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 February 12th, 2013, 10:31 AM #1 Newbie   Joined: Feb 2013 Posts: 21 Thanks: 0 Formulation of integral I need to formulate the following integral from first principle and evaluate it : volume of an ellipsoid described by equation x^2 + y^2 + (z^2)/4 = 1
 February 12th, 2013, 11:34 AM #2 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Formulation of integral Are you familiar with double integrals ? $2\int \int_{E}\, 2\sqrt{1-(x^2+y^2)}\, dA$ where E is the area to integrate over [a circle of radius 1].
 February 12th, 2013, 12:28 PM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Formulation of integral The first thing you have to do is decide on the order of integration you want to use. If you choose to use, say "$\int\int\int dzdydx$", then note that the smallest and largest possible values of x occur when y= z= 0: $x^2= 1$ so x= -1 and x= 1: $\int_{x=-1}^1 dx$. Now, for each x, the largest and smallest values of y occur when z= 0: $x^2+ y^2= 1$ so $y^2= 1- x^2$ and $y= \pm\sqrt{1- x^2}$: $\int_{x=-1}^1 \int_{y= -\sqrt{1- x^2}}^{\sqrt{1- x^2}} dydx$. Finally, for each x and y, $\frac{z^2}{4}= 1- x^2- y^2$ so $z= \pm 2\sqrt{1- x^2- y^2}$: $\int_{x= -1}^1\int_{y= -\sqrt{1- x^2}}^{\sqrt{1- x^2}}\int_{y= -2\sqrt{1- x^2- y^2}}^{2\sqrt{1- x^2- y^2}} dzdydx$. Of course, that first integral and gives the double integral zaidalyafey gives.

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