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Is my answer correct?

  1. Jan 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Evaluate the double integral 4x^3 dA, where R is the region bounded by the graphs of y=(x-1)^2 and y= -x+3.

    2. Relevant equations
    None.

    3. The attempt at a solution
    The answer in the book is 72/5. But I got 45 as the answer.
    Here's my work:
    (x-1)^2= -x+3
    x^2-2x+1= -x+3
    x=2, -1
    y=1, 4.
    double integral of 4x^3 dx dy from -1 to 2 for dx and from 1 to 4 for dy
    integral of x^4 (evaluate from -1 to 2) dy from 1 to 4 for dy
    =45
    Is my answer correct or the book's answer is right?
     
  2. jcsd
  3. Jan 30, 2015 #2

    Mark44

    Staff: Mentor

    No. The y values run from the parabola up to the line. They do not run from 1 to 4.
    For sure your answer is incorrect.
     
  4. Jan 30, 2015 #3
    So how do I find the limits of integration for y?
     
  5. Jan 30, 2015 #4

    Mark44

    Staff: Mentor

    You get the limits from these equations. Did you draw a picture of this region?

    Also, I got 72/5 as my answer, same as your book.
     
  6. Jan 30, 2015 #5
    Since it's a double integral, the x is from -1 to 2 for dx. What about y?
     
  7. Jan 30, 2015 #6

    O_o

    User Avatar

  8. Jan 31, 2015 #7
    I still don't get it. What are the limits of integration for y?
     
  9. Jan 31, 2015 #8

    O_o

    User Avatar

    When x is between -1 and 2 one of the functions is lower than the other. When you plot them can you see which it is? That is the lower bound. And the upper bound is the function that is higher.

    The upper and lower bound varies with x. That's why you can't just use constants...you need to write the bounds in terms of functions. Try watching these 2 videos:

    https://www.khanacademy.org/math/mu...tegrals/double_integrals/v/double-integrals-5

    https://www.khanacademy.org/math/mu...tegrals/double_integrals/v/double-integrals-6
     
    Last edited: Jan 31, 2015
  10. Jan 31, 2015 #9

    Mark44

    Staff: Mentor

    I asked before, but you didn't respond. Did you sketch a graph of the region over which you're integrating?
     
  11. Jan 31, 2015 #10
    You know what? I got it.
     
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