Can someone help me find the area enclosed by a given curve?

Discussion in 'Math' started by ElecNerd, May 1, 2011.

  1. ElecNerd

    Thread Starter New Member

    Jan 2, 2011
    Hi There,

    I need to find the area enclosed by given curves, the x axis and the given ordinates

    Y = sin3x
    X = 0
    X = ∏/2

    Thank you so much to those that take the time to explain this to me.
  2. Georacer


    Nov 25, 2009
    It's as simple as calculating the integral
     \int_0^{\frac{\pi}{2}} \sin (3 \cdot x) dx

    In general, when you want to find the area under a curve f(x) from the x-coordinate x1 to the x-coordinate x2, the answer is
     \int_{x_1}^{x_2} f(x) dx
  3. ElecNerd

    Thread Starter New Member

    Jan 2, 2011
    Hi thanks so much for your help. This is the answer that I got originally however did not think that it could be that simple and that I was meant to go further with it to actually get the area?
  4. DerStrom8

    Well-Known Member

    Feb 20, 2011
    If you are just finding the area between f(x) and the x-axis, it is this simple. What is more complicated is when you need to find the volume of a rotating solid between two or three functions. In that case, you would need to use one of several methods. The three most common ones I have heard of can all work, but which is easiest depends on the problem. The methods are called the "disc" method, in which you essentially "slice" the solid to be calculated into discs, and then add them all up, the "shell" method, in which you "unwrap" the solid (like an onion, as my calculus teacher used to say) and add up the layers, or the "hole" method, in which you take the integral of the entire thing to find the volume, then find the volume of the hole, and subtract that from the whole. All of them require taking the integral in order to find the volume. I don't blame you if you find this confusing right now. You will probably learn it soon, though.
    However, if you are just trying to find the AREA between the function and the x-axis, a simple integral is needed ( ∫ [from x1 to x2] f(x) dx ).
    Der Strom
  5. ian_gregg

    New Member

    Jan 19, 2011
    So the derivative of sin is cos, the derivative of cos is -sin. Integration is the inverse of differentiation. There is the factor of 3 to make a small adjustment. Check your answer by differentiating, evaluate at pi/2 and zero, find the difference, make sure you get the sign right.
  6. jegues

    Well-Known Member

    Sep 13, 2010
    It's almost as easy if you understand how to use double/triple integrals! But of course, you need a function with more than one variable! ;)