## 17Calculus Integrals - Volumes

This page gets you started on calculating volumes of two main types, volumes with known cross-section and volumes of revolution. For volumes of revolution, we use single integrals where an area is rotated about a vertical or horizontal line. The area is defined by equations in the form $$y=f(x)$$ or $$x=f(y)$$ and we use the washer (disc) method or the cylinder (shell) method. For other ways to calculate volume, see the links in the related topics panel.

Exam Preparation Suggestions

1. Volume integral problems are notoriously long and most will take more time than you are probably used to to work all the way through. So, once you are down to a few hours before your exam, just take the time to set up the integrals but not evaluate them completely. Most of your points will probably come from setting up the integral. By now, you know how to evaluate most integrals that you will see in class and evaluating them is the most time-consuming part. So focus on where you can get the most points and make sure you are confident setting up the integrals.
2. A day or two before your exam, go through the practice problems on the volumes-practice page. They will help you be better prepared to choose the correct technique without wasting a lot of time during your exam.

Volume With A Known Cross-Section built with GeoGebra

To get a feel for how to use integrals to find volumes, we will start with finding volumes when we know the equation of a cross-sectional area. Let's start with some examples that you are familiar with.
You know that the volume of a rectangular box is length times width times height or, in equation terms, if l is the length, w is the width and h is the height, the volume is $$V=lwh$$. Let's look at this from an integral calculus point of view. Let's take a cross-section parallel to the top of the box. The cross-section is the same no matter where we take it. In equation form, the area of the cross-section is $$A=lw$$.

The volume of a solid with cross-section $$A(x)$$ is $$V=\int_a^b{A(x)~dx}$$. In our example, our height goes from $$0$$ to $$h$$. So the volume is $$V=\int_0^h{lw~dx} =$$ $$[lwx]_0^h =$$ $$lwh$$, i.e. $$V=lwh$$ as we would expect.

In our example, the area, $$A=lw$$, was constant. However, usually, the cross-section will be different depending on where we take the cross-section. That's why we write the area as $$A(x)$$. Of course, if we take the cross-section perpendicular to the y-axis, the volume integral would be $$A(y)=\int_c^d{A(y)~dy}$$.

Use the volume integral to calculate the volume of a cylinder with radius r and height h.

$$V=\pi r^2 h$$

Problem Statement - Use the volume integral to calculate the volume of a cylinder with radius r and height h.
Solution - If you take the cross-section parallel to the base, the area is a circle, which is $$A=\pi r^2$$. We integrate from $$0$$ to $$h$$ to get $$V = \int_0^h{\pi r^2~dx} =$$ $$\pi r^2 x|_0^h =$$ $$\pi r^2 h$$.
For a more complete explanation including a 3dim graph, see this Pauls Online Notes page.

$$V=\pi r^2 h$$

Okay, so far the cross-section has been constant making the equation of the area quite simple. Let's look at an example where the cross-section is not constant.

Calculate the volume of the solid of a pyramid with a square base of side L and height h.

$$V=L^2h/3$$

Problem Statement - Calculate the volume of the solid of a pyramid with a square base of side L and height h.
Solution - We will outline the solution here but if you want a complete, well-written solution with 3dim plots, see this Pauls Online Notes page.
For this problem, it helps to think of the pyramid as upside-down with it's point sitting at the origin and the base up in the air. Since the cross-section is a square, we know the area of the cross-section is $$A=s^2$$, where $$s$$ is the length of side. However, $$s$$ is different depending on where the cross-section is taken. Using similar triangles, we can get $$s=Ly/h$$ where $$y$$ is the height above the origin.
The volume integral is then $$V=\int_0^h{L^2y^2/h^2~dy}$$ which solves to $$L^2h/3$$.

$$V=L^2h/3$$

That example is just one way to work the problem. Here are several videos showing other ways to solve it.

