Volumes of Revolution around the x-axis and y-axis

Master the concept of volumes of revolution with advanced worked examples.

📚 Further Maths 🎯 Difficulty: ⭐⭐ ⏱️ Reading time: 15 minutes 📋 Edexcel

📚 Introduction

The concept of volumes of revolution involves rotating a function around a specified axis to create a three-dimensional solid. Think of it like this: You have a fan with one blade. When you look at it from the side, it’s just a 2D shape. But when you turn it on, that blade spins around and creates a 3D shape. This is a volume of revolution, and we will see how we can revolve functions around the x and y axes.

You must know basic integration and finding the area under curves before this lesson.

🎯 Learning Objectives

Know the formula to calculate volumes of revolution around the x and y axes, and be able to apply it in questions.

Find the volume and area of 2D objects revolving around an axis
Understand and use the formula for the volume of revolution
Solve advanced problems involving calculus problem-solving

🔑 Key Concepts

Formula for Volume of Revolution:

The formula for the volume of revolution around the x-axis is given by:

V = \pi \int_{x_1}^{x_2} f(x)^2 \, dx$$ $$\text{Where:}$$ $$V = \text{Volume of the solid}$$ $$f(x) = \text{The function being revolved}$$ $$x_1, x_2 = \text{The limits of integration (the x-values)}

So all you do is square the function first, then integrate normally, then multiply by $\pi$. That is all. If you are revolving around the y-axis, just integrate with respect to y (so it’ll be $dy$ instead of $dx$).
This is the formula for the volume of revolution when it’s revolving around 360° (or $2\pi$ radians).

📝 Worked Examples

📋 Example 1: Simple Volume of Revolution:

Question: Below shows the curve $ f(x) = -x^2 + 8x + 2x-16$, which is being revolved around the x-axis. Find the exact volume of the solid formed.

Solution:

Step 1:

We’re given that $f(x) = -x^2 + 8x + 2x - 16$. Let’s simplify:

\underline{f(x) = -x^2 + 10x - 16} \text{ because } 8x + 2x = 10x

💡 Tip:

This lesson assumes previous integration knowledge; therefore, only the volume of revolution steps will be explained.

Step 2:

Let’s find the limits $x_1$ and $x_2$:

\text{The formula: } V = \pi \int_{x_1}^{x_2} f(x)^2 \, dx$$ $$\text{We need to find the limits of integration } x_1 \text{ and } x_2$$ $$\text{Use the quadratic formula to find the roots of } f(x) = 0$$ $$\text{So we set } -x^2 + 10x - 16 = 0$$ $$\text{Using the quadratic formula:}$$ $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-10 \pm \sqrt{10^2 - 4(-1)(-16)}}{2(-1)}$$ $$ = \frac{-10 \pm \sqrt{100 - 64}}{-2} = \frac{-10 \pm \sqrt{36}}{-2} = \frac{-10 \pm 6}{-2}$$ $$\text{So the two roots are: } \boxed{x_1 = 2 \text{ and } x_2 = 8}

Step 3:

Now we have everything to apply the formula:

\text{The formula is: } V = \pi \int_{2}^{8} (-x^2 + 10x - 16)^2 \, dx$$ $$\text{We have: } x_1 = 2, x_2 = 8, f(x) = -x^2 + 10x - 16$$ $$\text{Plug it in: } V = \pi \int_{2}^{8} (-x^2 + 10x - 16)^2 \, dx$$ $$\text{Now let’s expand: } \left(-x^2 + 10x - 16\right)^2$$ $$\left(-x^2 + 10x - 16\right)^2 = \underline{x^4 - 20x^3 + 132x^2 - 320x + 256}$$ $$\text{So now we have: }$$ $$ V = \pi \int_{2}^{8} (x^4 - 20x^3 + 132x^2 - 320x + 256) dx

Step 4:

Let’s now integrate:

