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Calc2: Volume

  1. Aug 24, 2004 #1
    I have this problem, that I have tried to understand, for quit ea while, and I simply can't see how they get the answer they get. some of my work is on this paper, on this image

    if it doesn't load, the problem is
    y=sqrt(25-x^2), y=3

    by the way, hi, I posted here a while ago, when I was in AP physics. Anyway, I hope to visit this site more this year, and I did pass the AP physics test, so yay! But I am having much trouble in Calc 2, so nay...
     
  2. jcsd
  3. Aug 24, 2004 #2

    mathwonk

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    that is not a problem, that is a statement. what is the problem?
     
  4. Aug 24, 2004 #3
    Are you not getting why the integral of pi*f(x)^2 would give the volume of a revolution, or do you not get the math they used?
     
  5. Aug 24, 2004 #4

    HallsofIvy

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    The only way I can make sense out of this is that you are asking for the volume of the figure created by rotating the area between y= 3 and y= sqrt(25- x2) around the x-axis.

    There are several ways of doing it:

    1. By "washers". Drawing a vertical line at a given x, it crosses the given area in a segment. Rotating that segment about the x axis sweeps out a "washer"- the area between two circles. The formula for such an area is just the difference between the two circles: πr12- πr22 which is the same as π(r12- r22). The "thickness" of each washer is dx so the volume is [itex]\pi\int_{-4}^{4}(r_1^2-r_2^2)dx[/itex].
    (The limits are -4 and 4 because (-4,3) and (4,3) are where the circle and line interesect.)
    In this problem r1 is the constant y= 3 and r2 is sqrt(25- x2dx so the volume is [itex]\pi\int_{-4}^{4}(25-x^2- 9)dx= \pi\int_{-4}^4(16- x^2)dx[/itex].

    2. By "cylinders". Drawing a horizontal line at a given y and rotating around the x-axis gives a cylinder with radius y and length x1- x2. The area of such a cylinder is the circumference of the cylinder times its length 2πy(x1-x2). The "thickness" of each cylinder is dy so the volume is the integral [itex]2\pi\int_3^5y(x_1-x_2)dy[/itex].
    Since y= sqrt(25- x2), x2+ y2=25 and
    x= +/- sqrt(25- y2) so x1- x2= 2sqrt(25- y2. The volume is [itex]4\pi\int_3^5y(\sqr{25-y^2})dy[/itex]. That can be integrated by the substitution u= 25- y2 and (much to my amazement!) gives the same volume as the first.
     
    Last edited: Aug 24, 2004
  6. Aug 24, 2004 #5
    oh, sorry, I forgot to mention what I was doing. But yeah HallsofIvy, you helped me a lot thanks!
     
  7. Aug 24, 2004 #6
    Well, I feel wierd saying this, but I am taking calc 1 over again, at my own choice. I feel I needed to before going into calc 2, and then into calc 3. I think it is called auditing the class at my comunity college, but I'm doing it!
     
  8. Aug 25, 2004 #7
    Yeah, could you explain why the integral of pi*f(x)^2 would give you the volume of the revolution?
     
  9. Aug 25, 2004 #8
    Your are estimating the volume using very thin cylinders. Called the height of the cylinders [itex]\Delta x[/itex], So a piece of the volume is [itex]V=\pi f(x)^2 \Delta x[/tex]. In this case f(x) replaces the radius of the cylinder. Take infinite many cylinders such that [itex]\lim\limits_{\Delta x \to 0} [/itex]. Then your add them all up so you get an integral.
     
  10. Aug 25, 2004 #9

    HallsofIvy

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    Good for you! You will start to grasp calculus the third or fourth time you take it but to really understand calculus, you need to teach it!
     
  11. Aug 26, 2004 #10
    An integral is really just adding up an infinite amount of infinitely small sections. Remember how we derived the integral? We added a bunch of squares that had infinitely small bases.

    It is kind of the same idea. Picture the curve y = x^2 rotated about x axis. Now pretend we have a bunch of cylinders with a height of that is really really small standing on their sides. Then gave them radius that it was always touching the edge of that rotated x^2. If we placed those cylinders next to each other inside the rotated x^2 wouldn’t those cylinders be a good approximation of that volume? (Much like a large amount of infinitely small rectangles gave us the area under a curve). Well how would we get that radius? And if we want a perfect approximation we will need to make the height of the cylinders infinitely small wont we?

    Now the real question is, how we get that radius for the cylinders so that they always touch the edge of our curve. Well isn’t the height of our curve at any given just f(x). And isn’t the distance from the x axis the height? So wouldn’t our radius be f(x).

    Remember the area of a cylinder is: Pi*r^2*h
    Where r = radius
    h = height

    If we want infinitely many infinitely small cylinders it would be [tex]\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \pi*f(x_i^*)^2 * \frac{a-b}{n} [/tex] where a and b is where you start and stop taking the area

    Isn’t that he same as [tex] \int \pi * f(x)^2 * dx [/tex]
     
    Last edited: Aug 26, 2004
  12. Aug 26, 2004 #11
    I see, thanks!
     
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