
#1
Apr2206, 05:45 PM

P: 20

Find the volume formed by rotating the area contained by y=sqrt(6x+4), the yaxis and the line y=2x about the yaxis. Set up, but do not evaluate the integral.
First I graphed it, then did the "washer" method of finding the area of the circle formed, and found that the radius is (y/2sqrt[(y^24)/6]. Is this right? I then found the height, which would be sqrt(y/2sqrt[(y^24)/6], wouldn't it? Then the answer is: the integral from 0 to 4 of pi times the radius squared, times the height times dy. I don't have an answer key, but could someone help me with this? 



#2
Apr2306, 05:26 AM

P: 113

Do you have to use the washer method? I think it would be easier to use cylindrical shells.
Secondly, there is no one "radius." The volume for a washer is given by, [tex]V = \pi[(\mbox{outer radius})^2  (\mbox{inner radius})^2)] * (\mbox{thickness}) [/tex] You want to find functions for the "outer radius" and the "inner radius." (Be careful, these might not be the same over your whole interval of integration) [tex]V = \pi[(\mbox{outer radius})^2  (\mbox{inner radius})^2)] * (\mbox{thickness}) [/tex] I think it would help if you went over some examples from your text book. Here is an example of the washer method when rotated about the xaxis. Go to example 5 under heading "washers" washer example your book should have a better example (hopefully) 



#3
Apr2306, 08:45 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

The volume you are looking for is [tex]\pi\int_0^2\frac{y^2}{4}dy+ \pi\int_2^4(\frac{y^2}{4}\frac{(y^24)^2}{36}dy[/tex] 


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