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Homework Help: Find the area enclosed by a curve and two lines.

  1. Jun 10, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the total area of the region enclosed by the curve x = y2/3 and the lines x = y and y = -1.

    2. The attempt at a solution

    I graphed x = y2/3 as y=x3/2 and y=-x3/2. Then I graphed the lines y=x and y=-1.

    I shaded the region in question, which appears off the cuff to have an area slightly larger than one. To find the area exactly I broke the problem up into two integration steps.

    1. [tex]\int[/tex] x dx (from x=-1 to x=0) - [tex]\int[/tex] -1 dx (in the same range of x=-1 to x=0)


    2. [tex]\int[/tex] -1 dx (from x=0 to x=1) - [tex]\int[/tex] -x3/2 dx

    I keep getting the whole area as 11/10. This seems close enough, seeing as how the area appears to be slightly greater than 1. However, my textbook's answer glossary tells me that the correct answer is actually 6/5. Can anyone tell me what i'm doing wrong?

    I attached a relevant graph (I did it in mspaint, lol, sorry). y=+x2/3 is left out of the graph because it is irrelevant. The shaded region, i'm assuming, is the area in question:

    Attached Files:

    Last edited: Jun 10, 2010
  2. jcsd
  3. Jun 10, 2010 #2
    EDIT: Let me think about this a little more. I was a little hasty, sorry.
  4. Jun 10, 2010 #3
    I was given the function x = y2/3. This is a function of y but not a function of x. By that I mean every x value has more than 1 y value. In order for me to graph the equation as a function of x, I had to break it into two equations.

    Graphing x = y2/3

    Is the same as graphing f(x)=x2/3 and g(x)=-x3/2 together.

    Does this make sense?
  5. Jun 10, 2010 #4


    Staff: Mentor

    It's much simpler to use horizontal slices with typical area element given by [itex]\Delta A[/itex] = (y2/3 - y)[itex]\Delta y[/itex], -1 <= y <= 1. Since the right and left endpoints of the typical area element don't change, a single integral will suffice.
  6. Jun 10, 2010 #5
    Well, I get the same answer as you, but I don't trust my integration skills. If anyone else can confirm or point out the error, that would be good.
  7. Jun 10, 2010 #6
    Mathematically speaking, your response makes sense, Mark. If i integrate this:

    (y2/3 - y)dy, -1<=y<=1

    I get 6/5.

    According to the book, this is the correct answer, and for that I thank you! However, I'm now a little confused. I guess my new question is, what made you choose that integrand? And why does my choice of integrands not work?
  8. Jun 10, 2010 #7
    Oh, I see now the mistake we both made. There is a tiny sliver of area sandwiched between y = x and the upper half of the graph you left out.
  9. Jun 10, 2010 #8


    Staff: Mentor

    If you are given the bounding functions as x = g(y) and x = h(y), the most natural thing to try first is horizontal slices, because you don't have to convert both to their inverses. It still might be the case that you want to use vertical slices if the integration with horizontal slices turns out to be too much work or too difficult.

    I noticed that your graph was missing that bit of area in the first quadrant, but didn't mention it. Its area just happens to be 1/10, which is why you were coming up with 11/10 instead of 12/10 = 6/5.
  10. Jun 10, 2010 #9
    I left out the curve in the first quadrant purposely because it didn't effect my shaded region in the 3rd/4th quadrants. I guess it is safe to assume that the area I've shaded is incorrect, right?

    I can do the problem pretty easily now on paper but I'm confusing myself a little by trying to graph everything.
  11. Jun 10, 2010 #10


    Staff: Mentor

    Yes, the region in your graph was incorrect because it didn't include the part in the first quadrant. Your graph needs to be an accurate depiction of the region as described in the problem statement.

    It's almost impossible to "do the problem on paper" unless you have a fairly accurate graph of the region you are going to integrate. These aren't separate, unrelated operations.
  12. Jun 11, 2010 #11
    I was saying that I can calculate the area based on the integrand you provided, but was having trouble deducing that integrand myself. Anyways, thanks for your help.
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