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Finding volume using integrals.

  1. May 22, 2013 #1
    Huhh First of all I'm sorry if this is the wrong question ( didn't know if this was considered pre cal. I have a gut feeling it is :P)

    1. The problem statement, all variables and given/known data

    Let R be the region bounded by y=tan x, y=0 and x = Pi/4
    Find the volume of the solid whose base is region R and whose cross-section is a square perpendicular to the x-axis

    2. Relevant equations

    Huh well I don't know whether its cylindrical shell:
    ab Pi(R)(Height)dx
    or disk method
    ab A(x) dx


    3. The attempt at a solution
    I can't even figure out what the graphs look like. The wording is really confusing. I tried using a pyramid shape with a square base because it says the cross section is a square and using similar triangle ( I think its called like that...) but I couldn't get the equation for the formula.
     
  2. jcsd
  3. May 22, 2013 #2

    Mark44

    Staff: Mentor

    No, this is calculus, not precalculus.
    First off, you need to figure out what region R looks like. Surely you have studied the trig functions before taking this class.
    The base isn't square. The vertical cross-sections are square.

    BTW, why do you start some of your sentences with "Huh" or "Huhh"?
     
  4. May 22, 2013 #3
    I don't know, force of habit when speaking out load transferred to my typing skills... thankfully not in essays. Plus English isn't my first or second language. And yeah i did study trig function I know what it looks like. I assumed the base was squared because all the other problems I did, whenever I took the cross section from the shape, it was always the same shape as the base ( so lets say a sphere, was a circle, a square-based pyramid was a square, etc...)

    Now for the graph It goes from 0 to pi/4 which is a curve on the first quadrant. However with my assumption from before, If i rotate that shape around the x-axis, I can only take a circle as the cross section as my shape looks something very similar to a cone, no?
     
  5. May 22, 2013 #4

    Mark44

    Staff: Mentor

    There is NO rotation going on!! The base of the shape is that graph of y = tan(x), 0 ≤ x ≤ ##\pi/4##.

    To draw the solid, you need to be able to draw something in three dimensions.
    1. Pick a point x in the interval.
    2. Draw a line from the x-axis straight over to the curve. (I'm assuming that your graph is in the x-y plane in 3D.
    3. Now draw a square whose vertical height is the same as its base. The square should be perpendicular to the x-axis.

    Pick another point in the interval, and repeat the steps above.
    Draw enough squares so that you have a good idea what the solid looks like.
     
  6. May 22, 2013 #5
    Ah! I see it now!! so my limits of integration would be from 0 to pi/4. and my Area would be tan^2(x)?
     
  7. May 22, 2013 #6

    Mark44

    Staff: Mentor

    If you mean this --
    $$ \int_0^{\pi/4} \tan^2(x)~dx$$

    then, yes.
     
  8. May 22, 2013 #7
    Thank you so much! and thanks for not giving me the answer. I like understanding what I'm doing! And that is what I meant!
     
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