This page gets you started on calculating volumes of the first type, volumes with known crosssection. 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 timeconsuming part. So focus on where you can get the most points and make sure you are confident setting up the integrals. 
Volume With A Known CrossSection
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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 crosssectional 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 crosssection parallel to the top of the box. The crosssection is the same no matter where we take it. In equation form, the area of the crosssection is \(A=lw\).
The volume of a solid with crosssection \(A(x)\) 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 crosssection will be different depending on where we take the crosssection. That's why we write the area as \(A(x)\). Of course, if we take the crosssection perpendicular to the yaxis, 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 crosssection 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.
Final Answer 

\(V=\pi r^2 h\) 
Okay, so far the crosssection has been constant making the equation of the area quite simple. Let's look at an example where the crosssection 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, wellwritten solution with 3dim plots, see this Pauls Online Notes page.
For this problem, it helps to think of the pyramid as upsidedown with it's point sitting at the origin and the base up in the air. Since the crosssection is a square, we know the area of the crosssection is \(A=s^2\), where \(s\) is the length of side. However, \(s\) is different depending on where the crosssection 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 \(L^2h/3\).
Final Answer 

\(V=L^2h/3\) 
That example is just one way to work the problem. Here are several videos showing other ways to solve it.
video by Michel vanBiezen 

video by MIP4U 

video by David Lippman 

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.
video by ProfRobBob 

Practice
The base of a solid is bounded by the curves \(f(x)=22x\) and \(g(x)=x^21\) and the cross sections perpendicular to the xaxis are perfect squares. Find the volume of the solid.
Problem Statement 

The base of a solid is bounded by the curves \(f(x)=22x\) and \(g(x)=x^21\) and the cross sections perpendicular to the xaxis 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 ydirection (perpendicular to the xaxis) and square it to get an equation for the area.
Problem Statement 

The base of a solid is bounded by the curves \(f(x)=22x\) and \(g(x)=x^21\) and the cross sections perpendicular to the xaxis are perfect squares. Find the volume of the solid.
Final Answer 

\(512/15\)
Problem Statement 

The base of a solid is bounded by the curves \(f(x)=22x\) and \(g(x)=x^21\) and the cross sections perpendicular to the xaxis 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 ydirection (perpendicular to the xaxis) and square it to get an equation for the area.
Solution 

video by PatrickJMT 

Final Answer 

\(512/15\) 
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Calculate the volume of the solid whose crosssection 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 crosssection is a rectangle with a height of \(y=\cos(x)\) and width of 2 over the interval \(0\leq x \leq \pi/2\).
Final Answer 

\(V=2\)
Problem Statement 

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

video by MIP4U 

Final Answer 

\(V=2\) 
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Calculate the volume of the solid where the crosssection 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 crosssection 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 xaxis 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.
Donâ€™t forget to give units in your answer.
Problem Statement 

Calculate the volume of the solid where the crosssection is a triangle with a base of 4 inches and height of \(y=x^2\) inches on the interval \([0,2]\).
Final Answer 

\(16/3\) in^{3}
Problem Statement 

Calculate the volume of the solid where the crosssection 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 xaxis 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.
Donâ€™t forget to give units in your answer.
Solution 

video by MIP4U 

Final Answer 

\(16/3\) in^{3} 
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Calculate the volume of the solid where the crosssection perpendicular to the xaxis is a circle with diameter \(y=\sqrt{x}\) on the interval \([0,4]\).
Problem Statement 

Calculate the volume of the solid where the crosssection perpendicular to the xaxis is a circle with diameter \(y=\sqrt{x}\) on the interval \([0,4]\).
Final Answer 

\(V=2\pi\)
Problem Statement 

Calculate the volume of the solid where the crosssection perpendicular to the xaxis is a circle with diameter \(y=\sqrt{x}\) on the interval \([0,4]\).
Solution 

video by MIP4U 

Final Answer 

\(V=2\pi\) 
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Calculate the volume of a solid with base bounded by the curves \(y=x^2\) and \(y=8x^2\) and crosssections parallel to the yaxis that are semicircles.
Problem Statement 

Calculate the volume of a solid with base bounded by the curves \(y=x^2\) and \(y=8x^2\) and crosssections parallel to the yaxis that are semicircles.
Hint 

The diameter of the semicircle is the distance between the curves.
Problem Statement 

