Volume of Revolution About the x-Axis with Examples

The volume of revolution about the x-axis can be found by evaluating a definite integral with respect to the square of the equation of the curve and multiplying the result by π. We can derive a formula to calculate this volume using limits and integrals.

In this article, we will learn how to calculate the volume of revolution with respect to the x-axis. We will learn how to derive its formula and use it in some practice problems.

CALCULUS
Formula for the volume of revolution around the x axis

Relevant for

Learning how to calculate the volume of revolution about the x-axis.

See examples

CALCULUS
Formula for the volume of revolution around the x axis

Relevant for

Learning how to calculate the volume of revolution about the x-axis.

See examples

How to find the volume of revolution about the x-axis

To find the volume of revolution of a solid with respect to the x-axis, we can apply the following formula:

$$V=\pi \int_{a}^{b} y^2 d x$$

Then, we can follow the following process to use the formula correctly:

1. Find the square of the equation of the curve that forms the volume of revolution.

That is, we start by finding $latex y^2$.

2. Find the value of the definite integral $latex \int_{a}^{b} y^2 d x$.

For this, we substitute the expression for $latex y^2$ from step 1.

3. Multiply the result from step 2 by π to find the volume of the solid formed.

If you need to review how to solve definite integrals, you can visit our article: How to calculate definite integrals.


Proof of the formula for the volume of revolution about the x-axis

To find a formula for the volume of revolution about the x-axis, we start by considering the area under the curve $latex y=x^2$ from $latex x=1$ to $latex x=2$, as shown in the diagram:

Diagram 1 for the volume of revolution with x axis

Now, we consider the solid formed when this area is rotated 2π radians around the x-axis, as shown in the diagram below. We can calculate its volume using integral calculus.

Diagram 2 for the volume of revolution with x axis

We take a very small strip of length $latex x$ under the curve $latex f(x)$, as shown in the following diagram:

Diagram 3 for the volume of revolution with x axis

When this small area is rotated by 2π radians around the x-axis, we form a disk with a radius equal to $latex y$ and a width of $latex \delta x$. The volume of this disk is:

$latex \delta V= \pi y^2 \delta x$.

To find the volume of the whole solid, we have to find the sum of all the disks from $latex x=a$ to $latex x=b$. Then, we have:

$$V=\sum_{x=a}^{b}\pi y^2 \delta x$$

As $latex \delta \to 0$, this summation approaches the value of $latex V$, so we have:

$$V=\lim_{\delta x \to 0} \pi \sum_{x=a}^{b} y^2 \delta x$$

which is equal to

$$V=\pi \int_{a}^{b} y^2 d x$$


Volume of revolution about the x-axis – Examples with answers

EXAMPLE 1

Find the volume generated when the area under $latex y=x$ from $latex x=0$ to $latex x=6$ is rotated about the x-axis.

We begin by writing the formula for the volume of revolution:

$$V=\pi \int_{a}^{b} y^2 d x$$

Now, we substitute the expression for $latex y$, square it and solve the definite integral:

$$V=\pi \int_{0}^{6} x^2 d x$$

$$=\pi \left[ \frac{x^3}{3} \right]_{0}^{6}$$

$$=\pi \left( \frac{216}{3} \right)-(0)$$

$latex V=72 \pi $

EXAMPLE 2

What is the volume generated when the area under $latex y=x^2$ from $latex x=0$ to $latex x=5$ is rotated with respect to the x-axis?

The formula for the volume of revolution with respect to the x-axis is:

$$V=\pi \int_{a}^{b} y^2 d x$$

In this case, we have $latex y=x^2$, so we substitute into the formula and evaluate the definite integral:

$$V=\pi \int_{0}^{5} x^4 d x$$

$$=\pi \left[ \frac{x^5}{5} \right]_{0}^{5}$$

$$=\pi \left( \frac{3125}{5} \right)-(0)$$

$latex V=625 \pi $

EXAMPLE 3

Find the volume of the solid formed when the area under $latex y=x^2+2$ from $latex x=1$ to $latex x=3$ is rotated about the x-axis.

The volume of the solid is given by the following formula:

$$V=\pi \int_{a}^{b} y^2 d x$$

By squaring $latex y$, we have $latex y^2=(x^2+2)^2$, which is equal to $latex y^2=x^4+4x^2+4$. Then, we have:

$$V=\pi \int_{1}^{3} (x^4+4x^2+4) d x$$

$$=\pi \left[ \frac{x^5}{5} + \frac{4x^3}{3}+4x \right]_{1}^{3}$$

$$=\pi \left( \frac{483}{5}- \frac{83}{15} \right)$$

$$ V=\frac{1366 \pi}{15} $$

EXAMPLE 4

Find the volume of revolution of the area under $latex y=3\sqrt{x}$ from $latex x=2$ to $latex x=4$ about the x-axis.

We start with the following formula to find the volume:

$$V=\pi \int_{a}^{b} y^2 d x$$

When we square $latex y$, we get $latex y^2=9x$. Then, we use this in the formula to solve the definite integral:

$$V=\pi \int_{2}^{4} 9x d x$$

$$=\pi \left[ \frac{9x^2}{2} \right]_{2}^{4}$$

$latex =\pi (72 )-( 18)$

$latex V=54\pi $

EXAMPLE 5

Find the volume of revolution when the area under $latex y=\sqrt{x^2+3x}$ from $latex x=2$ to $latex x=6$ is rotated about the x-axis.

We begin by writing the formula for the volume of revolution about the x-axis:

$$V=\pi \int_{a}^{b} y^2 d x$$

When we square $latex y$, we have $latex y^2=(\sqrt{x^2+3x})^2$, which is equal to $latex y^2=x^2+3x$. Then, we have:

$$V=\pi \int_{2}^{6} (x^2+3x) d x$$

$$=\pi \left[ \frac{x^3}{3} + \frac{3x^2}{2} \right]_{2}^{6}$$

$$=\pi [72 + 54 ]-\left[ \frac{8}{3} + 6 \right]$$

$$=\pi \left( \frac{352}{3} \right)$$

$$ V=\frac{352\pi}{3} $$

EXAMPLE 6

The area enclosed between the curve $latex y=4-x^2$ and the straight line $latex y=4-2x$ is rotated with respect to the x-axis. Find the volume of the generated solid.

In this case, we have an area between two curves. Then, we can plot a simple graph to visualize that area:

Example of the volume of revolution with area between two curves

Now, we can see that the area to be rotated is an area between two curves. Therefore, we solve as follows:

$$V=\pi \int_{0}^{2} (4-x^2)^2 d x-\pi \int_{0}^{2} (4-2x)^2 d x$$

$$=\pi \int_{0}^{2} (4-x^2)^2 – (4-2x)^2 d x$$

$$=\pi \int_{0}^{2} (x^4-12x^2+16x) d x$$

$$=\pi \left[ \frac{x^5}{5}-4x^3+8x^2 \right]_{0}^{2}$$

$$=\pi \left[ \frac{32}{5}-32+32 \right]-[0]$$

$latex V=\frac{32\pi}{5} $


Volume of revolution about the x-axis – Practice problems

Volume of revolution about the x-axis quiz
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You have completed the quiz!

What is the volume of revolution of the area under $latex y=\sqrt{3x^2+8}$ from $latex x=1$ to $latex x=3$ with respect to the x axis? ?

Write the answer in the input box.

$latex V=$ π

See also

Interested in learning more about integrals of functions? You can look at these pages:

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