To get the volume of revolution about the x axis of a function say f(x)=1/5(x-12)^3+2.95, I understand you calculate the definite integral of pi (f(x))^2. So if I integrate (pi*(1/5(x-12)^3+2.95)^2 from 7.15 to 12.38 I get 763.41 units cubed (from Wolfram Alpha). NOW If the function is shifted down in the y direction by say 0.1 then the Volume of revolution of the curve about the about the x axis should surely be less but if I again integrate (pi*(1/5(x-12)^3+2.85)^2 from 7.15 to 12.38 = 771.262 this is a bigger volume. Any idea where the mistake is?
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Still don't know why it is bigger on the inside.
- Thread starter tanon
- Start date
Explained the problem a little more this time... wondering if there is something going on like a negative component cancelling out a positive component like when you calculate areas under a curve that go below the x axis but I think this function is always positive and also because it is squared, it shouldn't matter except there is a ^6 term that when integrated is a ^7 so that is a negative. Just can't see why it works like this and what is the best way to get around it. Is it best to work out where it goes negative and then do two separate definite integrations?
You need to consider your actual function, not just how it is shifted. By shifting it down and integrating over that specific interval you are increasing the overall area between f(x) and the x-axis, and so the volume will be greater.
Here's a much simpler example: f(x)=0, and f(x)=-1 over any interval, say [0,1]. Revolving the first around the x-axis does nothing, the volume is zero. Revolving the second around you get a cylinder whose volume is 2pi.
Here's a much simpler example: f(x)=0, and f(x)=-1 over any interval, say [0,1]. Revolving the first around the x-axis does nothing, the volume is zero. Revolving the second around you get a cylinder whose volume is 2pi.
Thanks to anyone who posted or tried to help with this and sorry as well if I wasted your time. My specific problem was that I should be considering 1/50(x-12)^3+2.95 not 1/5(x-12)^3+2.95. 1/5 produces a cross over at the x axis and results in a "negative volume" in the range I am considering (like I was alluding too) but for the life of me never picked up the fact that I missed the 50 and was working on the assumption that the curve was always positive in the range I was looking at hence couldn't see why the volume was less if the curve was shifted down by 0.1.