Volumes of Revolution about the y-axis

In the previous lesson we rotated regions about the x-axis. Rotation about the y-axis is equally important in Further Mathematics and requires a slightly different approach. This lesson covers the disc/washer method adapted for y-axis rotation and introduces the shell method as an alternative.

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1 Disc Method about the y-axis

Derivation

When a curve x = g(y) (or equivalently y = f(x) rearranged) is rotated about the y-axis, horizontal slices at height y with thickness delta y form discs of radius x.

V = pi * integral from c to d of x^2 dy

where c and d are the y-limits of the region.

Key Step: You must express x in terms of y (i.e. x = g(y)) before using this formula. Rearranging y = f(x) to get x = f^(-1)(y) is often the first step.

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2 Worked Example 1 — Simple Rearrangement

Find the volume when the region bounded by y = x^2, the y-axis, and y = 4 is rotated about the y-axis.

Solution:

Rearrange: y = x^2 implies x^2 = y (taking x >= 0 since we are in the first quadrant).

The y-limits are from 0 to 4.

V = pi * integral from 0 to 4 of y dy

= pi * [y^2/2] from 0 to 4

= pi * (16/2 - 0)

= 8pi cubic units

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3 Worked Example 2 — Cubic Curve

Find the volume when y = x^3 (for 0 <= x <= 2, i.e. 0 <= y <= 8) is rotated about the y-axis.

Solution:

y = x^3 implies x = y^(1/3), so x^2 = y^(2/3).

V = pi * integral from 0 to 8 of y^(2/3) dy

= pi * [y^(5/3) / (5/3)] from 0 to 8

= pi * (3/5) * [8^(5/3) - 0]

Now 8^(5/3) = (8^(1/3))^5 = 2^5 = 32.

V = pi * (3/5) * 32 = 96pi/5 cubic units

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4 Worked Example 3 — Exponential

The region bounded by y = e^x, the y-axis, y = 1, and y = e^2 is rotated about the y-axis. Find the volume.

Solution:

y = e^x implies x = ln y, so x^2 = (ln y)^2.

V = pi * integral from 1 to e^2 of (ln y)^2 dy

This requires integration by parts. Let u = (ln y)^2, dv = dy.

Then du = 2(ln y)(1/y) dy, v = y.

integral of (ln y)^2 dy = y(ln y)^2 - integral of 2 ln y dy

For integral of ln y dy: use parts again with u = ln y, dv = dy.

= y ln y - y + C

So integral of (ln y)^2 dy = y(ln y)^2 - 2(y ln y - y) + C = y(ln y)^2 - 2y ln y + 2y + C

Evaluate from 1 to e^2:

At y = e^2: e^2(2)^2 - 2e^2(2) + 2e^2 = 4e^2 - 4e^2 + 2e^2 = 2e^2

At y = 1: 1(0) - 2(0) + 2 = 2

V = pi * (2e^2 - 2) = 2pi(e^2 - 1) cubic units

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5 The Shell Method (Introduction)

The shell method provides an alternative to the disc method. It is especially useful when:

• The function is given as y = f(x) and rearranging for x is difficult.
• The region is bounded by curves that are easier to describe in terms of x.

Derivation

Consider a thin vertical strip at position x, with height f(x) and width delta x. When this strip is rotated about the y-axis, it sweeps out a thin cylindrical shell with:

• Radius = x (distance from the y-axis)
• Height = f(x)
• Thickness = delta x

The volume of this shell is approximately:

delta V = 2pi * x * f(x) * delta x

(This is the circumference times the height times the thickness.)

Summing all shells:

V = 2pi * integral from a to b of x * f(x) dx

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6 Worked Example 4 — Shell Method

Use the shell method to find the volume when y = x^2 (for 0 <= x <= 2) is rotated about the y-axis.

Solution:

V = 2pi * integral from 0 to 2 of x * x^2 dx = 2pi * integral from 0 to 2 of x^3 dx

= 2pi * [x^4/4] from 0 to 2

= 2pi * (16/4)

= 8pi cubic units

Let us verify with the disc method: x^2 = y, limits y = 0 to y = 4.