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Riemann Sums and Integrals, feel lost without actual functions

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data

    At my old university, Calculus was taught much differently than it is where I am now. My old school focused on numerical things, which this school focuses much more on pictures, abstract, etc. and it's very difficult for me.

    At my old school, we were given a shape, bounded by functions, and told to build a Riemann sum for area and make it an integral.

    Now, I'm given shapes that aren't described at all by functions. They are just shapes.

    Right now they are slicing them into rectangular partitions, but it's hard for me to find a way to describe the other part of the rectangle (that isn't delta x, h, w, whatever.)

    2. Relevant equations



    3. The attempt at a solution

    Construct a Riemann Sum and integral for a circle of radius 3 using horizontal partitions.

    So, the height of the partition is delta h, and the width of the partition is completely unknown.

    My assumption is that I find it by not looking at the partition, but at the actual SHAPE created when h is at a set point, IE, finding a way to express w in terms of h.

    But how on a circle? w should get smaller the further I am from the center of the circle in either direction, so w grows, to a max of 3, and then shrinks to 0 again. How do I make this into actual math?
     
  2. jcsd
  3. Oct 1, 2011 #2

    vela

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    The only thing you're really missing is the equation for a circle. Can you figure out what that is? For simplicity, center it at the origin. A circle of radius r is all points a distance r from the origin, right? How do you write "distance from the origin" mathematically?
     
  4. Oct 1, 2011 #3
    Equation for a circle? No, I don't know what that is/means.
     
  5. Oct 1, 2011 #4

    vela

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    What's the distance of the point (x,y) from the origin?
     
  6. Oct 1, 2011 #5
    I have no idea what you are talking about. I would just say.. 3....

    The only thing I can figure out this way is if I make it a SEMI circle, create a triangle with a hypotenuse 3, which always has a height of 3-h, so then the width is just sqrt(9-(3-h)^2).

    But I have no idea how to make it work for a full circle. I need some type of periodic function apparently.
     
  7. Oct 1, 2011 #6
    Wait, if I set the center of circle as 0 height and start integration lower limit at -3, would that get me what I want?
     
  8. Oct 1, 2011 #7

    vela

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    I have no idea what that's supposed to mean.
     
  9. Oct 1, 2011 #8
    3.

    Message too short.
     
  10. Oct 1, 2011 #9

    vela

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    Really? All points in the (x,y) plane are a distance 3 from the origin?
     
  11. Oct 1, 2011 #10
    Sorry, I thought we were talking about x,y points on the edge of the circle. Why would we want to know the distance for any x/y point from the circle origin?
     
  12. Oct 1, 2011 #11

    vela

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    So you can deduce what the equation of a circle is.
     
  13. Oct 1, 2011 #12
    You only need to do a semi circle times 2. You came amazingly close to the equation here, I'm jealous of people with this much insight :P
    a circle is [itex] r^2=x^2+y^2 [/itex]

    So solve for x, since we're doing it sideways, and only use the positive side, because thats happier, and multiply by 2, and assume it's centered at the origin, so integrate from -1.5 to 15... I mean, not integrate, just, whatever they're making you do. Is it clearer now though?
     
  14. Oct 1, 2011 #13
    I would revert to physics mode and treat x and y as vector components and find the magnitude of the vector, sqrt(a^2 + b^2)
     
  15. Oct 1, 2011 #14
    no vectors.
     
  16. Oct 1, 2011 #15
    My response was to vela, we posted at the same time. Why is it wrong to treat any distance (line) between the origin and an x,y point as a vector with an x component and y component? Then the magnitude of the vector is just the hypotenuse of the triangle formed by x and y, how is that not distance from origin?

    Also, to your response, thanks. I really wanted to do it for a circle, not a semi-circle and then multiply by two. I knew it would have been correct, but at the same time it felt like it wasn't what they wanted.

    So, my integral should look like:

    2[itex]\int^{3}_{0} \sqrt{9 - (3-h)^{2}}dh[/itex]
     
  17. Oct 1, 2011 #16

    vela

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    Hmm, I get the impression you kind of learn how to solve certain problems by simply applying a recipe, without really understanding why the procedure works.

    The magnitude of a vector, the distance formula, the Pythagorean theorem...they're all essentially the same idea.

    So the distance of (x,y) from the origin is given by [itex]\sqrt{x^2+y^2}[/itex]. If you want only points a distance r from the origin, i.e. those on the circle, they satisfy the equation [itex]\sqrt{x^2+y^2} = r[/itex]. Do you see why? Now if you square that equation, you get what Arcana told you, [itex]x^2+y^2=r^2[/itex].
     
  18. Oct 1, 2011 #17
    If it's wrong, I don't know how, it's just not the way they're trying to make you think about it. Of course, relating other things you know to try and solve a problem is actually something to be commended, not quashed. Just make sure you know what they're saying too.

    Well if you add negative area plus positive the same area you get zero, so you kind of have no choice but to do positive times two, or move the circle off the origin, which just complicates the equations.

    well I think it should actually be [itex] \sqrt{9-y^2}dy [/itex]

    The bounds should be -3 to 3, I was wrong when I said 1.5, my circle was too small. You only found a quarter of a circle.
     
  19. Oct 1, 2011 #18
    I hate that they're making you do this circle sideways. Makes it much harder to deal with >_<
     
  20. Oct 1, 2011 #19
    Thanks!

    No, they are exactly the same idea.

    Yes, I used this 3 posts ago, when finding the integral of half of the circle.

    My problem was expressing the height of the triangle at any point h over the ENTIRE circle. 3-h is perfect for half of the circle, but not for the whole thing. I still don't really know how to do this, but if treating it as a semi-circle and multiplying by 2 really is the best solution, then the problem is solved.
     
  21. Oct 1, 2011 #20

    vela

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    It's not wrong, per se. It's just that introducing the concept of vectors at this point is completely unnecessary, so bringing them up tends to evoke the reaction "What? Why?"
     
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