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Math Help - [SOLVED] Volume of a graph

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    [SOLVED] Volume of a graph

    I've ran to this problem again, finding the volume of a graph that revolves around some axis. Whether it's sing cross sections, washer's method, or cylindrical shell's I do know know the difference and when to use them

    one question I ran by in my homework says to use cylindrical shells, but how?

    y = 1 + sin(x) and it revolves about y = -1

    I forgot to mention that there was a hint in the book saying that the integral of sine squared dx is x/2 - (sin2x)/4 + C
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  2. #2
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    Were you given endpoints?
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  3. #3
    Member >_<SHY_GUY>_<'s Avatar
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    from (0,1) to (pi,1)

    thanks
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  4. #4
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    That still doesn't define a region though. When doing a revolution problem like this, you should have a closed off area.
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  5. #5
    Member >_<SHY_GUY>_<'s Avatar
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    do you want me to post the picture of the graph instead?
    because i dont think Im being clear, i apologize
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    Is it a line connecting (0,1) and (pi,1)?
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  7. #7
    Member >_<SHY_GUY>_<'s Avatar
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    the graph of y = 1 + sin(x) rotates about the line y = -1

    and the grpah of sine (in terms of x) goes from 0 to pi

    i hope that makes sense
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  8. #8
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    the graph of y = 1 + sin(x) rotates about the line y = -1

    and the grpah of sine (in terms of x) goes from 0 to pi

    i hope that makes sense
    Does the region extend all the way down to y=-1?

    Or does it only extend down to the line y=1?

    That is what I am trying to understand. What is the "bottom" border of the region. It's bounded above by 1+sin(x), but it's not clear what it's bounded below by.
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  9. #9
    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by drumist View Post
    Does the region extend all the way down to y=-1?

    Or does it only extend down to the line y=1?

    That is what I am trying to understand. What is the "bottom" border of the region. It's bounded above by 1+sin(x), but it's not clear what it's bounded below by.
    the line y = -1 is just there as an axis of rotation...ill try attaching a picture of it
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  10. #10
    Member >_<SHY_GUY>_<'s Avatar
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    here it is, sorry for the double post
    Attached Thumbnails Attached Thumbnails [SOLVED] Volume of a graph-untitled.jpg  
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  11. #11
    Junior Member eddie2042's Avatar
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    here it is, sorry for the double post
    Ok here's when to use the methods.

    1) Use disks when the function is defined in terms of a variable who's line of rotation is PARALLEL to the axis of that variable.

    So for example; say you have a function that is defined in terms of x (i.e. y = 1 + sin (x)) and you're revolving it around a line (i.e. y = -1). notice, that the line y = -1 is parrelel to the x-axis. In this case, you use the method of disks which is defined as:

    V = \pi\int_a^b(f(x))^2dx

    where f(x) represents the RADIUS of the cross-sectional circle of the rotation. Sometimes, it may not be just f(x). In this case, we're revolving it around y = -1, so the radius isn't going to be f(x), it's going to be f(x) +1 (think about it).

    Another example. Say you have a function defined in terms of y (i.e. x = y^2 - 3y) and you want to rotate it around x = 4. This line is parallel to the y-axis (which is the variable the function is in terms of).

    2) The method of washers is simply an extension of the method of disk. It's when you have two functions and you're trying to rotate the volume enclosed/underneath them.:

    V = \pi\int_a^bf(x)^2 - g(x)^2dx

    over the interval in which the volume produced by f(x) is greater than the volume produced by g(x).

    DO NOT make the mistake of writing the integrand as
    (f(x) - g(x))^2 << this is simply wrong (try figuring out why).

    3) The method of cylindrical shells is used when the line you're rotating about is PERPENDICULAR to the axis of the variable the function is defined as. So, if you got a function, y = x^2, and you want to rotate it about y = 0 (the y axis), you can use the method of cylindrical shells:

    V = 2\pi\int_a^bx(f(x))dx

    Where the x represents the radius of the cylinder. Once again, it may not always be x on its own. It is if you're rotating it about the central axis, but if you're rotating, let's say, y = sqrt(x), from [4,9] about the line y = -1, then the radius will be (x+1) and that's what you put in place of the original x. The f(x) represents the height of the cylinder. You may have to change it if you're revolving the area enclosed by two graphs, so you'll have to put i.e. f(x) - g(x) in place of the original f(x)


    OK SO... to answer your original question.. what do we for y = 1 + sin (x)? Once again, use the method of disks.

