Area Between Curves | Volumes of Solids: The Washer Method |

Volumes of Solids by Cross Sections | Volumes of Solids: The Shell Method |

Volumes of Solids: The Disk Method | More Practice |

One very useful application of Integration is finding the **area and volume** of “curved” figures, that we couldn’t typically get without using Calculus. Since we already know that can use the integral to get the area between the $ x$- and $ y$-axis and a function, we can also get the volume of this figure by **rotating the figure** around either one of the axes.

## Area Between Curves

We learned how to get the **area under a curve** here in the **Definite Integrals** section, and actually, the area under a curve is technically the area between the curve (function) and the $ x$-axis, since the $ x$-axis is at $ y=0$. Generally, we can get the area between any two curves by **subtracting** the bottom curve from the top curve everywhere where the top curve is higher than the bottom curve. The cool thing about this is it even works if one of the curves is below the $ x$-axis, as long as the higher curve always stays above the lower curve in the integration interval.

Note that we may need to find out where the two curves intersect (and where they intersect the $ x$-axis) to get the limits of integration. And sometimes we have to divide up the integral if the functions cross over each other in the integration interval.

Here is the formal definition of the area between two curves:

Let’s try some problems:

Notice this next problem, where it’s much easier to **find the area with respect to **$ y$, since we don’t have to divide up the graph. When we integrate with respect to $ y$, we will have **horizontal rectangles** (parallel to the $ x$-axis) instead of **vertical rectangles **(perpendicular to the $ x$-axis), since we’ll use “$ dy$” instead of “$ dx$”. If we have the functions in terms of $ x$, we need to use **Inverse Functions** to get them in terms of $ y$.

**If you’re having trouble find the area with respect to **$ x$**, trying “tilting the graph” to see if it’s possible to integrate with respect to **$ y$**.**

Here are more problems where we take the area with respect to $ y$.

**Again, if you’re having trouble find the area with respect to **$ x$**, trying “tilting the graph” to see if it’s possible to integrate with respect to **$ y$**. For the entire interval, the top (right) curve must always be “above” (or to the right of) the bottom graph.**

## Volumes of Solids by Cross Sections

Now that we know how to get areas under and between curves, we can use this method to get the **volume** of a three-dimensional solid, either with cross sections, or by rotating a curve around a given axis. Think about it; every day engineers are busy at work trying to figure out how much material they’ll need for certain pieces of metal, for example, and they are using calculus to figure this stuff out!

Let’s first talk about getting the volume of **solids by cross-sections** of certain shapes. When doing these problems, think of the bottom of the solid being flat on your horizontal paper, and the 3-D part of it coming up from the paper. Cross sections might be squares, rectangles, triangles, semi-circles, trapezoids, or other shapes. (We’ll have to use some geometry to get these areas.)

Cross sections can either be perpendicular to the $ x$-axis or $ y$-axis; in our examples, they will be perpendicular to the $ x$-axis, which is what is we are used to.

Since **integration is “infinite summation”**, we can just integrate over the interval of cross sections to get a volume. Thus, given the cross-sectional area $ A(x)$ in interval $ [a,b]$, and cross sections are perpendicular to the $ x$-axis, the volume of this solid is $ \text{Volume = }\int\limits_{a}^{b}{{A\left( x \right)}}\,dx$

Here are examples of volumes of cross sections between curves. **Slices of the volume** are shown to better see how the volume is obtained:

Here’s one more that’s a little tricky:

## Volumes of Solids: The Disk Method

Now let’s talk about getting a volume by revolving a function or curve around a given axis to obtain a solid of **revolution**.

Since we know now how to get the area of a region using integration, we can get the volume of a solid by rotating the area around a line, which results in a right cylinder, or **disk**. (Remember that the formula for the volume of a cylinder is $ \pi {{r}^{2}}\cdot \text{height}$). The **radius** is the distance from the axis of revolution to the function, and the “height” of each disk, or slice is “$ dx$” or “$ dy$”, depending on the orientation. Again, since **integration is “infinite summation”**, we can just integrate over the interval to get a volume. Note that volumes may be different, depending on which axis is used for rotation!

Let’s try some problems:

## Volumes of Solids: The Washer Method

The **washer method** is similar to the disk method, but it covers solids of revolution that have “holes”, where we have inner and outer functions, thus inner and outer radii.

With washers, we have two revolving solids and we basically subtract the area of the inner solid from the area of the outer one. Note that for this to work, the middle function must be completely inside (or touching) the outer function over the integration interval.

Let’s try some problems:

## Volumes of Solids: The Shell Method

The **shell method** for finding volume of a solid of revolution uses integration along an axis **perpendicular** to the axis of revolution instead of **parallel**, as we’ve seen with the disk and washer methods. The nice thing about the shell method is that you can integrate around the $ y$-axis and not have to take the inverse of functions. Also, the rotational solid can have a hole in it (or not), so it’s a little more robust. It’s not intuitive though, since it deals with an infinite number of “surface areas” of rectangles in the shapes of cylinders (shells).

Here are the equations for the **shell method**:

| |

$ \displaystyle \text{Volume}=2\pi \int\limits_{a}^{b}{{x\,f\left( x \right)}}\,dx$ | Revolution around the $ \boldsymbol {x}$-axis: $ \displaystyle \text{Volume}=2\pi \int\limits_{a}^{b}{{y\,f\left( y \right)}}\,dy$ |

Since I believe the shell method is no longer required the Calculus AP tests (at least for the AB test), I will not be providing examples and pictures of this method. Please let me know if you want it discussed further.

**Learn these rules and practice, practice, practice!**

Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

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You can also go to the **Mathway** site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!

On to **Integration by Parts** — you are ready!