Triple Integral Help: Solving Equations and Finding Volume with Closed Curve C

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SUMMARY

The discussion focuses on solving the equations of two surfaces, z=f(x,y)=x²+y² and z=g(x,y)=20-(x-4)²-(y+2)², which intersect along a closed curve C that projects onto the xy-plane as a circle. The center of this circle is located at (2, -1) with a radius of √5. To find the volume of the region bounded by these surfaces, the discussion recommends using polar coordinates to set up the double integral, with the integrand being the difference g(x,y) - f(x,y). The integral is expressed as ∫∫(g(x,y)-f(x,y))r dr dθ.

PREREQUISITES
  • Understanding of triple integrals and volume calculations
  • Familiarity with polar coordinates in integration
  • Knowledge of surface equations and their intersections
  • Ability to manipulate equations of circles in the xy-plane
NEXT STEPS
  • Learn about polar coordinate transformations in integration
  • Study the method of setting up double integrals for volume calculations
  • Explore the properties of surface intersections in three-dimensional space
  • Review the derivation of equations for circles in the Cartesian plane
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Mathematicians, engineering students, and anyone involved in multivariable calculus or geometric volume calculations will benefit from this discussion.

Tarhead
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I have a group of problems that deals with the equations:

f(x,y)= x^2+y^2
g(x,y)=20-(x-4)^2-(y+2)^2.

I know that the surfaces z=f(x,y) and z=g(x,y) intersect in a closed curve, C, and the projection of C onto the xy-plane is a circle. However, I am having trouble finding its xy-equation, center, and radius. Additionally and more importantly, I am in the dark on setting up the double or triple integral for the volume of the region bounded by z=f(x,y) and z=g(x,y). Can anyone please help.
 
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The two surfaces intersect along a circle centered at (x, y) = (2, -1) with a radius of [itex]\sqrt 5[/itex] so you might consider a coordinate transformation placing (2, -1) at the new origin.
 


To find the xy-equation, center, and radius of the projection of C onto the xy-plane, we can use the fact that the projection of a circle onto the xy-plane is also a circle. This means that the xy-equation will also be a circle. We can write the equation as (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center of the circle and r is the radius.

To find the center, we can set both equations equal to each other and solve for x and y. This will give us the coordinates of the center. Once we have the center, we can find the radius by plugging in the coordinates of the center into either of the equations.

To set up the double or triple integral for the volume of the region bounded by z=f(x,y) and z=g(x,y), we first need to find the limits of integration. Since we know that the projection of C onto the xy-plane is a circle, we can use polar coordinates to set up the integral. The limits of integration for r would be from 0 to the radius of the circle, and for theta, it would be from 0 to 2π.

The integrand would be the difference between the two equations, g(x,y) and f(x,y). This will give us the height of the region at each point on the circle. Therefore, the integral would be ∫∫(g(x,y)-f(x,y))rdrdθ. This would give us the volume of the region bounded by z=f(x,y) and z=g(x,y).

I hope this helps. Let me know if you have any further questions. Good luck!
 

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