5.2. Volumes http://www.ck12.org
In this case, the radius is 3−xand the height isx^3 +^12 x+^14 −^14. Substituting,
V= 2 π∫ 1
0 (^3 −x)(
x^3 +^12 x+^14 −^14)
dx= 2 π∫ 1
0(
−x^4 + 3 x^3 −^12 x^2 +^32 x)
dx= 2 π[− 1
5 x(^5) +^3
4 x
(^4) −^1
6 x
(^3) +^3
4 x
2
] 1
0
= 2 π[− 1
5 +
3
4 −
1
6 +
3
4
]
= 2 π[ 17
15
]
=^3415 π.Multimedia Links
The following applet allows you to try out solids of revolution about the x-axis for any two functions. You can
try inputting the examples above to test it out, and then experiment with new functions and changing the bounds.
Volumes of Revolution Applet. In the following video the narrator walks trough the steps of setting up a volume
integration(14.0)(16.0). Khan Academy Solids of Revolution (10:05).
MEDIA
Click image to the left for use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/571Sometimes the same volume problem can be solved in two different ways(14.0)(16.0). In these two videos, the
narrator first finds a volume using shells Khan Academy Solid of Revolution (Part 5) (9:29)
MEDIA
Click image to the left for use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/572, and then he does the same volume problem using disks. Khan Academy Solid of Revolution (Part 6) (9:19).