### Michel vanBiezen - Calculus - Integration: Volume of a Pyramid - Square Base Pyramid [8mins-12secs]

video by Michel vanBiezen

video by MIP4U

### David Lippman - Volumes by integrating cross sections - square pyramid [8mins-16secs]

video by David Lippman

### Integrals and Volumes: Volume of a Pyramid [7mins-35secs]

There are several more examples on this Pauls Online Notes page that we recommend. Before we jump into the practice problems, take a few minutes to watch this video. It has a good explanation and several examples.

### ProfRobBob - Volumes of Solids with Known Cross Sections 3 Examples

video by ProfRobBob

The base of a solid is bounded by the curves $$f(x)=2-2x$$ and $$g(x)=x^2-1$$ and the cross sections perpendicular to the x-axis are perfect squares. Find the volume of the solid.

Problem Statement

The base of a solid is bounded by the curves $$f(x)=2-2x$$ and $$g(x)=x^2-1$$ and the cross sections perpendicular to the x-axis are perfect squares. Find the volume of the solid.

Hint

This solid is difficult to visualize. However, the cross sections are squares. So, just find an equation for the length in the y-direction (perpendicular to the x-axis) and square it to get an equation for the area.

Problem Statement

The base of a solid is bounded by the curves $$f(x)=2-2x$$ and $$g(x)=x^2-1$$ and the cross sections perpendicular to the x-axis are perfect squares. Find the volume of the solid.

$$512/15$$

Problem Statement

The base of a solid is bounded by the curves $$f(x)=2-2x$$ and $$g(x)=x^2-1$$ and the cross sections perpendicular to the x-axis are perfect squares. Find the volume of the solid.

Hint

This solid is difficult to visualize. However, the cross sections are squares. So, just find an equation for the length in the y-direction (perpendicular to the x-axis) and square it to get an equation for the area.

Solution

### 2262 video

video by PatrickJMT

$$512/15$$

Calculate the volume of the solid whose cross-section is a rectangle with a height of $$y=\cos(x)$$ and width of 2 over the interval $$0\leq x \leq \pi/2$$.

Problem Statement

Calculate the volume of the solid whose cross-section is a rectangle with a height of $$y=\cos(x)$$ and width of 2 over the interval $$0\leq x \leq \pi/2$$.

$$V=2$$

Problem Statement

Calculate the volume of the solid whose cross-section is a rectangle with a height of $$y=\cos(x)$$ and width of 2 over the interval $$0\leq x \leq \pi/2$$.

Solution

### 2263 video

video by MIP4U

$$V=2$$

Calculate the volume of the solid where the cross-section is a triangle with a base of 4 inches and height of $$y=x^2$$ inches on the interval $$[0,2]$$.

Problem Statement

Calculate the volume of the solid where the cross-section is a triangle with a base of 4 inches and height of $$y=x^2$$ inches on the interval $$[0,2]$$.

Hint

Plot the equation $$y=x^2$$ from $$x=0$$ to $$x=2$$. The height of the triangle is the distance from the x-axis to the curve. Remember the equation of the area of a triangle is $$A=(1/2)bh$$ where b is the base and h is the height.

Problem Statement

Calculate the volume of the solid where the cross-section is a triangle with a base of 4 inches and height of $$y=x^2$$ inches on the interval $$[0,2]$$.

$$16/3$$ in3

Problem Statement

Calculate the volume of the solid where the cross-section is a triangle with a base of 4 inches and height of $$y=x^2$$ inches on the interval $$[0,2]$$.

Hint

Plot the equation $$y=x^2$$ from $$x=0$$ to $$x=2$$. The height of the triangle is the distance from the x-axis to the curve. Remember the equation of the area of a triangle is $$A=(1/2)bh$$ where b is the base and h is the height.

Solution

### 2264 video

video by MIP4U

$$16/3$$ in3

Calculate the volume of the solid where the cross-section perpendicular to the x-axis is a circle with diameter $$y=\sqrt{x}$$ on the interval $$[0,4]$$.