\text{Remember the formula: } V = \pi \int_{x_1}^{x_2} f(x)^2 dx$$ $$\boxed{f(x) = x^4 - 20x^3 + 132x^2 - 320x + 256} \quad \boxed{x_1 = 2, x_2 = 8}$$ $$\text{Integrate first: }$$ $$ \int^8_2 (x^4 - 20x^3 + 132x^2 - 320x + 256) dx$$ $$\text{Which becomes: } \frac{x^5}{5} - 5x^4 + 44x^3 - 160x^2 + 256x$$ $$\text{Now we have: }$$ $$= \pi \left[ \frac{x^5}{5} - 5x^4 + 44x^3 - 160x^2 + 256x \right]_{2}^{8}$$ $$\text{Complete the integration normally:}$$ $$\text{Let } A = \left[ \frac{8^5}{5} - 5(8^4) + 44(8^3) - 160(8^2) + 256(8) \right] = \frac{16384}{5}$$ $$\text{Let } B = \left[ \frac{2^5}{5} - 5(2^4) + 44(2^3) - 160(2^2) + 256(2) \right] = \frac{96}{5}$$ $$\text{Do A - B: }$$ $$A - B = \frac{16384}{5} - \frac{96}{5} = \frac{16288}{5}$$ $$\text{Now that the integration is done, we have:}$$ $$V = \pi \int_{2}^{8} f(x)^2 dx = \pi (\frac{16288}{5}) $$ $$\text{So the final answer is: } \boxed{V = \frac{16288\pi}{5}}$$ $$\underline{\text{Keep in terms of } \pi \text{ when asked for exact volume}}

📋 Example 2: Advanced example with y axis and algebraic manipulation:

The graph below shows the function $f(x) = y^2 - 6y + 10$

Question: The region R is bounded by the curve $f(x)$, Find the exact volume of the solid formed when R is revolved around the y-axis.

Solution:

Step 1:

\text{Remember the formula for the y axis:}$$ $$V = \pi \int_{y_1}^{y_2} f(y)^2 \quad dy$$ $$\text{We have } f(y) = y^2 - 6y + 10$$ $$\text{We can see the limits on the graph: } y_1 = 1, y_2 = 5

Step 2:

Lets solve $f(y)^2$:

$$f(y)^2 = (y^2 - 6y + 10)^2 = y^4 - 12y^3 + 56y^2 - 120y + 100$$

Step 3:

Lets integrate now

\text{We have: } \pi \int_{1}^{5} (y^4 - 12y^3 + 56y^2 - 120y + 100) \, dy$$ $$\text{Evaluating the integral:}$$ $$=\left[ \frac{y^5}{5} - 3y^4 + \frac{56y^3}{3} - 60y^2 + 100y \right]_{1}^{5}$$ $$ \left[ \frac{5^5}{5} - 3(5^4) + \frac{56(5^3)}{3} - 60(5^2) + 100(5) \right]$$ $$ - \left[ \frac{1^5}{5} - 3(1^4) + \frac{56(1^3)}{3} - 60(1^2) + 100(1) \right]$$ $$= \frac{18190}{15}$$ $$\text{So now we multiply by } \pi : \boxed{V = \frac{18190\pi}{15}}

⚠️ Common Mistakes

🚫 Mistake 1: Forgetting to multiply by pi at the end

What students often do wrong: Its easy to get hyperfixated on the integral at first, then forget to multiply by $\pi$. Ensure you write out the entire equation at the start to avoid this.

💪 Practice Problems

Try these problems to test your understanding:

🎯 Practice Questions A

Evaluate the following:
(A) $x = \frac{1}{2}y + 1$ between $y = 2$ and $y = 5$
(B) $y = 2\sqrt{x}$ between $x = 0$ and $x = 1$
(C) $y = \frac{1}{x}$ between $x = 1$ and $x = 3$
(D) $y = 2x^2 - 4$ between $x = 5$ and $x = 11$

🔍 Show Solution

\text{(A) } V = \pi \int_{2}^{5} \left(\frac{1}{2}y + 1\right)^2 dy$$ $$= \pi \left[\frac{y^3}{12} + \frac{y^2}{2} + y\right]_{2}^{5}$$ $$= \pi \left(\frac{125}{12} + \frac{25}{2} + 5 - \frac{8}{12} - 2 - 2\right)$$ $$= \pi \left(\frac{117}{12} + \frac{21}{2} + 3\right) = \boxed{\frac{279\pi}{12}}

