## 17Calculus Integrals - Volumes

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This page gets you started on calculating volumes of the first type, volumes with known cross-section. 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.

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

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

Solution

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.

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

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

Solution

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

Practice

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

$$512/15$$

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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

$$V=2$$

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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

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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

$$V=2\pi$$

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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

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

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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

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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

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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

$$V=8$$

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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

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You CAN Ace Calculus

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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 Volume With A Known Cross-Section Practice

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