Problem Statement

Calculate the volume of the solid where the cross-section perpendicular to the x-axis is a circle with diameter $$y=\sqrt{x}$$ on the interval $$[0,4]$$.

$$V=2\pi$$

Problem Statement

Calculate the volume of the solid where the cross-section perpendicular to the x-axis is a circle with diameter $$y=\sqrt{x}$$ on the interval $$[0,4]$$.

Solution

### 2265 video

video by MIP4U

$$V=2\pi$$

Calculate the volume of a solid with base bounded by the curves $$y=x^2$$ and $$y=8-x^2$$ and cross-sections parallel to the y-axis that are semi-circles.

Problem Statement

Calculate the volume of a solid with base bounded by the curves $$y=x^2$$ and $$y=8-x^2$$ and cross-sections parallel to the y-axis that are semi-circles.

Hint

The diameter of the semi-circle is the distance between the curves.

Problem Statement

Calculate the volume of a solid with base bounded by the curves $$y=x^2$$ and $$y=8-x^2$$ and cross-sections parallel to the y-axis that are semi-circles.

$$\displaystyle{ V = \frac{256\pi}{15} }$$

Problem Statement

Calculate the volume of a solid with base bounded by the curves $$y=x^2$$ and $$y=8-x^2$$ and cross-sections parallel to the y-axis that are semi-circles.

Hint

The diameter of the semi-circle is the distance between the curves.

Solution

This video sets up the integral but does not evaluate it. Here is the remainder of the solution.

 $$\displaystyle{ V = \int_{-2}^{2}{ \frac{\pi}{2} (4-x^2)^2 ~dx } }$$ $$\displaystyle{ V = \frac{\pi}{2} \int_{-2}^2{ 16+x^4-8x^2 ~ dx } }$$ $$\displaystyle{ V = \frac{\pi}{2} \left[ 16x + \frac{x^5}{5} - \frac{8x^3}{3} \right]_{-2}^2 }$$ $$\displaystyle{ V = \frac{\pi}{2} \left[ 16(2)+ \frac{2^5}{5} - \frac{8}{3}(2^3) \right] - }$$ $$\displaystyle{ \frac{\pi}{2} \left[ 16(-2) + \frac{(-2)^5}{5} -\frac{8}{3}(-2)^3 \right] }$$ $$\displaystyle{ V = \frac{\pi}{2} \left[ 32+\frac{32}{5}-\frac{64}{3} \right] - }$$ $$\displaystyle{ \frac{\pi}{2} \left[ -32 - \frac{32}{5} + \frac{64}{3} \right] }$$ $$\displaystyle{ V = \frac{\pi}{2} \left[ 64+\frac{64}{5}- \frac{128}{3} \right] }$$ $$\displaystyle{ V = \frac{\pi}{2} \left[ \frac{960}{15} + \frac{192}{15} - \frac{640}{15} \right] }$$ $$\displaystyle{ V = \frac{\pi}{2}\left[ \frac{512}{15} \right] }$$ $$\displaystyle{ V = \frac{256\pi}{15} }$$

### 2266 video

video by David Lippman

$$\displaystyle{ V = \frac{256\pi}{15} }$$

Calculate the volume of a solid with an equilateral triangular cross-section parallel to the y-axis bounded by $$y=\pm (2-x^2/8)$$ and $$x=2$$. Units are in feet.

Problem Statement

Calculate the volume of a solid with an equilateral triangular cross-section parallel to the y-axis bounded by $$y=\pm (2-x^2/8)$$ and $$x=2$$. Units are in feet.

Hint

The equation of the area of equilateral triangle with side $$a$$ is $$A=\sqrt{3}(a^2/4)$$.

Problem Statement

Calculate the volume of a solid with an equilateral triangular cross-section parallel to the y-axis bounded by $$y=\pm (2-x^2/8)$$ and $$x=2$$. Units are in feet.

$$\displaystyle{\frac{153\sqrt{3}}{10}}$$ ft3

Problem Statement

Calculate the volume of a solid with an equilateral triangular cross-section parallel to the y-axis bounded by $$y=\pm (2-x^2/8)$$ and $$x=2$$. Units are in feet.

