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Thread: calc 3 question: 0=<x<=1 -x^2=<y<=x^2 for the function f(x,y) dydx

  1. #1

    calc 3 question: 0=<x<=1 -x^2=<y<=x^2 for the function f(x,y) dydx

    i couldnt get it to let me upload a picture so ill do my best to write it out

    I have to rewrite a double integral

    original bounds are 0=<x<=1 -x^2=<y<=x^2 for the function f(x,y) dydx



    what i got when i rewrote it is -1=<y<=1 sqrt(-y)=<x<=sqrt(y) dxdy

    i want to say this is right because i can't think of any other way to do this but i knew that a negative under a square root is an imaginary number so it shouldnt work. Please let me know if im right or wrong!

    the follow up question to this ask what the original double integral represents

  2. #2
    Elite Member
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    Quote Originally Posted by BlitzCannon View Post
    i couldnt get it to let me upload a picture so ill do my best to write it out

    I have to rewrite a double integral

    original bounds are 0=<x<=1 -x^2=<y<=x^2 for the function f(x,y) dydx



    what i got when i rewrote it is -1=<y<=1 sqrt(-y)=<x<=sqrt(y) dxdy

    i want to say this is right because i can't think of any other way to do this but i knew that a negative under a square root is an imaginary number so it shouldnt work. Please let me know if im right or wrong!

    the follow up question to this ask what the original double integral represents
    Your response is not quite correct.

    Did you draw an approximate sketch of the region over which you are integrating?

    If not, do that first.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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