Solids of Revolution


by Mr. Snookums
Tags: revolution, solids
Mr. Snookums
Mr. Snookums is offline
#1
Apr22-06, 05:45 PM
P: 20
Find the volume formed by rotating the area contained by y=sqrt(6x+4), the y-axis and the line y=2x about the y-axis. 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/2-sqrt[(y^2-4)/6]. Is this right?

I then found the height, which would be sqrt(y/2-sqrt[(y^2-4)/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?
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nocturnal
nocturnal is offline
#2
Apr23-06, 05:26 AM
P: 113
Do you have to use the washer method? I think it would be easier to use cylindrical shells.

Quote Quote by Mr. Snookums
Find the volume formed by rotating the area contained by y=sqrt(6x+4), the y-axis and the line y=2x about the y-axis. 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/2-sqrt[(y^2-4)/6]. Is this right?
Wrong. First of all, if your solving [itex] y = \sqrt{6x + 4}[/itex] for x, the answer is not [itex] x = \sqrt{\frac{y^2 -4}{6}}[/itex], (where did the square root come from?)

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)

I then found the height, which would be sqrt(y/2-sqrt[(y^2-4)/6], wouldn't it?
Where do you get "height" from. The formula for the volume of a washer is
[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 x-axis.
Go to example 5 under heading "washers" washer example
your book should have a better example (hopefully)
HallsofIvy
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#3
Apr23-06, 08:45 AM
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Quote Quote by Mr. Snookums
Find the volume formed by rotating the area contained by y=sqrt(6x+4), the y-axis and the line y=2x about the y-axis. 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/2-sqrt[(y^2-4)/6]. Is this right?
No, it's not. The graph of y= sqrt(6x+4) does not cross the x-axis until y=sqrt(6(0)+ 4)= 2. From y= 0 to y= 2, the radius will be just x= y/2. From y=2 to y= 4, the [area] is given by [itex]\pi ((y/2)^2- (y^2-4)^2/6^2) (no squareroot as nocturnal said).

I then found the height, which would be sqrt(y/2-sqrt[(y^2-4)/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?
The height? Did you look at the units on your formula? "Radius squared" would have units of distance^2, "times the height time dy" since "height" (I'm not sure which height you mean here) and dy both have units of distance, you "area" would have units of distance^4 !

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^2-4)^2}{36}dy[/tex]


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