Hint

The equation of the area of equilateral triangle with side $$a$$ is $$A=\sqrt{3}(a^2/4)$$.

Solution

He writes his answer as $$15\frac{3}{10}\sqrt{3}$$ which is equal to $$(15+3/10)\sqrt{3}$$. Do not write it this way. This is only done in basic algebra. After you reach algebra 2, trig and especially calculus, this is interpreted as $$15(3/10)\sqrt{3}$$ which is not what he means.

### 2267 video

$$\displaystyle{\frac{153\sqrt{3}}{10}}$$ ft3

For $$1 \leq x \leq 4$$, determine the volume of the solid where each face is a triangle with base $$(x+1)$$ ft and height $$\sqrt{x}$$ ft.

Problem Statement

For $$1 \leq x \leq 4$$, determine the volume of the solid where each face is a triangle with base $$(x+1)$$ ft and height $$\sqrt{x}$$ ft.

$$128/15$$ ft3

Problem Statement

For $$1 \leq x \leq 4$$, determine the volume of the solid where each face is a triangle with base $$(x+1)$$ ft and height $$\sqrt{x}$$ ft.

Solution

### 2268 video

video by MIP4U

$$128/15$$ ft3

Calculate the volume of the solid where each face at x is a square with a height of $$\sqrt{x}$$ on $$[0,4]$$.

Problem Statement

Calculate the volume of the solid where each face at x is a square with a height of $$\sqrt{x}$$ on $$[0,4]$$.

$$V=8$$

Problem Statement

Calculate the volume of the solid where each face at x is a square with a height of $$\sqrt{x}$$ on $$[0,4]$$.

Solution

### 2269 video

video by MIP4U

$$V=8$$

Calculate the volume of the solid whose base is bounded by $$y=6-2x^2/3$$ and $$y=0$$ with cross-section parallel to the x-axis is a triangle whose height and base are equal.

Problem Statement

Calculate the volume of the solid whose base is bounded by $$y=6-2x^2/3$$ and $$y=0$$ with cross-section parallel to the x-axis is a triangle whose height and base are equal.

$$V=54$$ cubic units

Problem Statement

Calculate the volume of the solid whose base is bounded by $$y=6-2x^2/3$$ and $$y=0$$ with cross-section parallel to the x-axis is a triangle whose height and base are equal.

Solution

### 2270 video

video by MIP4U

$$V=54$$ cubic units

Volume of Rotation

When calculating the volume of rotation, there are 3 factors that determine how to set up the integral.
1. method (washer-disc or cylinder-shell)
2. axis of rotation
3. function (graph and form of the equations)

On this page, the axis of rotation will always be either an axis or a straight line that is parallel to one of the axes. However, before we discuss the rotation of an area, we need to know how to describe an area in the plane, which is explained in this first section.

Describing A Region In The xy-Plane

To describe an area in the xy-plane, the first step is to plot the boundaries and determine the actual region that needs to be described. There are several graphing utilities listed on the tools page. Our preference is to use the free program winplot (used to plot these graphs; we used gimp to add labels and other graphics). However, graphing by hand is usually the best and quickest way. We use the graph to the right to facilitate this discussion. A common way to describe this area is the area bounded by $$f(x)$$ (red line), $$g(x)$$ (blue line) and $$x=a$$ (black line).
[Remember that an equation like $$x=a$$ can be interpreted two ways, either the point x whose value is a or the vertical line. You should be able to tell what is meant by the context.]

Okay, so we plotted the boundaries and shaded the area to be described. Now, we need to choose a direction to start, either vertically or horizontally. We will show both ways, starting with vertically, since it is more natural and what you are probably used to seeing. Also, this area is easier to describe vertically than horizontally (you will see why as you read on). Vertically

Our first step is to draw a vertical arrow on the graph somewhere within the shaded area, like we have done here. Some books draw an example rectangle with the top on the upper graph and the bottom on the lower graph. That is the same idea as we have done with the arrow.

