# Volume of Revolution

(Difference between revisions)
 Revision as of 16:42, 14 June 2011 (edit)← Previous diff Current revision (16:31, 21 June 2011) (edit) (undo) (14 intermediate revisions not shown.) Line 17: Line 17: - If we revolve this area about the x axis ($y=0$), then we get the main image on the right hand side of the page[[Image:Revolution.gif|Right|thumb|This image shows a plane area being revolved to create a solid http://curvebank.calstatela.edu/volrev/volrev.htm]] + If we revolve this area about the x axis ($y=0$), then we get the image below to the left. [[Image:Revolution.gif|Right|thumb|This image shows a plane area being revolved to create a solid http://curvebank.calstatela.edu/volrev/volrev.htm]] To find the volume of the solid using the disc method: To find the volume of the solid using the disc method: Line 31: Line 31: Volume of all discs = ${\sum}{\pi}y^2{\Delta x}$, with $X$ ranging from 0 to 1 Volume of all discs = ${\sum}{\pi}y^2{\Delta x}$, with $X$ ranging from 0 to 1 - If we make the slices infinitesmally thick, the Riemann sum becomes the same as: + If we make the slices infinitesimally thick, the Riemann sum becomes the same as: $\int_0^1 {\pi}y^2\,dx ={\pi}\int_0^1 (x^2)^2\, dx$ $\int_0^1 {\pi}y^2\,dx ={\pi}\int_0^1 (x^2)^2\, dx$ Line 37: Line 37:

- Evaluating this intergral, + Evaluating this integral, Line 51: Line 51: In the example we discussed, the area is revolved about the $x$-axis. This does not always have to be the case. A function can be revolved about any fixed axis. Also, given a different function, to find the volume of revolution about the $x$-axis, we can substitute it in the place of $x^2$. Note: we would also need to change the bounds as per the given information. The method discussed in the example works for all functions that have bounds and are revolved about the $x$-axis. In the example we discussed, the area is revolved about the $x$-axis. This does not always have to be the case. A function can be revolved about any fixed axis. Also, given a different function, to find the volume of revolution about the $x$-axis, we can substitute it in the place of $x^2$. Note: we would also need to change the bounds as per the given information. The method discussed in the example works for all functions that have bounds and are revolved about the $x$-axis. + + ==Washer Method== + + [[Image:cross_section.jpg|Cross Section|right]] + The washer method can be used when the rotated plane does not touch the axis around which it is being rotated. One instance in which the plane isn't touching the rotational axis is when the plane is not just bounded by one function, but instead two. For now we'll assume that one function is consistently smaller than the other, so there is a 'smaller function' and a 'larger function.' The main image on this page is an example of when the washer method is used. The top curve (which we will call f(x) ) is proportional to the square root of x, and the bottom curve (which we will call g(x) )is linear. The boundaries for the functions are x =2 and x = 10. A cross section is shown to the right. + + The basic philosophy behind the washer method is the same as behind the disk method. We still must integrate around the rotational axis. The difference is that we cannot just use one radius (ie Radius = R - r). This wouldn't work because then two sections with the same area would necessarily have to have the same volume, but this is not the case. If two circles of the same radius are rotated around the same axis, if one is farther away, it will create more volume, as demonstrated in the animation below. Thus, instead of subtracting the radius of the smaller function from the bigger function, we subtract the volume the rotated smaller function would create from the volume the bigger function creates. The formula for the washer method is:
+ $V = \pi \times \int(f(x)^2 - g(x)^2)dx$ + + Rotating_circles.gif + + If one function is not consistently smaller than the other, we can break up the problem into two smaller problems. If the functions f(x) and g(x) cross at some arbitrary value c, we use f(x) as the larger function from our start value to c, but as the smaller function from c to the end value. If our start and end values are a and b respectively, the formula is: + $V = \pi \times \Big( \int_a^c (f(x)^2 - g(x)^2)dx + \int_c^b (g(x)^2 - f(x)^2)dx \Big)$ ==References== ==References== Bread image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html
Revolving image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html Revolving image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html - + |AuthorName=Nordhr - |AuthorName=Lizah Masis + |AuthorDesc=made in OpenGL - |Field=Algebra + |SiteName=Nordhr - |InProgress=No + |SiteURL=http://wikis.swarthmore.edu/miwiki/index.php/User:Nordhr - volume of solid= [itex]{\pi\over 5} units^3 + |Field=Calculus - + |Pre-K=No |Pre-K=No |Elementary=No |Elementary=No Line 66: Line 78: |HighSchool=Yes |HighSchool=Yes |HigherEd=Yes |HigherEd=Yes - |other=Pre-calculus and elementary calculus - |AuthorName=Lizah Masis - |SiteURL=http://wikis.swarthmore.edu/miwiki/index.php/User:Lmasis1 - |Field=Calculus - |References= - - |AuthorName=Nordhr - |AuthorDesc=made with OpenGL - |Field=Calculus - |InProgress=No - } - |Field=Algebra - |InProgress=No }} }}

