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Homework Help: Finding Area by an Iterated Integral

  1. Oct 15, 2014 #1
    1. The problem statement, all variables and given/known data
    Use an iterated integral to find the area of the region bounded by the graphs of:
    [itex] f(x) = sin(x) [/itex]
    [itex] g(x) = cos(x) [/itex]
    [itex] x = \frac{\pi}{4} and x = \frac{5\pi}{4} [/itex]
    Find two solutions, one using a vertical representative rectangle and another using a horizontal representative rectangle.

    2. Relevant equations
    Definite Integral Equation

    3. The attempt at a solution
    I had no issue finding the solution using the vertical representative rectangle, here is my solution:
    [itex] A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \int_{cos(x)}^{sin(x)} dy dx [/itex]
    [itex] A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (sin(x) - cos(x)) dx [/itex]
    [itex] A = 2\sqrt{2} [/itex]
    The problem I am having is trying to solve this problem using a horizontal representative rectangle. I understand that the solution will be the same, but I can't even seem to set up the problem properly. Would I have to define the two fractions:
    [itex] \frac{\pi}{4} and \frac{5\pi}{4} [/itex]
    as functions of y and then make the trigonometric functions constant and then integrate?
    Last edited: Oct 15, 2014
  2. jcsd
  3. Oct 15, 2014 #2
    Another idea I had was to make the two trigonometric functions functions of 'y' instead of 'x', but I do not know if that is the right approach...
  4. Oct 15, 2014 #3


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    Re problem statement: Do you mean between ##x = \frac{\pi}{4}## and ## {\bf x} = \frac{5\pi}{4}## ? I really can't see y = 5/4 ##\pi## as a sensible bound...

    Re horizontal rectangles: If I make a drawing (you should do so too!) and give it a quarter turn, it looks as if I need to split up in three pieces: one from y = 1 to ##\cos{\pi \over 4}##, one from there to ##\cos{5\pi \over 4}## and one from there to -1. bounds on x are then ## \arcsin y## and ##\pi -\arcsin y## for the first, etc.
  5. Oct 15, 2014 #4
    wow I'm sorry, I messed up entering the equations, it should be
    [itex] x = \frac{5\pi}{4} [/itex]
  6. Oct 15, 2014 #5
    ok, so the integral would end up looking something like:
    [itex] A = \int_{\cos\frac{pi}{4}}^{\cos\frac{5\pi}{4}} \int_{arcsin(y)}^{{\pi}-arcsin(y)} dx dy [/itex]
    or would I need to make 3 seperate integrals?
  7. Oct 15, 2014 #6


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    You need three integrals but you can just do the top half with two integrals and double the answer using symmetry. But remember that using a horizontal "##dy##" element your integral must have the form$$
    \int_{y_{smallest}}^{y_{largest}}~x_{right}- x_{left}~dy$$And you can read the ##y## limits from your graph.

    I notice that you are giving your answers as double integrals, which has the same effect once you work out the inner integral.
  8. Oct 15, 2014 #7
    yes, a double integral is just a definite integral with an integral in the integrand, no?
  9. Oct 15, 2014 #8
    Thank you, I've done enough calc for one night but ill give this a shot tomorrow and post my solution
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