Now we need to think of this arrow as starting at the left boundary and sweeping across to the right boundary of the area. This sweeping action is important since it will sweep out the area. As we think about this sweeping, we need to think about where the arrow enters and leaves the shaded area. Let's look our example graph to demonstrate. Think about the arrow sweeping left to right. Notice that it always enters the area by crossing $$g(x)$$, no matter where we draw it. Similarly, the arrow always exits the area by crossing $$f(x)$$, no matter where we draw it. Do you see that?

But wait, how far to the right does it go? We are not given that information. What we need to do is find the x-value where the functions $$f(x)$$ and $$g(x)$$ intersect. You should be able to do that. We will call that point $$(b,f(b))$$. Also, we will call the left boundary $$x=a$$. So now we have everything we need to describe this area. We give the final results below.

Vertical Arrow

$$g(x) \leq y \leq f(x)$$

arrow leaves through $$f(x)$$ and enters through $$g(x)$$

$$a \leq x \leq b$$

arrow sweeps from left ($$x=a$$) to right ($$x=b$$) Horizontally

We can also describe this area horizontally (or using a horizontal arrow). We will assume that we can write the equations of $$f(x)$$ and $$g(x)$$ in terms of $$y$$. ( This is not always possible, in which case we cannot describe the area in this way. ) For the sake of this discussion, we will call the corresponding equations $$f(x) \to F(y)$$ and $$g(x) \to G(y)$$.

Let's look at the graph. Notice we have drawn a horizontal arrow. Just like we did with the vertical arrow, we need to determine where the arrow enters and leaves the shaded area. In this case, the arrow sweeps from the bottom up. As it sweeps, we can see that it always crosses the vertical line $$x=a$$. However, there is something strange going on at the point $$(b,f(b))$$. Notice that when the arrow is below $$f(b)$$, the arrow exits through $$g(x)$$ but when the arrow is above $$f(b)$$, the arrow exits through $$f(x)$$. This is a problem. To overcome this, we need to break the area into two parts at $$f(b)$$.

Lower Section - - This section is described by the arrow leaving through $$g(x)$$. So the arrow sweeps from $$g(a)$$ to $$g(b)$$.
Upper Section - - This section is described by the arrow leaving through $$f(x)$$. The arrow sweeps from $$f(b)$$ to $$f(a)$$.
The total area is the combination of these two areas. The results are summarized below.

Horizontal Arrow

lower section

$$a \leq x \leq G(y)$$

arrow leaves through $$G(y)$$ and enters through $$x=a$$

$$g(a) \leq y \leq g(b)$$

arrow sweeps from bottom ($$y=g(a)$$) to top ($$y=g(b)$$)

upper section

$$a \leq x \leq F(y)$$

arrow leaves through $$F(y)$$ and enters through $$x=a$$

$$f(b) \leq y \leq f(a)$$

arrow sweeps from bottom ($$y=f(b)$$) to top ($$y=f(a)$$)

Type 1 and Type 2 Regions

Some instructors may describe regions in the plane as either Type 1 or Type 2 (you may see II instead of 2). As you know from the above discussion, some regions are better described vertically or horizontally. Type 1 regions are regions that are better described vertically, while Type 2 regions are better described horizontally. The example above was a Type 1 region.

Here is a quick video clip going into more detail on Type 1 and Type 2 regions.

### Krista King Math - type I and type 2 regions [1min-39secs]

video by Krista King Math

washer-disc method

x-axis rotation y-axis rotation cylinder-shell method

x-axis rotation y-axis rotation Getting Started

Here are some key things that you need to do and know to get started.
1. Draw a rough plot of the area that is being rotated. This is usually best done by hand since you will need to label it.
2. Decide what method you will use, washer-disc or cylinder-shell.
3. On the rough plot from point 1, label the axis of rotation and draw a representative rectangle somewhere in the area.
4. Label R and r or p and h (depending on your method; details on the next two pages).

Once those steps are done, you are ready to set up your integral. We suggest the washer-disc method first, followed by the cylinder-shell method. Once you have successfully completed the practice problems on those two pages, go to the volume practice problems page to hone your skills before your exam.

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