arcsin is the closest branch of the “inverse functions” of sin⁡\sin. So if we just care about the part of the given graph from (1,0)(1, 0) to (π2+1,1)(\frac{\pi}{2} + 1, 1), then we can simply use

x=arcsin⁡y+1x = \arcsin y + 1

through finding the inverse.

Vy=π∫01(arcsin⁡y+1)2−02 dy≈2.609π≈8.20■\begin{align*}V_y &= \pi\int_{0}^{1}\left(\arcsin y + 1\right)^2 - 0^2\d y \\ &\approx 2.609\pi \approx 8.20 \qed\end{align*}

Example: (Extension of previous problem) Find the volume of the solid formed when the region bounded by the graph of y=sin⁡(x−1),1≤x≤π+1y=\sin(x−1), 1 \leq x \leq \pi + 1 and the line y=0y = 0 is rotated by 2π2π about the yy-axis.

The left curve is x=arcsin⁡y+1x = \arcsin y + 1, the right curve is bit tricky, but it’s arcsin⁡\arcsin flipped, and would pass through (π2+1,1)(\frac{\pi}{2} + 1, 1) and (π+1,0)(\pi + 1, 0). This is x=−arcsin⁡y+π+1x = -\arcsin y + \pi + 1 or equivalently x=arcsin⁡(−y)+π+1x = \arcsin (-y) + \pi + 1. It can be thought as flipping the arcsin⁡\arcsin function and move (0,0)(0, 0) to (π+1,0)(\pi + 1, 0). See inverse of trig functions.

Vy=π∫01(−arcsin⁡y+π+1)2−(arcsin⁡y+1)2 dy≈10.28π≈32.3■\begin{align*}V_y &= \pi\int_{0}^{1}(-\arcsin y + \pi + 1)^2 - (\arcsin y + 1)^2\d y \\ &\approx 10.28\pi \approx 32.3 \qed\end{align*}

Note that the same region revolving around xx-axis is

Vx=π∫1π+1sin⁡2(x−1) dx≈4.93\begin{align*}V_x &= \pi\int_{1}^{\pi + 1}\sin^2(x-1)\d x \approx 4.93\end{align*}

which is a lot smaller, as it’s a lot closer to the axis of rotation.

Variant: rotating less than a full revolution

One revolution (full turn) is 2π2\pi radians, so the rotating only θ\theta radians is θ2π\frac{\theta}{2\pi} times the formulas above. or θ2×integral\frac{\theta}{2} \times \text{integral}, as a π\pi cancel out. If the turn is given in degrees, use θ°360°\frac{\theta\degree}{360\degree} times π\pi times the integral.

Tips

  1. Always quickly sketch the functions or curves using function transformations to have visualize the area to revolve.
  2. The axis of rotation determines whether to integrate with respect to xx or yy.
  3. Use a function or a difference of functions on one side of the axis of rotation, with intersecting the axis.
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