Volume of a Paraboloid SectionDate: 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, Lawrence 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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