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### Volume of a Paraboloid Section

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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

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,
Lawrence
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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
http://mathforum.org/dr.math/
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Associated Topics:
College Calculus

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