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Washer Method

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Suppose you're tasked with finding the area of a two-dimensional donut. You know how to find the area of a circle, but how can you account for the hole in the middle of the circle?

One way to find the area of the donut would be to see it as a circle within a circle! This way you can find the area of the smaller circle and subtract if from the area of the bigger circle.

You can use a similar idea for finding the volume of a solid of revolution! Through a simple generalization of the disk method, you can remove a portion of the disk to find the volume of a hollow solid of revolution.

The **washer method** is an integration method that slices a solid of revolution up into a series of three-dimensional washers, or flat donuts. The volume of each washer is added to find the total volume of the solid.

The **washer method **is a method for calculating the volume of a solid of revolution that is formed by rotating a region that is not bound by the axis of rotation.

Usually, the region bound between two curves is rotated. Like the disk method, the integration is **parallel** to the axis of rotation.

This method is a modification of the Disk Method for solids that are hollow in a certain way. It is called the "washer method" because the cross-sections look like *washers*.

Figure 1. A cross-section of a washer as seen from above

The washer method is particularly useful in finding the volume of a solid of revolution when the axis of rotation does **not **bound the revolving region. Because the axis of revolution is not a boundary of the region to be rotated, the resulting solid of revolution is hollow.

In order to better understand the washer method, take a look at the following image.

Figure 2. A three-dimensional washer

The cross-section of a washer consists of two concentric circles. To find the **area of this cross-section**, \( A_\text{washer} \), you need to subtract the area of the smaller circle from the area of the bigger circle, that is

\[ \begin{align} A_{\text{washer}} &= \pi R^2 - \pi r^2 \\ \\ &= \pi (R^2-r^2), \end{align}\]

where \( R \) is the radius of the bigger circle, and \( r \) is the radius of the smaller circle.

Figure 3. The cross-section of a washer can be found by removing the area of the inner circle from the area of the bigger circle

To find the **volume of the washer**, \( V_\text{washer} \), you multiply the area of the above cross-section by its thickness, so

\[ V_{\text{washer}} = \pi (R^2-r^2) \, \Delta x, \]

where \( \Delta x \) is the thickness of the washer.

Now that you found the volume of an individual washer, the next step is to add up the volumes of all the washers.

To get the exact volume of the solid some considerations have to be made. First, suppose that the solid of revolution is obtained by rotating the area bound between \( f(x) \) and \( g(x),\) where \( |f(x)| > |g(x)|.\)

Both the big and small radii of the cross-section of a washer become functions, so \(R\) becomes \( f(x) \) and \( r \) becomes \( g(x).\)

The washers become very thin, so \( \Delta x \) becomes \( \mathrm{d}x.\)

Instead of adding all the washers, you integrate, which means summation, \( \Sigma, \) becomes integration, \( \int.\)

You can find which function represents the inner circles by looking at which function is closer to the \(x-\)axis.

As the washers become very thin, their length becomes a differential. These differentials are added up by means of integration. This is where calculus comes to play!

You can obtain the formula for finding the volume of a solid of revolution obtained with the washer method by following the above considerations.

Suppose that the region bounded by two functions, \( f(x) \) and \( g(x),\) is revolved around the \(x-\)axis on an interval \( [a,b]. \) The **formula for finding the volume of the solid of revolution with the washer method** is

\[V=\int_a^{b} \pi \left[ \left( f(x) \right) ^2 - \left( g(x) \right)^2 \right] \, \mathrm{d}x, \]

where \( |f(x)| > |g(x)| \) in the interval of integration.

If the revolution is done around the \(y-\)axis, then the formula is adapted as

\[ V =\int_a^b \pi \left[ \left( f(y) \right) ^2 - \left( g(y) \right) ^2 \right] \, \mathrm{d}y. \]

In this case, \( g(y) \) is the function that is closer to the \( y-\)axis.

You can use the properties of integrals to write the above formulas as

\[ V = \int_a^b \pi \left[ f(x) \right]^2 \, \mathrm{d}x - \int_a^b \pi \left[ g(x) \right]^2 \, \mathrm{d}x, \]

and

\[ V = \int_a^b \pi \left[ f(y) \right]^2 \, \mathrm{d}y - \int_a^b \pi \left[ g(y) \right]^2 \, \mathrm{d}y, \]

essentially making the formulas as if you did the disk method for two functions, and then subtracted the inner volume from the outer volume.

Though the washer method builds upon the disk method, it is slightly more involved. Here are the general steps you need to follow to obtain the volume of a solid obtained through the disk method.

To ensure that you're visualizing the region correctly, graph the region while identifying the axis of rotation. This will help you with the next step.

Next, you need to find which function corresponds to the outer radius, and which corresponds to the inner radius. The radii are determined based on the distance between the functions and the axis of rotation, so the function that is closer to the axis should be the inner radius, \( r(x),\) and the other should be the outer radius, \( R(x).\)

This means that the outer radius will be the function that is greater in **absolute value,** which means that \( |R(x)| > |r(x)|.\) The absolute value is only to identify which function is which! There is no need to actually calculate the absolute value in the integration.

All that's left to do is plug in bounds, the two radii, and you are left with a definite integral which you can solve using any method of your choice.

Here are some examples of the washer method to find the volume of solids of revolution.

Consider the functions

\[ f(x) = -x^2 + 3 \quad \text{for} \quad 0 \leq x \leq 1,\]

and

\[ g(x) = -x^2 + 2 \quad \text{for} \quad 0 \leq x \leq 1.\]

Find the volume of the solid of revolution obtained by rotating the area bound between the two curves along the \(x-\)axis.

