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Math Help - Evaluate the double integral Questions

  1. #1
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    Evaluate the double integral Questions

    Question # 01:

    Evaluate the double integral


    Question # 02:

    Prove that

    Question # 03:
    Use double integral to find the volume under the surface given by


    over the region bounded by the curves


    plz help me.


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    Last edited by sajjad002; April 16th 2009 at 11:13 PM. Reason: pictures not display
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  2. #2
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    1. \int_{-1}^1{\int_{x^2}^1{(x^2 + y^2)\,dy}\,dx}

     = \int_{-1}^1{\left[x^2y + \frac{1}{3}y^3\right]_{y=x^2}^{y =1}\,dx}

     = \int_{-1}^1{\left[x^2 + \frac{1}{3}\right] - \left[x^4 + \frac{1}{3}x^6\right]\,dx}

     = \int_{-1}^1{\frac{1}{3} + x^2 - x^4 - \frac{1}{3}x^6\,dx}

     = \left[\frac{1}{3}x + \frac{1}{3}x^3 - \frac{1}{5}x^5 - \frac{1}{21}x^7\right]_{-1}^1

     = \left[\frac{1}{3} + \frac{1}{3} - \frac{1}{5} - \frac{1}{21}\right] - \left[-\frac{1}{3} - \frac{1}{3} + \frac{1}{5} + \frac{1}{21}\right]

     = \frac{4}{3} - \frac{2}{5} - \frac{2}{21}.


    I'll let you do the simplifying.
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  3. #3
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    Quote Originally Posted by sajjad002 View Post
    Question # 02:

    Prove that

    2. is easy.

    Since the limits on the integration are all constants it doesn't make any difference when you reverse the order of the integration.
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  4. #4
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    thanks
    Last edited by Jhevon; April 17th 2009 at 08:20 AM. Reason: removed chat speak
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    So wat abt the third question ... any body could solve that one .. or just give some hints in that.. and in Q2 also!
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  6. #6
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    Quote Originally Posted by Angel Rox View Post
    So wat abt the third question ... any body could solve that one .. or just give some hints in that.. and in Q2 also!
    I already GAVE you the answer to question 2.

    If you're reversing the order of integration, you only need to change the terminals on the integrals when the terminals are functions of a variable.

    Since they're constants, they stay the same.
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