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Thread: multiple integrals

  1. #1
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    multiple integrals

    hello,

    I would like to find the volume in the first octant bounded by the
    parabloid z= 1 - x^2 - y^2 and the plane x+y=1

    the limits of the integral w.r.t. y are y = 0 to y=1
    ... .... .... w.r.t. x are x= 1-y to x= (1-y^2)^(1/2)
    ... ... .... w.r.t z are z=0 to z= 1 - x^2 - y^2

    I can do the differentiation w.r.t z and x but it gets to difficult for me to do it w.rt. y.

    can someone help me? much appreciated!
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  2. #2
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    Quote Originally Posted by pepsi View Post
    hello,

    I would like to find the volume in the first octant bounded by the
    parabloid z= 1 - x^2 - y^2 and the plane x+y=1

    the limits of the integral w.r.t. y are y = 0 to y=1
    ... .... .... w.r.t. x are x= 1-y to x= (1-y^2)^(1/2)
    ... ... .... w.r.t z are z=0 to z= 1 - x^2 - y^2

    I can do the differentiation w.r.t z and x but it gets to difficult for me to do it w.rt. y.

    can someone help me? much appreciated!
    $\displaystyle z = 1 - {x^2} - {y^2},{\text{ }}x + y = 1,{\text{ }}x \geqslant 0,{\text{ }}y \geqslant 0.$

    $\displaystyle D = \left\{ {\left( {x,y} \right):{\text{ }}1 - x \leqslant y \leqslant \sqrt {1 - {x^2}} ,{\text{ }}0 \leqslant x \leqslant 1} \right\}.$

    $\displaystyle V = \iint\limits_D {\int\limits_0^{1 - {x^2} - {y^2}} {dzdydx} } = \int\limits_0^1 {\int\limits_{1 - x}^{\sqrt {1 - {x^2}} } {\left( {1 - {x^2} - {y^2}} \right)dydx} } =$

    $\displaystyle = \int\limits_0^1 {\left. {\left( {y - {x^2}y - \frac{{{y^3}}}
    {3}} \right)} \right|_{y = 1 - x}^{y = \sqrt {1 - {x^2}} }dx} = \int\limits_0^1 {\left( {\frac{2}
    {3}\sqrt {{{\left( {1 - {x^2}} \right)}^3}} - \frac{4}
    {3}{x^3} + 2{x^2} - \frac{2}
    {3}} \right)dx} =$

    $\displaystyle = \frac{2}
    {3}\int\limits_0^1 {\sqrt {{{\left( {1 - {x^2}} \right)}^3}} dx} - \left. {\left( {\frac{{{x^4}}}
    {3} - \frac{{2{x^3}}}
    {3} + \frac{{2x}}
    {3}} \right)} \right|_0^1 = \frac{2}
    {3}\int\limits_0^1 {\sqrt {{{\left( {1 - {x^2}} \right)}^3}} dx} - \frac{1}
    {3} =$

    $\displaystyle = \left\{ \begin{gathered}
    x = \sin t, \hfill \\
    dx = \cos t\,dt, \hfill \\
    0 \leqslant t \leqslant \pi /2 \hfill \\
    \end{gathered} \right\} = \frac{2}
    {3}\int\limits_0^{\pi /2} {{{\cos }^4}t\,dt} - \frac{1}
    {3} = \frac{2}
    {3}\int\limits_0^{\pi /2} {{{\left( {\frac{{1 + \cos 2t}}
    {2}} \right)}^2}dt} - \frac{1}
    {3} =$

    $\displaystyle = \frac{1}
    {6}\int\limits_0^{\pi /2} {\left( {1 + 2\cos 2t + {{\cos }^2}2t} \right)dt} - \frac{1}
    {3} = \frac{1}
    {6}\int\limits_0^{\pi /2} {\left( {1 + 2\cos 2t + \frac{{1 + \cos 4t}}
    {2}} \right)dt} - \frac{1}
    {3} =$

    $\displaystyle = \frac{1}
    {{12}}\int\limits_0^{\pi /2} {\left( {3 + 4\cos 2t + \cos 4t} \right)dt} - \frac{1}
    {3} = \frac{1}
    {{12}}\left. {\left( {3t + 2\sin 2t + \frac{1}
    {4}\sin 4t} \right)} \right|_0^{\pi /2} - \frac{1}
    {3} =$

    $\displaystyle = \frac{1}
    {{12}} \cdot \frac{{3\pi }}
    {2} - \frac{1}
    {3} = \frac{\pi }
    {8} - \frac{1}
    {3}{\text{ }}\left( {{\text{cubic units}}} \right).$
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