\text{(B) } V = \pi \int_{0}^{1} (2\sqrt{x})^2 dx = \pi \int_{0}^{1} 4x dx$$ $$= 4\pi \left[\frac{x^2}{2}\right]_{0}^{1} = 4\pi \left(\frac{1}{2} - 0\right) = \boxed{2\pi}

\text{(C) } V = \pi \int_{1}^{3} \left(\frac{1}{x}\right)^2 dx = \pi \int_{1}^{3} x^{-2} dx$$ $$= \pi [-x^{-1}]_{1}^{3} = \pi \left(-\frac{1}{3} + 1\right) = \boxed{\frac{2\pi}{3}}

\text{(D) } V = \pi \int_{5}^{11} (2x^2 - 4)^2 dx$$ $$= \pi \int_{5}^{11} (4x^4 - 16x^2 + 16) dx$$ $$= \pi \left[\frac{4x^5}{5} - \frac{16x^3}{3} + 16x\right]_{5}^{11} = \boxed{\frac{2048\pi}{15}}

🎯 Practice Questions B

(A) The curve C with equation $y = 2x^2 + 5$. The region bounded by the y-axis, the curve C, and the line $y = 10$ is rotated $360°$ about the y-axis. Find the exact volume of the solid generated. (6 marks)

(B) The curve $C$ with equation $x = \frac{1}{2}y^2 + 1$. The region $R$ is bounded by the lines $y=1$, $y=4$, the $y$-axis and the curve $C$. The region is rotated through $2\pi$ radians about the $y$-axis. Find the volume of the solid generated. (6 marks)

(C) The finite region $R$ is bounded by the curve $x = \sqrt{y} + \frac{1}{y^2}$, the lines $y=4$, $y=9$, and the $y$-axis.
(i) Find the exact area of the region. (3 marks)
(ii) The region $R$ is rotated through $2\pi$ radians about the $y$-axis. Use integration to find the volume of the solid generated. Round your answer to 2 decimal places. (5 marks)

🔍 Show Solution

(A) \text{ } V = \pi \int_{0}^{10} \left(\frac{y-5}{2}\right)^2 dy = \pi \int_{0}^{10} \frac{(y-5)^2}{4} dy$$ $$= \frac{\pi}{4} \left[\frac{y^3}{3} - 5y^2 + 25y\right]_{0}^{10} = \boxed{\frac{250\pi}{3}}

(B) \text{ } V = \pi \int_{1}^{4} \left(\frac{1}{2}y^2 + 1\right)^2 dy$$ $$= \pi \int_{1}^{4} \left(\frac{1}{4}y^4 + y^2 + 1\right) dy$$ $$= \pi \left[\frac{y^5}{20} + \frac{y^3}{3} + y\right]_{1}^{4} = \boxed{\frac{293\pi}{15}}

(C)(i) \text{ } A = \int_{4}^{9} \left(\sqrt{y} + \frac{1}{y^2}\right) dy$$ $$= \left[\frac{2}{3}y^{3/2} - y^{-1}\right]_{4}^{9} = \boxed{\frac{38}{3}}

(C)(ii) \text{ } V = \pi \int_{4}^{9} \left(\sqrt{y} + \frac{1}{y^2}\right)^2 dy$$ $$= \pi \int_{4}^{9} \left(y + 2y^{-3/2} + y^{-4}\right) dy$$ $$= \pi \left[\frac{y^2}{2} - 4y^{-1/2} - \frac{y^{-3}}{3}\right]_{4}^{9}$$ $$= \boxed{37.70} \text{ (to 2 decimal places)}

📋 Summary

🎯 Key Takeaways

The volume of revolution around the x-axis is given by $V = \pi \int_{x_1}^{x_2} f(x)^2 \, dx$
The volume of revolution around the y-axis is given by $V = \pi \int_{y_1}^{y_2} f(y)^2 \, dy$
Always square the function first, then integrate, then multiply by $\pi$
Find the limits of integration by determining where the curve intersects the axis or given boundaries
Remember to expand squared expressions carefully and integrate term by term

📚 What's Next?

Practice more volume problems with different types of functions. Next, explore more complex volumes involving parametric equations, and applications in engineering and physics contexts.

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