Calculate the volume of a solid with base bounded by the curves \(y=x^2\) and \(y=8x^2\) and crosssections parallel to the yaxis that are semicircles.
Final Answer 

\(\displaystyle{ V = \frac{256\pi}{15} }\)
Problem Statement 

Calculate the volume of a solid with base bounded by the curves \(y=x^2\) and \(y=8x^2\) and crosssections parallel to the yaxis that are semicircles.
Hint 

The diameter of the semicircle 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.
\(\displaystyle{ V = \int_{2}^{2}{ \frac{\pi}{2} (4x^2)^2 ~dx } }\) 
\(\displaystyle{ V = \frac{\pi}{2} \int_{2}^2{ 16+x^48x^2 ~ dx } }\) 
\(\displaystyle{ V = \frac{\pi}{2} \left[ 16x + \frac{x^5}{5}  \frac{8x^3}{3} \right]_{2}^2 }\) 
\(\displaystyle{ V = \frac{\pi}{2} \left[ 16(2)+ \frac{2^5}{5}  \frac{8}{3}(2^3) \right]  }\) \(\displaystyle{ \frac{\pi}{2} \left[ 16(2) + \frac{(2)^5}{5} \frac{8}{3}(2)^3 \right] }\) 
\(\displaystyle{ V = \frac{\pi}{2} \left[ 32+\frac{32}{5}\frac{64}{3} \right]  }\) \(\displaystyle{ \frac{\pi}{2} \left[ 32  \frac{32}{5} + \frac{64}{3} \right] }\) 
\(\displaystyle{ V = \frac{\pi}{2} \left[ 64+\frac{64}{5} \frac{128}{3} \right] }\) 
\(\displaystyle{ V = \frac{\pi}{2} \left[ \frac{960}{15} + \frac{192}{15}  \frac{640}{15} \right] }\) 
\(\displaystyle{ V = \frac{\pi}{2}\left[ \frac{512}{15} \right] }\) 
\(\displaystyle{ V = \frac{256\pi}{15} }\) 
video by David Lippman 

Final Answer 

\(\displaystyle{ V = \frac{256\pi}{15} }\) 
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Calculate the volume of a solid with an equilateral triangular crosssection parallel to the yaxis bounded by \(y=\pm (2x^2/8)\) and \(x=2\). Units are in feet.
Problem Statement 

Calculate the volume of a solid with an equilateral triangular crosssection parallel to the yaxis bounded by \(y=\pm (2x^2/8)\) 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 crosssection parallel to the yaxis bounded by \(y=\pm (2x^2/8)\) and \(x=2\). Units are in feet.
Final Answer 

\(\displaystyle{\frac{153\sqrt{3}}{10}}\) ft^{3}
Problem Statement 

Calculate the volume of a solid with an equilateral triangular crosssection parallel to the yaxis bounded by \(y=\pm (2x^2/8)\) 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.
Final Answer 

\(\displaystyle{\frac{153\sqrt{3}}{10}}\) ft^{3} 
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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.
Final Answer 

\(128/15\) ft^{3}
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 

video by MIP4U 

Final Answer 

\(128/15\) ft^{3} 
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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 \([0,4]\).
Final Answer 

\(V=8\)
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 

video by MIP4U 

Final Answer 

\(V=8\) 
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Calculate the volume of the solid whose base is bounded by \(y=62x^2/3\) and \(y=0\) with crosssection parallel to the xaxis is a triangle whose height and base are equal.
Problem Statement 

Calculate the volume of the solid whose base is bounded by \(y=62x^2/3\) and \(y=0\) with crosssection parallel to the xaxis is a triangle whose height and base are equal.
Final Answer 

\(V=54\) cubic units
Problem Statement 

Calculate the volume of the solid whose base is bounded by \(y=62x^2/3\) and \(y=0\) with crosssection parallel to the xaxis is a triangle whose height and base are equal.
Solution 

video by MIP4U 

Final Answer 

\(V=54\) cubic units 
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You CAN Ace Calculus
related topics on other pages 

external links you may find helpful 
Larson Calculus (Google books) Volume  Disk, Washer Methods and Shell, Cylinder Methods pgs 458477 
Interactive Mathematics  Volume of Solid of Revolution by Integration (Disk method) 
Pauls Online Notes  Volumes of Solids of Revolution / Method of Rings 
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  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle 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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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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