1. 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!

2. Originally Posted by pepsi
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!
$\displaystyle z = 1 - {x^2} - {y^2},{\text{ }}x + y = 1,{\text{ }}x \geqslant 0,{\text{ }}y \geqslant 0.$

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

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

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

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

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

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

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

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