**Answer:**

1. *Graph the region.*

Begin by graphing the two functions. \( f(x) \) is a parabola with vertex at \( (0,3) \) and \( g(x) \) is a parabola with vertex at \( (0,2).\) Both parabolas open downwards.

The solid of revolution will be obtained by rotation of the following area bounded between both curves.

The resulting solid looks as follows.

* *2. *Find the outer and inner radii.*

Note that \( f(x) \) is above \( g(x) \) in the given interval, so \(f(x)=R\) and \( g(x)=r.\)

* *3. *Use the formula of the washer method.*

You will need to square both functions, so

\[ \begin{align} R^2 &= ( f(x) )^2 \\ \\ &= (-x^2+3)^2 \\ \\ &= x^4-6x^2+9 \end{align} \]

and

\[ \begin{align} r^2 &=( g(x) )^2 \\ \\ &= (-x^2+2)^2 \\ \\ &= x^4-4x^2+4. \end{align} \]

Next, you subtract the square of the function of the inner radius, \( r^2,\) from the square of the function of the outer radius, \( R^2,\) that is

\[ \begin{align} R^2-r^2 &= \left( x^4-6x^2+9 \right) - \left( x^4-4x^2+4\right) \\ \\ &= -2x^2+5. \end{align}\]

This way you can use the formula for the washer method, so

\[ \begin{align} V &= \int_a^b \pi \left( R^2-r^2 \right) \, \mathrm{d}x \\ \\ &= \int_0^2 \pi (-2x^2+5) \, \mathrm{d}x \\ \\ &= \pi \int_0^1 (-2x^2+5) \, \mathrm{d}x. \end{align} \]

You can evaluate the resulting definite integral by first finding the indefinite integral using the Power Rule, that is

\[ \int (-2x^2+5) \, \mathrm{d}x = -\frac{2}{3}x^3+5x,\]

and then use the Fundamental Theorem of Calculus, which will give you

\[ \begin{align} \int_0^1 (-2x^2+5) \, \mathrm{d}x &= \left( -\frac{2}{3}(1)^3+5(1) \right) - \left( -\frac{2}{3}(0)^3+5(0)\right) \\ \\ &= \frac{13}{3}. \end{align}\]

Finally, you multiply the above definite integral by \( \pi, \) obtaining the volume of the solid, that is

\[ V= \frac{13\pi}{3}.\]

How about an exponential function?

Consider the functions

\[ p(x) = e^x \quad \text{for} \quad 0 \leq x \leq 2 \]

and

\[ q(x) = 1 \quad \text{for} \quad 0 \leq x \leq 2.\]

Find the volume of the solid of revolution obtained by rotating the area bound between the two curves along the \(x-\)axis.

**Answer:**

1. *Graph the region.*

Once again, you need to graph the given functions. For \( p(x),\) note that it is an exponential function, and \( q(x) \) is a constant function, so it is a horizontal line.

This way, the region to be revolved is given as follows.

And the solid of revolution is shown in the next picture.

* *2. *Find the outer and inner radii.*

From the previous graph you can note that \( q(x) \) is closer to the \(x-\)axis in the given interval, so \( q(x)=r.\) The remaining function is then \( p(x)=R.\)

* *3. *Use the formula of the washer method.*

You will need to take the difference of the squares of both functions, so first square \(R,\) obtaining

\[ \begin{align} R^2 &= (e^x)^2 \\ &= e^{2x}. \end{align}\]

The square of the constant function 1 its just

\[ \begin{align} r^2 &=1^2 \\ &=1, \end{align}\]

this way, the difference of the squares becomes

\[ R^2-r^2=e^{2x}-1. \]

Now you can plug in the above expression into the washer method formula, that is

\[ \int_a^b \pi (R^2-r^2)\,\mathrm{d}x = \int_0^2 \pi \left( e^{2x}-1 \right) \, \mathrm{d}x.\]

To evaluate the definite integral begin by finding the indefinite integral

\[ \int (e^{2x}-1) \, \mathrm{d}x = \frac{1}{2}e^{2x}-x,\]

and then use the Fundamental Theorem of Calculus to evaluate both ends, that is

\[ \begin{align} \int_0^2 (e^{2x}-1)\,\mathrm{d}x &= \left(\frac{1}{2}e^{2(2)}-2\right) - \left( \frac{1}{2}e^{2(0)}-0\right) \\ \\ &= \left( \frac{1}{2}e^4-2\right) - \left(\frac{1}{2}\right) \\ \\ &= \frac{1}{2}e^4-\frac{5}{2} \\ \\ &= \frac{1}{2}(e^4-5) .\end{align}\]

Finally, by multiplying by \(\pi\) the above result, you get the volume of the solid, so

\[ V = \frac{\pi}{2}(e^4-5).\]

Note how the solids of revolution are hollow in both examples!

- The
**washer method**is a method for calculating the volume of a solid of revolution when the region to be rotated is not bounded by the axis of rotation.- It is a modification of the Disk Method for solids with a hole in the middle
- It is called the "washer method" because the cross-sections look like washers

- The
**formula**for the washer method is\[V=\int_a^{b}(R^{2}-r^{2})dx\]

where \(R\) is the outer radius of the solid and \(r\) is the inner radius of the solid

If a region is revolved around the \(x\)-axis or any other horizontal line, then the slices are vertical - we should integrate with respect to \(x\)

If a region is revolved around the \(y\)-axis or any other vertical line, then the slices are horizontal - we should integrate with respect to \(y\)

The washer method is used when the axis of revolution does **not** bound the revolving region.

It is called the "washer method" because the cross-sections look like washers.

**perpendicular** to the axis of revolution.

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