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I need help finding the area of a function using integrals and logs

  1. Jan 16, 2005 #1
    Here is the problem:

    Find the volume of the solid formed by rotating the region enclosed by
    y=e^(3x)+1,y=0,x=0,x=0.7 about the y-axis.

    What I know is that I am supposed to solve for x since it rotates on the y axis. So I have


    Now what? I am not sure how I am supposed to solve this or what I do now. I am starting Calc II so this is new to me and i've never used logs before for this. What do I do, and how does it work for me to solve this?
  2. jcsd
  3. Jan 16, 2005 #2


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    Do you know the general formulae for finding the area of volume generated by rotating around an axis?
  4. Jan 16, 2005 #3
    My understanding is this, it is probably a basic understanding.

    Find out the axis it rotates on, then solve for the opposing variable. If it rotates on X-axis, solve for Y, if it rotates on Y solve for X.

    If there are two functions find out which function forms the outside of the rotation and which is the inside. The outside function comes first in the subtraction of the two functions.

    Square both functions. If you have y=x^2, y=x and it rotates on the x-axis then you get (x^2)^2-(x)^2=x^4-x^2.

    Integrate the function(s) to get F(x), and do an F(b)-F(a) calculation.

    I have an idea how to do it with simple functions but I have never done it with logs before, so I am lost.
  5. Jan 16, 2005 #4


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    It is simpler than that, I don't know how to explain this very well, I think of it as a matter of an infinite number of circles but although this is my intuitive understanding it is not mathematically rigorous.

    But if you have the function y=f(x) and you rotate it [itex]2\pi ^c[/itex] around the x-axis assuming the function is continuous and integrable from x=a to x=b the general formulae holds:

    [tex]V_x = \int_{x=a}^{x=b} 2 \pi y^2 \text{d}x[/tex]

    Conversely it holds true when rotating around the y-axis, assuming the function is invertible bound within the region x=a, x=b, y=f(a), y=f(b) that the general formulae is:

    [tex]V_y = \int_{y=f(a)}^{y=f(b)} 2 \pi x^2 \text{d}y[/tex]

    So assuming I remember my maths right (big assumption probably best to get it checked lol) all you have to do is solve:

    [tex]\int_2^{e^{2.1}+1} 2 \pi \left( \frac{ \ln{(y-1)}}{3} \right)^2 \text{d}y[/tex]
    Last edited: Jan 16, 2005
  6. Jan 16, 2005 #5


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    Yes,Zurtex,it's alright.Let's tell him that the integration needs a simple substitution and 2 times partial integration.

  7. Jan 17, 2005 #6
    I tried it and I got 11.2031 and that didn't work. I entered it as nInt(2pi(ln(y-1)/3)^2,y,2,e^(2.1)+1) and the equation I got on my calculator looks right but its not working. I also tried it on the interval 0,0.7 and It said 'non real result' which is understandable because when you graph ln(x-1)/3 you get a graph that doesn't start until about 1 on the x-axis.
    Last edited: Jan 17, 2005
  8. Jan 17, 2005 #7


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    Lets look at the integration without bounds for a moment:

    [tex]I = \int 2 \pi \left( \frac{ \ln{(y-1)}}{3} \right)^2 \text{d}y[/tex]

    Let t=y-1, dt/dy = 1

    [tex]I = \frac{2 \pi}{9} \int \ln (t) \ln (t) \text{d}t[/tex]

    Then using by parts (and assuming you know how to integrate ln(t) with respect to t)

    u= ln(t)

    v'= ln(t)
    v = t ln(t) - t

    [tex]\frac{9I}{2\pi} = t \ln(t) \left( \ln(t) - 1 \right) - \int \ln(t) - 1 dt[/tex]

    [tex]\frac{9I}{2\pi} = t \ln(t) \left( \ln(t) - 1 \right) - t \ln(t) + 2t[/tex]

    Please note, I've ignored the constant of integration, I can't be bothered to change the variable back so I am just going to adjust the bounds accordingly:

    [tex]\frac{9I}{2\pi} = \left[t \ln(t) \left( \ln(t) - 1 \right) - t \ln(t) + 2t \right]_1^{e^{2.1}}[/tex]

