Volume of Revolution

(Difference between revisions)
 Revision as of 15:42, 14 June 2011 (edit)← Previous diff Revision as of 15:58, 14 June 2011 (edit) (undo)Next diff → 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 76: Line 76: |AuthorDesc=made with OpenGL |AuthorDesc=made with OpenGL |Field=Calculus |Field=Calculus + |InProgress=No + } + |Field=Algebra |InProgress=No |InProgress=No } }

Revision as of 15:58, 14 June 2011

Solid of revolution
Field: Algebra
Image Created By: Nordhr
Website: [1]

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

Note: understanding of this explanation requires: *Pre-calculus and elementary calculus

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 infinitesmally 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 intergral,

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

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

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

=${\pi}\over 5$

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.