5.1.1 Volumes of Revolution

Volumes of Revolution

This is the volume covered when an area is rotated around an axis.

📑For example: $y = x^2$

When this area is rotated 360° around the x-axis, a solid shape is formed.

The shape formed looks like this.

Proposition:

The formula for working out the volume of such a solid formed by rotating a curve around the x-axis is:

\boxed {V = \pi \int_a^b y^2 \, dx}

Proof (Sort of):

Consider the volume generated to be approximated by a number of discs.

\therefore \quad V \approx \pi y_1^2 w + \pi y_2^2 w + \ldots + \pi y_n^2 w

= \pi \sum_{i=1}^n y_i^2 w

We can make this a better approximation by using more and thinner discs, i.e., by making w smaller.

Letting $w \to 0$ , we get:

\sum_{i=1}^n y_i^2 w \to \int_a^b y^2 \, dx

\therefore \quad V = \pi \int_a^b y^2 \, dx

The above becomes an integral as $w \to 0$ , the discs become infinitesimal, thus integration is the summation of the infinitesimals.

infoNote

For example: Find the volume of the solid formed when the following area is rotated 360° around the x-axis.

Using $V = \pi \int_a^b y^2 \, dx$ :

y = x^3 \Rightarrow y^2 = x^6

\therefore \quad V = \pi \int_2^4 x^6 \, dx = \pi \left[ \frac{1}{7}x^7 \right]_2^4

= \pi \left[ \frac{1}{7}(4)^7 - \frac{1}{7}(2)^7 \right] = \frac{16256}{7} \pi

NOTE: Must not forget $\pi$

lightbulbExample

Example: Find the volume of the solid formed when $y = \frac{1}{\sqrt{x-3}}$ is rotated 360° considering the area between $x = 5, x = 6$ , and the x-axis.

y = \frac{1}{\sqrt{x-3}} \quad \Rightarrow \quad y^2 = \frac{1}{x-3}

\therefore \quad V = \pi \int_5^6 \frac{1}{x-3} \, dx = \pi \bigg[\ln | x-3|\bigg]_5^6

= \pi (\ln 3 - \ln 2)

BAD:

$1.273806205$

(using calculator result for $\pi (\ln 3 - \ln 2))$

GOOD:

\pi \ln \left( \frac{3}{2} \right)

Keep answers exact when irrational!!!

infoNote

Q2 (OCR 4723, Jan 2007, Q6) The diagram shows the curve with equation $y = \frac{1}{\sqrt{3x+2}}$ . The shaded region is bounded by the curve and the lines $x = 0, x = 2$ , and $y = 0$ .

(i) Find the exact area of the shaded region.

(ii) The shaded region is rotated completely about the x-axis. Find the exact volume of the solid formed, simplifying your answer.

Solution: (i)

\int_0^2 \left( {3x + 2} \right)^{-\frac{1}{2}} dx = \left[ \frac{1}{3}(3x + 2)^{\frac{1}{2}} \right]_0^2 = \bigg[\frac{2}{3} (3x + 2)^{\frac{1}{2}} \bigg ]_0^2

= \frac{2}{3} (8)^{\frac{1}{2}} - \frac{2}{3}(2)^{\frac{1}{2}} = 0.9428090416 \quad \color {red}\text {BAD}

\frac{2}{3} \sqrt 8 - \frac{2}{3} \sqrt 2 \\= \frac{4}{3} \sqrt 2 - \frac{2}{3} \sqrt 2

=\boxed {\frac{2}{3} \sqrt 2}

Solution: (ii)

y = \frac{1}{\sqrt{3x + 2}} \Rightarrow y^2 = \frac{1}{3x + 2}

Thus:

V = \pi \int_0^2 \frac{1}{3x + 2} dx = \pi \left[ \frac{1}{3} \ln(3x + 2) \right]_0^2

= \pi \left[ \frac{1}{3} \ln 8 - \frac{1}{3} \ln 2 \right]= \frac{1}{3}\pi \left( \ln 8 - \ln 2 \right)

= \frac{1}{3}\pi \ln \left( \frac{8}{2} \right) = \boxed {\frac{1}{3}\pi \ln(4)}

Volume Generated When Rotating Around the y-Axis

lightbulbExample

Example: Find the volume generated when the following area is rotated around the y-axis 360°.

Formula: $V = \pi \int_a^b x^2 \, dy$

(y-limits = b, a)

Rearrange to say $x = f(y)$ :

y = x^2 \Rightarrow x = \sqrt{y}

(Not $\pm \sqrt{y}$ since $x > 0$ in the diagram)

Use the above volume formula, being sure to use y-limits, not x-limits:

x = 3 \Rightarrow y = 3^2 = 9

x = 5 \Rightarrow y = 5^2 = 25

x^2 = y

V = \pi \int_9^{25} x^2 \, dy = \pi \int_9^{25} y \, dy = \pi \left[ \frac{y^2}{2} \right]_9^{25}

= \pi \left[ \frac{25^2}{2} - \frac{9^2}{2} \right] = \pi \times 272 = \boxed {272\pi}

Volumes of Revolution Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

5.1.1 Volumes of Revolution

Volumes of Revolution

Proposition:

Proof (Sort of):

Volume Generated When Rotating Around the y-Axis

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