## 8.9 Volume with Disc Method: Revolving Around the x- or y-Axis

In AP Calculus, we dive deep into the concept of calculating volume through the , where we revolve shapes around either the x- or . This method is fundamental in solving problems involving volumes of solids of revolution. Letβs get started!

## πͺοΈ Volumes of Solids of Revolution

When we talk about finding the volume of a solid of revolution, we're essentially determining how much *space* the 3D shape occupies. To do this, we take a curve and rotate it around a particular axis, which creates a solid shape. The disc method allows us to calculate the volume of these solids accurately.

### π₯ The Disc Method: X-Axis

The disc method involves slicing the solid into infinitely thin discs perpendicular to the axis of rotation. By summing up the volumes of these discs using definite integrals, we obtain the total volume of the solid.

A Rotated Solid with a cut-out cross section that is disk-shaped with a width of

Courtesy of Cochran Math

To find the volume of a solid rotated around the , as shown above, we add together the volume of many thin cross-sections! These cross sections of the rotate volume have a width of $dx$dx (or approaching 0) and a of $f(x)$f(x). As shown above, each cross-section is a super flat cylinder, with a volume of $w \times \pi r^2$wΓΟr2, where $w$w is the width and $r$r is the radius. Plugging in our formulas for width and radius, we get:

$\pi (f(x))^2 dx$Ο(f(x))2dx

To add many of these volumes together, we use an integral! Our final integral is:

$\int_{a}^{b}\pi (f(x))^2dx$β«abβΟ(f(x))2dx

Where $a$a and $b$b are the stated boundaries or endpoints for the given equation $f(x)$f(x), given as $x = a$x=a and $x=b$x=b.

### π₯ The Disc Method: Y-Axis

Sometimes, you will be asked to rotate an object around the y-axis.

Rotated Solid around y-axis

Courtesy of MathScene

The process is pretty much the same, except you replace $f(x)$f(x) and $dx$dx with $f(y)$f(y) and $[dy](https://www.fiveableKeyTerm:dy)$[dy](https://www.fiveableKeyTerm:dy). The new equation to use is:

$\int_{c}^{d}\pi (f(y))^2dy$β«cdβΟ(f(y))2dy

Where the boundaries $c$c and $d$d are given as $y=c$y=c and $y=d$y=d.

### βοΈ Solving Using the Disc Method

To solve any rotated solid problem involving the x- or y-axis and only 1 equation, use these steps!

- π§
**Determine the Axis**: Identify whether the solid is being revolved around the x-axis or y-axis. This determines how we set up our integral. - βοΈ
**Slice the Solid**: Visualize slicing the solid into thin discs perpendicular to the chosen axis. Each disc represents an infinitesimally small volume element. - π§±
**Setup Integral**: Using definite integrals, set up the integral to sum up the volumes of all these discs along the given interval. Use the equation above! - π
**Evaluate the Integral**: Once the integral is set up, evaluate it to find the total volume of the solid.

π‘

Important! The equation given above *only* works with solids rotated around the x- or y-axis, and when only *one* $f(x)$f(x) or $f(y)$f(y) equation is given. For more information on rotating around other axes or using multiple equations with the washer method, check out our 8.10, 8.11, and 8.12 guides!

## πΆοΈ Practicing with the Disc Method

Letβs get into how to apply these steps to practice problems!

### βοΈ Disc Method: Practice Question 1

Calculate the volume of the solid obtained by the bounded by the curves $y=x^2$y=x2, $x=1$x=1 , and the y-axis about the x-axis.

*π§ Step 1) Determine the Axis*

As stated in the question, we need to find the volume of the solid obtained by revolving the region bounded by the curves about the x-axis.

*βοΈ Step 2) Slice the Solid*

Visualize slicing the region into thin discs perpendicular to the x-axis. At this point, you should be graphing all of your equations and shading the area to be revolved!