## Current revision

Solid of revolution
Field: Calculus
Image Created By: Nordhr
Website: Nordhr

Solid of revolution

This image is a solid of revolution

# Basic Description

When finding the volume of revolution of solids, in many cases the problem is not with the calculus, but with actually visualizing the solid. To find the volume of a solid like a cylinder, usually we use the formula ${\pi} {r^2} h$. Alternatively we can imagine chopping up the cylinder into thin cylindrical plates, much like slicing up bread, computing the volume of each thin slice, then summing up the volumes of all the slices.
The disc method is much like slicing up bread and computing the volume of each slice http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html

# A More Mathematical Explanation

## Disk Method

In general, given a function, we can graph it then revolve the area under the curve b [...]

## Disk Method

In general, given a function, we can graph it then revolve the area under the curve between two specific coordinates about a fixed axis to obtain a solid called the solid of revolution. The volume of the solid can then be computed using the disc method.

Note: There are other ways of computing the volumes of complicated solids other than the disc method.

In the disc method, we imagine chopping up the solid into thin cylindrical plates calculating the volume of each plate, then summing up the volumes of all plates.

For example, let's consider a region bounded by $y=x^2$, $y=0$,$x=0$ and $x=1$

<-------Plotting the graph of this area,

If we revolve this area about the x axis ($y=0$), then we get the image below to the left.
This image shows a plane area being revolved to create a solid http://curvebank.calstatela.edu/volrev/volrev.htm

To find the volume of the solid using the disc method:

Volume of one disc = ${\pi} y^2{\Delta x}$ where $y$- which is the function- is the radius of the circular cross-section and $\Delta x$ is the thickness of each disc. Using the analogy of the bread, computing the volume of one disc would correspond to computing the volume of one slice of bread. With this in mind, the area of one disc would correspond to the area of a slice of bread, while the thickness of a disc would correspond to the thickness of a slice of bread. To find the total volume of the bread, we would have to sum up the volumes of each of the slices.

Volume of all discs:

Volume of all discs = ${\sum}{\pi}y^2{\Delta x}$, with $X$ ranging from 0 to 1

If we make the slices infinitesimally thick, the Riemann sum becomes the same as:

$\int_0^1 {\pi}y^2\,dx ={\pi}\int_0^1 (x^2)^2\, dx$

Evaluating this integral,

${\pi}\int_0^1 x^4 dx$

=$[{{x^5\over 5} + C|}_0^1] {\pi}$

=$[{1\over 5} + {0\over 5}] {\pi}$

=${\pi}\over 5$

volume of solid= ${\pi\over 5} units^3$

In the example we discussed, the area is revolved about the $x$-axis. This does not always have to be the case. A function can be revolved about any fixed axis. Also, given a different function, to find the volume of revolution about the $x$-axis, we can substitute it in the place of $x^2$. Note: we would also need to change the bounds as per the given information. The method discussed in the example works for all functions that have bounds and are revolved about the $x$-axis.

## Washer Method

The washer method can be used when the rotated plane does not touch the axis around which it is being rotated. One instance in which the plane isn't touching the rotational axis is when the plane is not just bounded by one function, but instead two. For now we'll assume that one function is consistently smaller than the other, so there is a 'smaller function' and a 'larger function.' The main image on this page is an example of when the washer method is used. The top curve (which we will call f(x) ) is proportional to the square root of x, and the bottom curve (which we will call g(x) )is linear. The boundaries for the functions are x =2 and x = 10. A cross section is shown to the right.

The basic philosophy behind the washer method is the same as behind the disk method. We still must integrate around the rotational axis. The difference is that we cannot just use one radius (ie Radius = R - r). This wouldn't work because then two sections with the same area would necessarily have to have the same volume, but this is not the case. If two circles of the same radius are rotated around the same axis, if one is farther away, it will create more volume, as demonstrated in the animation below. Thus, instead of subtracting the radius of the smaller function from the bigger function, we subtract the volume the rotated smaller function would create from the volume the bigger function creates. The formula for the washer method is:
$V = \pi \times \int(f(x)^2 - g(x)^2)dx$

Click to stop animation.

If one function is not consistently smaller than the other, we can break up the problem into two smaller problems. If the functions f(x) and g(x) cross at some arbitrary value c, we use f(x) as the larger function from our start value to c, but as the smaller function from c to the end value. If our start and end values are a and b respectively, the formula is: $V = \pi \times \Big( \int_a^c (f(x)^2 - g(x)^2)dx + \int_c^b (g(x)^2 - f(x)^2)dx \Big)$