# Volume of Revolution

(Difference between revisions)
 Revision as of 10:54, 10 July 2009 (edit)← Previous diff Revision as of 10:58, 10 July 2009 (edit) (undo)Next diff → Line 4: Line 4: |ImageIntro=This image is a solid of revolution |ImageIntro=This image is a solid of revolution |ImageDescElem=This image shows the solid formed after revolving the region bounded by $y=x^2$, $y=0$,$x=0$ and $x=1$, about the $x$-axis |ImageDescElem=This image shows the solid formed after revolving the region bounded by $y=x^2$, $y=0$,$x=0$ and $x=1$, about the $x$-axis - |ImageDesc=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 slice, each of which is ${\Delta x }$ units thick , then summing up the volumes of all the slices.[[Image:bread.gif|left|thumb| The disc method is much like slicing up bread and computing the volume of each slice http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html]] + |ImageDesc=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.[[Image:bread.gif|left|thumb| The disc method is much like slicing up bread and computing the volume of each slice http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html]] 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'''.
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'''.
Line 46: Line 46: volume of solid= ${\pi\over 5} units^3$ 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 funtion can be revolved about any fixed axis. Also, given a different function, to find the volume of revolution, we can substitute it in the place of $x^2$ when evaluating the intergral. 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 funtion 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 $x^2$ when evaluating the intergral. 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. ==References== ==References==

## Revision as of 10:58, 10 July 2009

Solid of revolution
This image is a solid of revolution

# Basic Description

This image shows the solid formed after revolving the region bounded by $y=x^2$, $y=0$,$x=0$ and $x=1$, about the $x$-axis

# A More Mathematical Explanation

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

When finding the volume of revolution of solids, in many cases the problem is not with the calculus, [...]

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

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 main image on the right hand side of the page
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 funtion 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 $x^2$ when evaluating the intergral. 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.