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Math Help - Double Integration

  1. #1
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    Double Integration

    Use double integration to calculate the area that liew between the curves

     \sqrt{x} + \sqrt{y} = \sqrt{a}  and x + y = a, a > 0This is what i have done , between the limits on the y = a-x and y = 0 and for x = a and 0  \iint dy dx \int a- x dx = ax - \frac{{x}^{ 2} }{2 } = {a}^{ 2} - \frac{{a}^{ 2} }{2 } = \frac{{a}^{ 2} }{2 } But according to my text book the answer is \frac{{a}^{ 2} }{3 }
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  2. #2
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    Quote Originally Posted by adam_leeds View Post
    Use double integration to calculate the area that liew between the curves

     <br />
\sqrt{x} + \sqrt{y} = \sqrt{a}<br />

     <br />
and x + y = a, a > 0<br />

    This is what i have done , between the limits on the y = a-x and y = 0 and for x = a and 0

     <br />
\iint dy dx<br /> <br />
\int a- x dx<br /> <br />
= ax - \frac{{x}^{ 2} }{2 } = {a}^{ 2} - \frac{{a}^{ 2} }{2 } = \frac{{a}^{ 2} }{2 } <br />

    But according to my text book the answer is \frac{{a}^{ 2} }{3 } <br />
    Your limits of integration are incorrect.

    If you sketch the region you will see that the line

    y=a-x

    lies above the curves

    y=(\sqrt{a}-\sqrt{x})^2

    and they both have x intercept

    (a,0)

    This gives

    \int_{0}^{a}\int_{(\sqrt{a}-\sqrt{x})^2}^{a-x}1 dydx
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  3. #3
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    Where did you get y = 0? Why not use the first equation?

    We probably should establish, due ot the first constraint, that a >= 0.
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    This gives

    \int_{0}^{a}\int_{(\sqrt{a}-\sqrt{x})^2}^{a-x}1 dydx
    That doesnt get a^2/3
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  5. #5
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    Quote Originally Posted by adam_leeds View Post
    That doesnt get a^2/3
    ummm no.

    (\sqrt{a}-\sqrt{x})^2=a-2\sqrt{ax}+x

    This gives

    \int_{0}^{a} \int_{a-2\sqrt{ax}+x}^{a-x}dydx =\int_{0}^{a}(a-x)-(a-2\sqrt{ax}+x)dx

    \int_{0}^{a} -2x+2\sqrt{a}\sqrt{x}dx=-x^2+\frac{4}{3}\sqrt{a}x^{\frac{3}{2}}\bigg|_{0}^{  a}=\frac{a^2}{3}
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  6. #6
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    Missed a negative, thanks.
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