    [tex]I = \frac{2 \pi}{9} \left[\left(2.1*e^{2.1}*(2.1 - 1) - 2.1*e^{2.1} + 2*e^{2.1} \right) - (0 - 0 + 2)\right][/tex]

    [tex]I = \frac{2 \pi}{9} \left[2.21*e^{2.1} - 2 \right]\approx 11.2030838174462[/tex]

    I've done a lot of that in my head so that doesn't show every step, I'd highly recommend you try it yourself to see every step and make sure I haven't made a mistake.
    Last edited: Jan 17, 2005
  9. Jan 17, 2005 #8
    11.2030838174462 is essentially the same answer as the 11.2031 the calculator gave me. I think there is some other way to do this problem but I do not know what it is.
  10. Jan 17, 2005 #9


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    I do not understand what you mean, this is the answer and it does not change if you do the problem in a diffrent way.
  11. Jan 17, 2005 #10


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    I believe this is the easiest way to get to the result...The initial formula is the simplest and the integration is not really that complicated.After all,it could have been a double integral there to start you off... :wink:

  12. Jan 18, 2005 #11
    We have a system where you turn in your homework online, it gives you the question and you put your answer in and click submit (you get limitless tries). The answer I got from plugging in the equation you made in post 3 was 11.2031, which is more or less identical to the 11.20308 you got in post 6. However neither answer worked, so there must be some other way to find the area to get a different answer.
  13. Jan 18, 2005 #12


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    I get 8.5086804 by using 'vertical slabs'.
    Can you check if it's correct?

    EDIT: BTW; The more conventional way of integration integration gives the same answer.
    Last edited: Jan 18, 2005
  14. Jan 18, 2005 #13
    Yeah that was correct. Can you give me a step by step explanation of how you got to that answer? With 'regular' functions I can integrate on axis's of rotation but I do not know how to do things like square the logs.
  15. Jan 19, 2005 #14


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    There are two approaches to this problem.
    In any case, draw a graph of the function to get a picture. It's easy to work with.

    Instead of integrating directly and splitting the integral into 2 pieces, notice that if we take the area of the rectangle with vertices (0,0),(7/10,0),(0,exp(2.1)) and (7/10, exp(2.1)) and subtract the area of part we don't need we get the area under the curve form x=0 to x=0.7.
    In the same way, if we rotate the rectangle around the y-axis we get a cylinder. To get the right volume we must subtract the volume we get from the part where y goes from 2 to exp(2.1); it looks something like a bowl.
    So: Volume we need = volume of cylinder - volume of bowl.

    The volume of the cylinder is clearly: [itex]\pi(0.7)^2\exp(2.1)[/itex].
    The volume of the bowl is:

    [tex]V_{bowl}=\int_2^{\exp(2.1)}\pi \left(\frac{\ln(y-1)}{3}\right)^2dy \approx 1.32\pi[/tex]
    (integrating this is a little tedious. You need to do integration by parts)

    So the volume V is about [itex]\pi(0.7)^2*\exp(2.1)-1.32\pi=8.4239[/tex].
    (It differs somewhat from my previous answer, so I must have made some stupid mistake somewhere).

    (II). I like this method better, but it's hard to describe without a drawing. Perhaps you're familiar with it though.
    Instead of approximating the volume by cutting the area into horizontal rectangles (with width [itex]\Delta y[/itex]) before rotating each rectangle about the y-axis, we divide the area into vertical rectangles (with width [itex]\Delta x[/itex]) and then rotate them about the y-axis.
    Then each rotated rectangle will be a hollow cylinder with radius x, height f(x) and thickness [itex]\Delta x[/itex]. So its volume is: [itex]2\pi x f(x)\Delta x[/itex].
    Integrating this expression to get the total volume we get:

    [tex]V=2\pi\int_0^{0.7}x(1+\exp(3x))dx \approx 8.5086[/tex]

    You need integration by parts for [itex]\int x\exp(3x)dx[/itex] as well.

    Try working out both ways.

    EDIT: Corrected 'partial integration' to 'integration by parts'. I`m mixing these up too often, since in Dutch we refer to integration by parts as partial integration. :yuck:
    Last edited: Jan 19, 2005
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