Graph of the equations given above

Graph Courtesy of Julianna Fontanilla

*π§± Step 3) Setup Integral*

Since we're revolving around the x-axis, we integrate with respect to x and use the first general equation. The volume can be found using the formula below where $f(x)$f(x) represents the defining the region and $a$a and $b$b represent the interval of .

$V= \int_{a}^{b}\pi (f(x))^2dx=\pi\int_{a}^{b}(f(x)^2)\, dx$V=β«abβΟ(f(x))2dx=Οβ«abβ(f(x)2)dx

*π Step 4) Evaluate the Integral*

Integrate the function $y=x^2$y=x2 from $x=0$x=0 to $x=1$x=1 using definite integration. As you can see on the graph, the lower bound of this integral occur first where the function $y=x^2$y=x2 intersects with the y-axis, or when $x=0$x=0. The upper bound occurs when $x=1$x=1.

$V=\pi\int_{0}^{1}(x^2)^2\, dx$V=Οβ«01β(x2)2dx$V=\pi\int_{0}^{1}x^4\, dx$V=Οβ«01βx4dx$V=\pi[\frac15x^5]\vert_0^1$V=Ο[51βx5]β£01β$V=\pi(\frac15(1)^5-\frac15(0)^5)$V=Ο(51β(1)5β51β(0)5)$V=\pi(\frac15)=\boxed{\frac{\pi}{5}}$V=Ο(51β)=5Οββ

Amazing work! Letβs give another one a try.

### βοΈ Disc Method: Practice Question 2

Find the area of the solid found by rotating the region bounded by $y = 1$y=1, $y=8$y=8, and $y = x^3$y=x3 around the y-axis.

*π§ Step 1) Determine the Axis*

The problem states that we are rotating around the y-axis. This means that our integral should be in terms of $f(y)$f(y) instead of $f(x)$f(x).

*βοΈ Step 2) Slice the Solid*

Again, we need to draw out the integral!

Graph of listed functions

Graph Courtesy of Julianna Fontanilla

As you can see here, we are rotating the area between the purple function and the y-axis which is also between the red and black horizontal lines of $y = 8$y=8 and $y = 1$y=1 respectively.

*π§± Step 3) Setup Integral*

Be careful here! Remember that for rotating around the y-axis, you need to have your main function to rotate in terms of $f(y)$f(y), or look like $x = f(y)$x=f(y). Our current equation given is:

$y = x^3$y=x3

To change this to terms of $f(y)$f(y), we would take the cube root of both sides to get:

$\sqrt[3] {y}=x$3yβ=x

This is the correct function to use in your general integral formula!

*π Step 4) Evaluate the Integral*

Again, the general integral format is

$V= \int_{a}^{b}\pi (f(y))^2dy=\pi\int_{a}^{b}(f(y)^2)\, dy$V=β«abβΟ(f(y))2dy=Οβ«abβ(f(y)2)dy

We are told that our bounds are $y = 1$y=1, $y=8$y=8, and $y = x^3$y=x3. Plugging this information into the equation, we get:

$V= \int_{1}^{8}\pi (\sqrt[3]{y})^2dy=\pi\int_{1}^{8}(\sqrt[3]{y}^2)\, dy=\pi\int_{1}^{8}({y}^{2/3}) dy$V=β«18βΟ(3yβ)2dy=Οβ«18β(3yβ2)dy=Οβ«18β(y2/3)dy

This simplifies toβ¦

$V=\pi[\frac35x^{5/3}]\vert_1^8$V=Ο[53βx5/3]β£18β$V=\pi(\frac35(8)^{5/3}-\frac35(1)^{5/3})$V=Ο(53β(8)5/3β53β(1)5/3)$V=\pi(\frac35(32)-\frac35(1))$V=Ο(53β(32)β53β(1))$V=\pi(\frac35(31))$V=Ο(53β(31))$V=\boxed{\frac{93}5 \pi}$V=593βΟβ

You did it! π

## π Conclusion

Mastering the disc method for finding volumes of solids of revolution expands our problem-solving toolkit in calculus. Letβs move on to the next guide for wider applications of this important technique!