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Finding volume of area revolved around x axis 
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#1
Jul812, 07:00 PM

P: 106

1. The problem statement, all variables and given/known data
The area bounded by the xaxis and curve y = 4x^2 is rotated about xaxis, find the volume. 2. Relevant equations [integral] pi y^2 dx 3. The attempt at a solution the answer is supposed to be 512/15 pi 


#2
Jul812, 07:18 PM

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That's 1/2 of the given correct answer. Where does y = 4x^{2} intercept the xaxis ? 


#3
Jul812, 08:53 PM

P: 57

[If you know how to find the volume of solid of revolution then you can skip to the equations at the bottom of my post.]
Ok, first go through this website: http://curvebank.calstatela.edu/volrev/volrev.htm As you can see, when you revolve a curve about the xaxis, you get a bunch of circles stacked on top of each other. The radius of each circle is equal to the height from the xaxis to the edge of the circle, which is the yvalue at that point of the curve. Thus, the formula becomes pi * f(x)^2 with an integral having limits which depend on how much of the curve you revolve. If you graph the curve, (I cheated and used the grapher app on mac), you can find and locate the region bounded by the curve and the x axis. You will be able to see that the points where the curve touches the xaxis are none other than the roots of the quadratic equation. To recap: For f(x), use the equation 4x^2 For the limits, solve the quadratic equation for the x intercepts, and use those as the limits. Finally, solve the integral with the limits as the x intercepts, and the function as 4x^2. Your final equation should be: [itex]\int _{ intercept1 }^{ intercept2 }{ \pi (4{ x }^{ 2 })\quad dx } [/itex] [ 


#4
Jul912, 08:14 AM

P: 106

Finding volume of area revolved around x axis
I could have answered that easily. Zeros are 2 and 2.



#5
Jul912, 08:15 AM

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You appear to have integrated from 0 to 2. But nothing is said about the yaxis being a boundary.
Also, you leave [itex]\pi[/itex] out until the last step. That's very confusing. 


#6
Jul912, 08:20 AM

P: 106

Understood. I did integrate with 2 only. So I should be integrating from 2 to 2. I quickly did that calculation and came up with 48.93333 using the 2. That can't be right.
so you're saying that 2 and 2 shouldn't be used since the y axis is not a boundary? EDIT: I got it, integrating 2 answer 17.067 


#7
Jul912, 11:22 AM

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