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Volume of a Paraboloid Section

Date: 05/19/2000 at 09:10:07
From: Lawrence Chiu Shun Man
Subject: Triple integral

Hello Dr. Math,

I'm having difficulties with the following problem; I can't find the 
limits and boundary. Could you please help me to solve it?

The problem is:

     triple integral(z  dx dy dz)

where omega is bounded from below by z = x^2 + y^2, and bounded from 
above by z = 10 - 2x - y.

Thanks very much,

Date: 05/19/2000 at 12:02:52
From: Doctor Rob
Subject: Re: Triple integral

Thanks for writing to Ask Dr. Math, Lawrence.

The elliptic paraboloid z = x^2 + y^2 is what you get when you rotate 
the parabola z = x^2 around the z-axis. The equation z = 10 - 2*x - y 
is a plane, which cuts off a section of the paraboloid. It is over 
that region which you have to integrate. The bounds on z are given to 
you. To find the bounds on x and y, you eliminate z from these two 
equations, x^2 + y^2 = 10 - 2*x - y, and figure out what this equation 
represents in the xy-plane.

     x^2 + 2*x + y^2 + y = 10,
     (x+1)^2 + (y+1/2)^2 = 45/4,

which is a circle with center (-1,-1/2), and radius 3*sqrt(5)/2. 
That means that x ranges from:

     -1-3*sqrt(5)/2 to -1+3*sqrt(5)/2

and for fixed x, y ranges over:

     -1/2 - sqrt(45/4-(x+1)^2) to -1/2 + sqrt(45/4-(x+1)^2)

Thus you have to integrate

        -1+3*sqrt(5)/2   -1/2+sqrt(45/4-(x+1)^2)   10-2*x-y
     INT              INT                       INT        z dz dy dx.
        -1-3*sqrt(5)/2   -1/2-sqrt(45/4-(x+1)^2)   x^2+y^2

Carrying out this integration I leave to you. HINT: You might try to 
substitute x = u*3*sqrt(5)/2 - 1 and y = v*3*sqrt(5)/2 - 1/2 after 
finishing the integration with respect to z.

- Doctor Rob, The Math Forum
Associated Topics:
College Calculus

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