    V = \pi\int_0^\pi(1 + sin(x) (+1))^2dx << the +1 is there because the radius isn't just f(x) (remember?)

    V = \pi\int_0^\pi(2 + sin(x))^2dx

    V = \pi\int_0^\pi sin^2(x) + 4sin(x) + 4dx
    here's where your hint comes in handy, though i will write it slightly differently..

    V = \pi\left[-\frac{sin(2x) - 2x}{4} - 4cos(x) + 4x\right]_0^\pi

    V = \pi\left[(\frac{\pi}{2} + 4 + 4\pi)-(-4)\right]

    V = \frac{9\pi^2 + 16\pi}{2}\approx69.54596103

    I could be wrong...
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  12. #12
    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by eddie2042 View Post
    Ok here's when to use the methods.

    1) Use disks when the function is defined in terms of a variable who's line of rotation is PARALLEL to the axis of that variable.

    So for example; say you have a function that is defined in terms of x (i.e. y = 1 + sin (x)) and you're revolving it around a line (i.e. y = -1). notice, that the line y = -1 is parrelel to the x-axis. In this case, you use the method of disks which is defined as:

    V = \pi\int_a^b(f(x))^2dx

    where f(x) represents the RADIUS of the cross-sectional circle of the rotation. Sometimes, it may not be just f(x). In this case, we're revolving it around y = -1, so the radius isn't going to be f(x), it's going to be f(x) +1 (think about it).

    Another example. Say you have a function defined in terms of y (i.e. x = y^2 - 3y) and you want to rotate it around x = 4. This line is parallel to the y-axis (which is the variable the function is in terms of).

    2) The method of washers is simply an extension of the method of disk. It's when you have two functions and you're trying to rotate the volume enclosed/underneath them.:

    V = \pi\int_a^bf(x)^2 - g(x)^2dx

    over the interval in which the volume produced by f(x) is greater than the volume produced by g(x).

    DO NOT make the mistake of writing the integrand as
    (f(x) - g(x))^2 << this is simply wrong (try figuring out why).

    3) The method of cylindrical shells is used when the line you're rotating about is PERPENDICULAR to the axis of the variable the function is defined as. So, if you got a function, y = x^2, and you want to rotate it about y = 0 (the y axis), you can use the method of cylindrical shells:

    V = 2\pi\int_a^bx(f(x))dx

    Where the x represents the radius of the cylinder. Once again, it may not always be x on its own. It is if you're rotating it about the central axis, but if you're rotating, let's say, y = sqrt(x), from [4,9] about the line y = -1, then the radius will be (x+1) and that's what you put in place of the original x. The f(x) represents the height of the cylinder. You may have to change it if you're revolving the area enclosed by two graphs, so you'll have to put i.e. f(x) - g(x) in place of the original f(x)


    OK SO... to answer your original question.. what do we for y = 1 + sin (x)? Once again, use the method of disks.

    V = \pi\int_0^\pi(1 + sin(x) (+1))^2dx << the +1 is there because the radius isn't just f(x) (remember?)

    V = \pi\int_0^\pi(2 + sin(x))^2dx

    V = \pi\int_0^\pi sin^2(x) + 4sin(x) + 4dx
    here's where your hint comes in handy, though i will write it slightly differently..

    V = \pi\left[-\frac{sin(2x) - 2x}{4} - 4cos(x) + 4x\right]_0^\pi

    V = \pi\left[(\frac{\pi}{2} + 4 + 4\pi)-(-4)\right]

    V = \frac{9\pi^2 + 16\pi}{2}

    I could be wrong...
    you have no idea how much i appreciate your help Sir. I really appreciate that you took that much time to write all of that...thank you so much, i cannot thank you enough
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  13. #13
    Junior Member eddie2042's Avatar
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    you have no idea how much i appreciate your help Sir. I really appreciate that you took that much time to write all of that...thank you so much, i cannot thank you enough
    well at least tell me you understand it LOL.. do you have the answers at the back of the book? ..ooh plzzz tell me it's right
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  14. #14
    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by eddie2042 View Post
    well at least tell me you understand it LOL.. do you have the answers at the back of the book? ..ooh plzzz tell me it's right
    i do, i actually understand it...i quickly reviewed your definition with other practice problems nd i get it a bit better

    it gave me the approximation, and yours was right
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  15. #15
    Junior Member eddie2042's Avatar
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    i do, i actually understand it...i quickly reviewed your definition with other practice problems nd i get it a bit better

    it gave me the approximation, and yours was right
    woo-hoo!!! glad i can help someone out =D
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